Matter, Chemical Bonding, and Naming

Background: Elements

Which of the following is not a metal? Solution

Sodium

Gallium

Silicon

Tin

Magnesium

Silicon (Si) is a non-metal

Background: Atomic Numbers

Determine the number of neutrons in the isotope of Copper: Solution $^{64}_{29}\text{Cu}$

Hint Clear Info

Incorrect Attempts:

CHECK

neutrons

Hint Unavailable

$\begin{align} \text{Number of neutrons} & = \text{Mass number} - \text{Atomic number} \\ \\ & = 64 - 29 \\ \\ & = 35 \ \text{Neutrons} \\ \\ \end{align}$

Background: Substances

A compound is defined as Solution

A molecule that contains atoms

A substance that contains one or more atoms

A molecule that has one or more ionic bond

A substance that is pure

A substance that can be chemically separated into fix ratios of two or more pure elements.

Compounds are molecules that consist of two or more different elements

Bohr Nuclear Model

The following three terms are equivalent: Orbit, Shell, Energy Level

True

False

Electron Transitions

In an atom, energy is released in the form of light in certain wavelengths when:

Electrons transition or fall to a lower energy level

Electrons leave the ground state

Electrons change direction of travel on an energy level

Electrons transition to a higher energy level

Protons and neutrons collide

Electrons on Shells (Energy Levels)

The maximum number of electrons on the third energy level is: Solution

Hint Clear Info

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electrons

Hint Unavailable

The maximum number of electrons on an energy level equals 2(n)².

2(3)² = 18.

2 electrons in the first, 8 in the second, and actually 18 in the third.

Periodic Trends

Atomic radius increases across a period because of the greater atomic number. Solution

True

False

Atomic radius decreases across a period (row) because more protons attracts more electrons and squeezes the radius smaller. The greater positive nuclear charge of protons attracts the greater number of negative valence electrons.

Which element will have the largest atomic radius? Solution

Nitrogen, N

Phosphorus, P

Lithium, Li

Rubidium, Rb

Tin, Sn

Atomic radius increases down a group, and from right-to-left across a period. The lower-left elements have the largest atomic radius (size).

Periodic Trends: Atomic Radius of Anions

Predict the reason neutral elemental oxygen (O) will have a larger or smaller atomic radius compared to its oxide (O^2-) ion. Solution

An oxide (O^2-) ion has a smaller radius because more valence electrons pull the radius closer to the positive nuclear charge

An oxide (O^2-) ion has a larger radius because more protons and electrons increase the volume of the atom

An oxide (O^2-) ion has a smaller radius because the number of protons is constant while the number of electrons increases

An oxide (O^2-) ion has a larger radius because the extra electrons cause more repulsions in the electron cloud and expand the size

None of the above

Most people think the greater valence electron charge on the oxide (O^2-) ion causes a stronger attraction with the positive nuclear charge resulting in a smaller radius - but this is not correct.

Negative ions actually have a larger atomic radius than the neutral atom because the extra electrons cause greater electron-electron repulsions, increasing the size of the electron cloud.

(Positive ions have a lower atomic radius then the elemental state)

Periodic Trends: Atomic Radius of Cations

The atomic radius of a cation is always smaller than the original neutral element. Solution

True

False

Because removing an electron decreases some of the electron-electron repulsions in a neutral element causing the cation to be smaller than its neutral atom.

Periodic Trends: Ionization & Electron Affinity

Ionization energy is defined as the energy required to add an electron to an atom or ion. Electron affinity is the energy change that occurs when an electron is added to an atom. Solution

True

False

Ionization energy is defined as the energy required to remove an electron to an atom or ion.

Periodic Trends

The ionization energy and electron affinity of a group 2 element would be higher than that of a group 17 (7A) element. Solution

True

False

Ionization energy is defined as the energy required to remove an electron from an atom.

Both ionization energy and electron affinity increase from:
- bottom-to-top, up a group.
- left-to-right, across a period.

Ionization energy generally increases from left-to-right across a period because it is harder (requires more energy) to remove an electron from elements that are close to having a full valence octet.

The ionization energy of a group 2 element would be lower than that of a group 17 (7A) element, in a similar period.

Periodic Trends

Which element has the highest electronegativity? Solution

Zinc, Zn

Mercury, Mg

Potassium, K

Bromine, Br

Cesium, Cs

Electronegativity is the ability of an atom to attract electrons when bonded.

Electronegativity increases left-to-right across a period, and bottom-to-top up a group.

An element with a low electron affinity will most likely also have

High ionization energy

High polarity

High electronegativity

Large atomic radius

Low melting point

Background: Compounds

Which of the following would form a molecular compound? Solution

Metal + Metal

Metal + Nonmetal

Nonmetal + Nonmetal

Crystal Lattice Ions

None of the above

Molecular compounds form by combining two nonmetals. Ionic compounds are formed by combining a metal + a nonmetal.

Background: Ionic Versus Covalent

An ionic bond forms with atoms that Solution

Share electrons

Share electrons and neutrons

Transfer electrons

Transfer protons

None of the above

Ionic bonds form when electrons are transferred from the positive cation to the negative anion.

Electron Dot Diagrams with Ionic Compounds

Draw the following molecules with electron dot diagrams: Solution Potassium Bromide, Calcium Dichloride, and Disodium Oxide
Potassium Bromide
Lewis Diagram: Potassium loses 1 electron as K⁺, bromine gains 1 electron as Br^-.

Calcium Dichloride
Lewis Diagram: Calcium loses 2 electrons as Ca²⁺, each chlorine atom gains 1 electron as Cl^-.

Disodium Oxide
Lewis Diagram: Each sodium loses 1 electron as Na⁺, oxygen gains 2 electrons as O^2-.

Forming Covalent Compounds

Which of the following elements is most likely to form a covalent, molecular compound? Solution

Oxygen

Sodium

Calcium

Neon

Carbonate

Covalent, molecular compounds form between two non-metals. Sodium and calcium are metals. Carbonate is a charged, polyatomic ion. Neon is a noble gas and doesn't bond... leaving Oxygen.

Identifying Covalent Compounds

Which of the following is a covalent compound? Solution Video

KF

MgCl₂

NH₃

Na⁺OH^-

NaNO₃

Covalent is two non-metals...

Lewis Structures: Intro

A Lewis structure for an element shows an arrangement of dots around the symbol of the element. These dots indicate the:

Nuclear charge

Atomic Number

Total number of electrons

Number of electrons in the outer, valence shell

Charge on the element

Lewis Structures of Covalent Compounds

Carbon would form the central atom in the molecule below. Solution CH₂BrF

True

False

Lewis Structures of Covalent Compounds

Draw the Lewis Structures of the following molecules using the rules below.

Step	What to Do
1	Arrange the atoms with the element that forms most bonds in the central position.
2	Add the total number of valence electrons to the central atom clockwise or counterclockwise.
3	Distribute the other atoms around the central atom and place one bonding pair of electrons between each atom.
4	Put remaining electrons as lone pairs on all atoms except the central atom, to a maximum octet of 8 electrons per atom.
5	If octet on central atom is incomplete, move peripheral lone pairs into bonding electrons that are shared with the central atom.
6	Make sure central and peripheral atoms have complete octets. Now if there are extra electrons, place these as lone pairs on the central atom (*exception to octet rule).

N₂H₂ Solution

Molecular compounds have a high melting point and are poor conductors of electricity when dissolved in water. Solution

True

False

Molecular compounds have a low melting point and are poor conductors of electricity when dissolved in water.

Ionic compounds have a high melting and boiling point.

Molecular Forces

Which of the following is not an intermolecular force? Solution

London Dispersion

Van der Waals

Dipole-Dipole

Covalent

Hydrogen Bonding

Covalent is an intra-molecular bond...

The melting point and boiling point of ionic and covalent compounds can be explained with intramolecular forces. Solution

True

False

The melting point of ionic and covalent compounds can be explained with intermolecular forces.

Intramolecular forces - within molecules - determine things like if a compound is likely to break apart and dissolve in water.

Intermolecular forces - between molecules - determine things like how much energy is required to melt or boil a substance because this requires energy to pull the different molecules away from adjacent molecules.

The van der Waals intermolecular force is composed of two forces: Dipole-dipole and London. Solution

True

False

London forces are caused by the induced attraction of the electrons from one molecule to the positive centre in an adjacent molecule.

Dipole-dipole forces are caused by permanent dipoles of the molecules based on electronegativity differences. The van der Waals forces include: London and Dipole-dipole, but not Hydrogen bonding.

Which of the following molecules would only have London Dispersion Forces between the molecules. Solution

HCl

LiF

MgCl₂

H₂O

N₂

Calculating the electronegativity difference between the atoms, you can see that N₂ has the lowest (zero), so this molecule is non-polar covalent. Non-polar molecules do not have dipole-dipole forces, only London Dispersion.

Match the following intermolecular force properties... Solution

Higher melting and boiling point

Lower melting and boiling point

Stronger force

Weaker force

Non-polar covalent

Polar covalent

Non-polar covalent has weaker intermolecular forces (London dispersion) vs polar covalent which has stronger intermolecular forces (London + Dipole + maybe even Hydrogen bonding...)

PERIODIC TRENDS

Electronegativity

Rank the elements in terms of increasing electronegativity (from bottom to top). Solution H, Br, C, F, P, Mg, K, I

Mg

Br

F

I

K

H

P

C

Electronegativity increases up and to-the-right on the periodic table.

Which of the following bonds has the highest electronegativity difference? Solution Video

Br — Br

C — O

Ca — I

Mg — O

N — H

Electronegativity difference is estimated by the distance across the periodic table, or calculated by subtracting the electronegativity of just one of each atom across a bond.

Br — Br = 0
C — O = 1.0
Ca — I = 1.66
Mg — O = 2.3
N — H = 0.9

Neon has a higher electronegativity than Fluorine. Solution

True

False

Rule notwithstanding, noble gases have extremely low electronegativity because of their full valence octet.

Electronegativity Difference (∆EN)

Complete the following table using the given electronegativities. Solution Video

Element Electronegativity

O 3.44

H 2.2

C 2.55

Al 1.61

Cl 3.16

Li 0.98

S 2.58

Na 0.93

F 3.98

Hint Clear Info

Compound Electronegativity Difference (∆EN)

H₂O 1.4

CH₄

AlCl₃

Li₂S

NaF

Incorrect Attempts:

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Hint Unavailable

Compound Electronegativity Difference (∆EN)

H₂O 3.44 - 2.20 = 1.4

CH₄ 2.55 - 2.20 = 0.35

AlCl₃ 3.16 - 1.61 = 1.55

Li₂S 2.58 - 0.98 = 1.6

NaF 3.98 - 0.93 = 3.05

Polarity Based on Electronegativity

Determine the polarity of the following molecules using the following guidelines. (* Indicates an exception to the guidelines).

Electronegativity Difference Range:	0 - 0.5	0.5 - 1.7	1.7 +
Bonding Result:	Non-polar Covalent	Polar Covalent	Ionic

H₂ Solution

Hint Clear Info

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Hint Unavailable

Non-polar Covalent

$\begin{align} \Delta \text{EN} & = 2.20 - 2.20 \\ \\ & = 0 \\ \\ \end{align}$

CH₄ Solution

Hint Clear Info

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Hint Unavailable

Non-polar Covalent

$\begin{align} \Delta \text{EN} & = 2.55 - 2.20 \\ \\ & = 0.35 \\ \\ \end{align}$

LiF Solution

Hint Clear Info

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Hint Unavailable

Ionic

$\begin{align} \Delta \text{EN} & = 3.98 - 0.98 \\ \\ & = 3.00 \\ \\ \end{align}$

H₂O Solution

Hint Clear Info

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Hint Unavailable

Polar Covalent

$\begin{align} \Delta \text{EN} & = 3.44 - 2.20 \\ \\ & = 1.24 \\ \\ \end{align}$

MgI₂ Solution

Hint Clear Info

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Hint Unavailable

Ionic

$\begin{align} \Delta \text{EN} & = 2.66 - 1.31 \\ \\ & = 1.35 \\ \\ \end{align}$ This molecule is ionic because it is made of a metal plus a non-metal, and it can dissolve into Mg²⁺ and 2I^- in water as an electrolyte.

CO₂ Solution Video

Hint Clear Info

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Hint Unavailable

Non-polar Covalent

$\begin{align} \Delta \text{EN} & = 3.44 - 2.55 \\ \\ & = 0.89 \\ \\ \end{align}$ This molecule is not polar covalent as the electronegativity would suggest (EN between 0.5 - 1.7). Instead it is non-polar because the molecule is symmetrical! Symmetry outranks the electronegativity difference. There is no net pull of the electrons shared in the covalent bonds in any direction...

Exceptions to Polarity Based on Electronegativity

Determine the polarity of the molecules that form between the metal Beryllium (Be) and the non-metals Hydrogen (H) and Carbon (C). Solution

Hint Clear Info

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Hint Unavailable

Covalent compounds typically occur between two non-metals... but Beryllium (Be) is an exception. Even though it is a metal, it can form covalent compounds with some non-metals...

Beryllium Hydride BeH₂: Beryllium (Be) & Hydrogen (H)... $\begin{align} \Delta \text{EN} & = 2.1 - 1.6 \\ \\ & = 0.5 \\ \\ \end{align}$ ... is Polar Covalent

Beryllium Carbide BeC₂: Beryllium (Be) & Carbon (C)... $\begin{align} \Delta \text{EN} & = 2.5 - 1.6 \\ \\ & = 0.9 \\ \\ \end{align}$ ...is Polar Covalent.

Polarity

Hydrochloric acid (HCl) is a polar covalent molecule, rather than ionic. Solution

True

False

The ∆EN between 0.5 to 1.7 indicates the molecule is polar covalent. The fact that this molecule physically dissolves into H⁺ and Cl^- ions in water and conducts electricity does not make it ionic. As you see, it is possible for polar covalent molecules to dissociate into ions in solution. $\begin{align} \Delta \text{EN} & = 3.16 - 2.20 \\ \\ & = 0.96 \\ \\ \end{align}$

Physical Properties, Boiling and Melting Point Based on Polarity

Using the polarity of the following molecules as a guide, order the molecules from low to high boiling points (from bottom to top). Solution

CH₄

H₂O

CO₂

LiF

Non-polar covalent < Polar covalent < Ionic.

Rank from low ∆EN to high ∆EN:
0.35 (CH₄), 0.89 (CO₂), 1.24 (H₂O), 3.00 (LiF)

Ionic Compounds

Which of the following ions would fit into this space indicated by 'Z' below? Solution MgZ₂

Na²⁺

Li⁺

Ca²⁺

O^2-

Br^-

Br^-

Chemical Formulas

Complete the following table on paper, with the correct chemical formula for each compound. Solution

Hint Clear Info

Ag⁺ Cu²⁺ Fe³⁺ Ti⁴⁺

Cl^-

O^2-

PO₄^3- ( ) ( )

Incorrect Attempts:

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Balance the charges using the criss-cross method. Look up the charges for the polyatomic ions. Remember to reduce the subscripts by the greatest common factor (if any).

Ag⁺ Cu²⁺ Fe³⁺ Ti⁴⁺

Cl^- AgCl CuCl₂ FeCl₃ TiCl₄

O^2- Ag₂O CuO Fe₂O₃ TiO₂

PO₄^3- Ag₃PO₄ Cu₃(PO₄)₂ FePO₄ Ti₃(PO₄)₄

Use the criss-cross method to balance the charges with subscripts on the ions. Subscripts with a common factor reduce. For example... Cu₂O₂ --> CuO Or... Ti₂O₄ --> TiO₂

NAMING

When to Use All the Different Naming Systems

Compounds use different naming systems and it is important to know when to use each one.

Explain when the stock naming system is used, and how to use it. [3] Solution Video

Hint Clear Info

Incorrect Attempts:

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Look at the flow chart, given above

The STOCK system is used for multivalent (more than one possible charge like Cu¹⁺/Cu²⁺, or Au¹⁺/Au³⁺) metals in ionic compounds (ionic is a metal + a nonmetal). The second word does not use roman numerals, only the first word does. (Sometimes, rarely, the classical system is used instead of stock IUPAC naming)

(Side-note: the diatomic gases HOFBrINCl do not use any naming system, just name the element)

Explain when the prefix naming system is used, and how to use it. [3] Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Look at the flow chart, given above

The PREFIX system is used for covalent compounds (covalent is two non-metals), not ionic compounds. A prefix is only used on the first word if 2 or more, and a prefix is always used on the last word. The second word ends with -ide. (Note that mono is only used in the second part of the name, never in the first part. CO₂ is not monocarbon dioxide...)

Explain when the general naming system is used, and how to use it. [3] Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Look at the flow chart, given above

The GENERAL system is used if the ionic compound is not multivalent. Just name the first word like it appears on the periodic table and the second word ends with -ide (for example: Na₂0 is sodium oxide).

Naming Molecular/Covalent

Prefix Naming

Order the prefixes, with lowest at the top, by dragging them around. Solution

deca

tri

penta

di

hepta

octa

tetra

hexa

nona

mono

1) Mono
2) di
3) tri
4) tetra
5) penta
6) hexa
7) hepta
8) octa
9) nona
10) deca

Naming Covalent Molecules

Name the molecule using the prefix naming system: H₂O Solution

dihydrogen monoxide

dihydrogen oxide

hydrogen oxide

hydrogen(II) oxide

The prefix naming system is used in covalent compounds (between two non-metals). The prefix comes from the subscript number indicating number of each atom in the molecule. _ + _ Tee first word uses a prefix if 2+ atoms...
The second word uses a prefix if 1+ atoms (always)...

Practice Naming Compounds: Molecular/Covalent

Name the following compounds using the prefix system.

CO Solution Video

Hint Clear Info

Incorrect Attempts:

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Determine the number of each element, and don't forget to add the suffix.

carbon monoxide

(not monocarbon monoxide - because mono never goes on the 1^st word.)

NO Solution

Hint Clear Info

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Determine the number of each element, and don't forget to add the suffix.

nitrogen monoxide

(Remember to only use mono on the second word, never the first).

N₂O Solution

Hint Clear Info

Incorrect Attempts:

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Determine the number of each element, and don't forget to add the suffix.

dinitrogen monoxide

(drop the 'o' on the mono suffix)

B₂H₆ Solution

Hint Clear Info

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Determine the number of each element, and don't forget to add the suffix.

diboron hexahydride

(Still add the suffix -ide to the end of hydrogen = hydride)

CH₄ Solution

Hint Clear Info

Incorrect Attempts:

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Determine the number of each element, and don't forget to add the suffix.

carbon tetrahydride

PCl₅ Solution

Hint Clear Info

Incorrect Attempts:

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Determine the number of each element, and don't forget to add the suffix.

phosphorus pentachloride

CO₂ Solution

Hint Clear Info

Incorrect Attempts:

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Determine the number of each element, and don't forget to add the suffix.

carbon dioxide

N₂F₄ Solution

Hint Clear Info

Incorrect Attempts:

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Determine the number of each element, and don't forget to add the suffix.

dinitrogen tetrafluoride

P₂O₅ Solution

Hint Clear Info

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Determine the number of each element, and don't forget to add the suffix.

diphosphorus pentoxide

(drop the 'a' on the penta suffix)

Practice Writing Formulas: Molecular/Covalent

Write the chemical formula for the following compounds.

Dihydrogen monosulfide Solution

⁰
¹
²
³
⁴
⁵
⁶
⁷
⁸
⁹
⁻
⁺
⁽
⁾
₀
₁
₂
₃
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₆
₇
₈
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₋
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₎

Incorrect Attempts:

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The prefix in-front of each element indicates the amount.

H₂S

Nitrogen trihydride Solution

⁰
¹
²
³
⁴
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⁾
₀
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Incorrect Attempts:

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The prefix in-front of each element indicates the amount.

NH₃

Diphosphorus trisulfide Solution

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¹
²
³
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The prefix in-front of each element indicates the amount.

P₂S₃

Silicon tetrachloride Solution

⁰
¹
²
³
⁴
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Incorrect Attempts:

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The prefix in-front of each element indicates the amount.

SiCl₄

Nitrogen triiodide Solution

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¹
²
³
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Incorrect Attempts:

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The prefix in-front of each element indicates the amount.

NI₃

Dinitrogen trioxide Solution

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¹
²
³
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Incorrect Attempts:

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The prefix in-front of each element indicates the amount.

N₂O₃

Sulfur trioxide Solution

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¹
²
³
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Incorrect Attempts:

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The prefix in-front of each element indicates the amount.

SO₃

Trioxygen Solution

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¹
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³
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Incorrect Attempts:

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The prefix in-front of each element indicates the amount.

O₃

(common name: ozone)

Correcting Common Mistakes

Find what's wrong with the following names or formulas, and correct them.

Mononitrogen monoxide Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

Where should prefixes be used?

Covalent compounds using the prefix naming system never have mono in the first name.
The correct name for NO: nitrogen monoxide (This is like carbon monoxide, which you may know is CO).

Dihydrogen sulfide Solution

Hint Clear Info

Incorrect Attempts:

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Where should prefixes be used?

The last name always has a prefix, even if it's just one.
The correct name for H₂S: dihydrogen monosulfide

C₁F₄ Solution

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¹
²
³
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Incorrect Attempts:

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What subscripts should not be used?

Don't write subscripts of 1.
The correct formula is just: CF₄

Naming Ionic

Practice Naming Compounds: Ionic

Name the following ionic compounds. Remember that ionic compounds consist of a metal plus a non-metal.

MgO Solution Video

Hint Clear Info

Incorrect Attempts:

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Hint Unavailable

Mg²⁺O^2-

Magnesium is not multivalent, so no roman numerals are required. Don't use prefix naming with ionic compounds (metal plus non-metal). magnesium oxide

AlCl₃ Solution

Hint Clear Info

Incorrect Attempts:

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Hint Unavailable

Al³⁺Cl^1-

Aluminum is not multivalent, so no roman numerals are required. aluminum chloride

InF₃ Solution

Hint Clear Info

Incorrect Attempts:

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Hint Unavailable

indium fluoride

Na₂O Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

sodium oxide

CaSe Solution

Hint Clear Info

Incorrect Attempts:

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Hint Unavailable

calcium selenide

BaI₂ Solution

Hint Clear Info

Incorrect Attempts:

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Hint Unavailable

barium iodide

ScF₃ Solution

Hint Clear Info

Incorrect Attempts:

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Hint Unavailable

scandium fluoride

ZrBr₄ Solution

Hint Clear Info

Incorrect Attempts:

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Hint Unavailable

zirconium bromide

Practice Writing Formulas: IONIC

Write the chemical formula for the following compounds.

Sodium sulfide Solution

⁰
¹
²
³
⁴
⁵
⁶
⁷
⁸
⁹
⁻
⁺
⁽
⁾
₀
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₎

Incorrect Attempts:

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Start with the charges on the ions, and then use the criss-cross method to balance the charges

C.C. Credit: CC BY-SA 3.0, W. Oelen, 2005

Na₂S

Calcium sulfide Solution

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¹
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³
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Incorrect Attempts:

CHECK

Start with the charges on the ions, and then use the criss-cross method to balance the charges

Using criss-cross with the ions... Ca²⁺ S^2-

The 2s cancel, so there's one of each...

CaS

Silver fluoride Solution

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¹
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³
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Incorrect Attempts:

CHECK

Start with the charges on the ions, and then use the criss-cross method to balance the charges

Using criss-cross with the ions... Ag¹⁺ F^1-

AgF

Zinc bromide Solution

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Incorrect Attempts:

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Start with the charges on the ions, and then use the criss-cross method to balance the charges

Using criss-cross with the ions... Zn²⁺ Br^1-

ZnBr₂

Gallium oxide Solution

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Incorrect Attempts:

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Start with the charges on the ions, and then use the criss-cross method to balance the charges

Ga₂O₃

Lithium iodide Solution

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Incorrect Attempts:

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Start with the charges on the ions, and then use the criss-cross method to balance the charges

LiI

Germanium oxide Solution

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Incorrect Attempts:

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Start with the charges on the ions, and then use the criss-cross method to balance the charges

GeO₂

Zirconium nitride Solution

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Incorrect Attempts:

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Start with the charges on the ions, and then use the criss-cross method to balance the charges. Use the most common ion charge.

Zr⁴⁺ + N^3-...
Zr₃N₄

Aluminum chloride Solution

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Incorrect Attempts:

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Start with the charges on the ions, and then use the criss-cross method to balance the charges. Use the most common ion charge.

AlCl₃

Correcting Common Mistakes

Find what's wrong with the following names or formulas and correct them.

disodium monoxide Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Ionic compounds don't use the prefix naming system. The correct name would be: sodium oxide

The roman numeral in, copper(II) sulphate, also known as copper(II) sulfate/sulphate, indicates that there are two atoms of copper per molecule. Solution

True

False

The roman numeral indicates the charge (or oxidation number) of the multivalent cation.

scandium(III) phosphide Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Don't use the stock roman numeral numbering system for monovalent elements - elements that only have one valence or possible charge. Scandium can only ever have a 3³⁺ charge and is implied in its name.

Correct name: scandium phosphide

Naming Polyatomics

Polyatomic Ions

How many atoms of Oxygen are in 2 molecules of aluminum nitrate? Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

The subscript, 3 multiplies to the 3 oxygens so 1 molecule of aluminum nitrate has 9 oxygens.
2 × Al(NO₃)₃ 2 molecules of aluminum nitrate have 18 oxygens.

Polyatomic Anions

Which of the following polyatomic anions is sulphate (aka sulfate)? Solution

HS^-

SO₃^2-

SO₂^-

S^2-

SO₄^2-

Sulphate/Sulfate is SO₄^2-
(Sulphite/Sulfite is SO₃^2-)

The suffix (ending) with '-ate' has more oxygens because it ate more...

The oxides of chlorine and bromine form four different polyatomic anions with different numbers of oxygen atoms, x. Rank the naming system in the order of decreasing number of oxygens from top to bottom. Solution ClO_x^- BrO_x^-

_____-ite

per-_-ate

hypo-_-ite

_____-ate

In terms of the number of oxygens:
"Per-" is highest
"Hypo-" is lowest
"-ate" is higher
"-ite" is lower

ClO₄^– perchlorate

ClO₃^– chlorate

ClO₂^– chlorite

ClO^– hypochlorite

BrO₄^– perbromate

BrO₃^– bromate

BrO₂^– bromite

BrO^– hypobromite

Making Ionic Compounds from Ions using the Criss-Cross Method (Including Polyatomic)

Compose the following ions into their salt/ionic compound by balancing the charges, and then name the compound. Put the formula of the compound in the box provided below, using proper capitalization and subscripts. [2]

Ca²⁺, PO₄^3- Solution Video

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Hint Unavailable

NH₄⁺, CO₃^2- Solution

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Hint Clear Info

Incorrect Attempts:

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Hint Unavailable

Incorrect Attempts:

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Hint Unavailable

Practice Naming Compounds: POLYATOMICS

Name the following compounds containing polyatomic ions.

Ca(NO₃)₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

calcium nitrate

(Note it's not calcium dinitrate because we don't use the prefix naming system on polyatomic ions).

Al₂(SO₃)₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

aluminum sulphite (aka aluminum sulfite)

(Don't use prefix naming system: aluminum trisulfite, aluminum trisulphite)

(NH₄)SO₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

ammonium sulphate (aka ammonium sulfate)

This is an ionic compound with polyatomic ions.

AgCN Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

silver cyanide

Zn(CH₃COO)₂, a.k.a. Zn(C₂H₃O₂)₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

zinc acetate

Al(HCO₃)₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

aluminum bicarbonate

Be₃(PO₃)₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

beryllium phosphite

Zn(ClO₂) Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

zinc chlorite

Mg(OH)₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

magnesium hydroxide

KMnO₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

potassium permanganate

Practice Writing Formulas: POLYATOMICS

Write the chemical formula for the following compounds with polyatomic ions.

Sodium bicarbonate Solution

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Incorrect Attempts:

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Hint Unavailable

Na⁺ + HCO₃^-

NaHCO₃

Magnesium perchlorate Solution

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Incorrect Attempts:

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Hint Unavailable

Mg²⁺ + ClO₄^-

Mg(ClO₄)₂

Diammonium sulfide Solution

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Incorrect Attempts:

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Hint Unavailable

Diammonium is two (di) ammonium ions.

NH₄⁺ + S^2-

(NH₄)₂S

Tin(II) sulphite, aka Tin(II) sulfite Solution

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Incorrect Attempts:

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Hint Unavailable

Sn²⁺ + SO₃^2-

SnSO₃

Strontium cyanide Solution

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Incorrect Attempts:

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Hint Unavailable

Sr(CN)₂

Silver phosphite Solution

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Incorrect Attempts:

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Hint Unavailable

Ag₃PO₃

Potassium hydroxide Solution

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Incorrect Attempts:

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Hint Unavailable

KOH

Aluminum acetate (C₂H₃O₂) Solution

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Incorrect Attempts:

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Hint Unavailable

Al(C₂H₃O₂)₃

Correcting Common Mistakes

Find what's wrong with the following names or formulas and correct them.

nitrogen-tetrahydride bromide Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Don't try to name polyatomic ions as molecular/covalent compounds. Use the name of the polyatomic ion. NH₄Br is just ammonium bromide.

Sulphite (aka Sulfite): SO₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

This may be a bit tricky because SO₃ is not polyatomic when it's by itself. You can tell because it is not bound to a metal cation in-front, like __SO₃.

SO₃ is a covalent compound by itself - so you name it with the prefix system.
= sulfur trioxide

Sodium Sulfur Oxide: Na₂SO₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Na₂SO₃ is an ionic compound with a monovalent metal (and a polyatomic anion), so it is named using the general naming system.
= sodium sulphite (aka sodium sulfite).

Naming Multivalent

Naming Molecules: Ionic → Multivalent

Name the molecule using the stock naming system: Ni₂O₃ Solution Video

Dinickel Trioxygen

Dinickel Trioxide

Nickel Trioxide

Nickel(II) Trioxide

Nickel(III) Oxide

Multivalent metals (metals that could have different charges like Ni²⁺ or Ni³⁺) need to have the charge represented by roman numerals without a space between the cation and roman numeral.

The oxygen anion oxygen begins with a prefix representing the number of atoms, and it ends with the suffix -ide indicating that it is an ion.

Stock: Nickel(III) Oxide

(Classical system: Nicklic Oxide)

Practice Naming Compounds: Multivalent

Name the following compounds using the stock system.

CuS Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Cu²⁺S^2-

Copper is multivalent (Cu¹⁺, Cu²⁺) so roman numerals are required.

copper(II) sulfide

CuCl₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

copper(II) chloride

FeI₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Iron is multivalent (Fe²⁺, Fe³⁺) so use the stock naming system:

iron(III) iodide

SnH₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Sn⁴⁺ + H^-1

tin(IV) hydride

CoPO₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Co³⁺ + PO₄^3-

cobalt(III) phosphate

AuP Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Au³⁺ + P_3-

gold(III) phosphide

TiS₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Ti⁴⁺ + S^2-

titanium(IV) sulfide

Hg₂O Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Hg^1- + O^2-

mercury(I) oxide

Practice Writing Formulas: MULTIVALENT

Write the chemical formula for the following compounds with multivalent ions.

mercury(II) fluoride Solution

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Incorrect Attempts:

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Hint Unavailable

HgF₂

copper(I) oxide Solution

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Hint Unavailable

Using criss-cross with the ions... Cu¹⁺ O^2-

Cu₂O

copper(I) sulfide Solution

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Hint Unavailable

Using criss-cross with the ions... Cu¹⁺ S^2-

Cu₂S

manganese(II) nitride Solution

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Incorrect Attempts:

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Hint Unavailable

Mn²⁺ + N^3-

Mn₃N₂

platinum(II) oxide Solution

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Incorrect Attempts:

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Hint Unavailable

Pt²⁺ + O^2-

PtO

nickel(III) oxide Solution

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Incorrect Attempts:

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Hint Unavailable

Ni³⁺ + O^2-

Ni₂O₃

lead(IV) chloride Solution

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Incorrect Attempts:

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Hint Unavailable

Pb⁴⁺ + Cl^-

PbCl₄

antimony(V) phosphide Solution

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Incorrect Attempts:

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Hint Unavailable

Sb⁵⁺ + P^3-

Sb₃P₅

chromium(VI) oxide Solution

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Incorrect Attempts:

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Hint Unavailable

Cr⁶⁺ + O^2-

Cr₂O₆

= CrO₃

Correcting Common Mistakes

Find what's wrong with the following names or formulas and correct them.

iron(III) triiodide Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

The stock naming system is used without using prefixes. The correct name: iron(III) iodide We know there must be 3 Iodide atoms for each Fe³⁺ because the charge on each iodide is only I^1-.

Sn₂O₂ Solution

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Incorrect Attempts:

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Hint Unavailable

Sn²⁺ + O^2-

Using the criss-cross method, the subscripts will reduce to the lowest form:

= SnO

Naming Multivalent and Polyatomic

Practice Naming Compounds: Multivalent & Polyatomic

Name the following compounds using the systematic stock naming system.

Cu²⁺, SO₄^2- Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

CuSO₄

copper(II) sulphate, [aka copper(II) sulfate]

HgHCO₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Hg⁺¹ + HCO₃^-

mercury(I) bicarbonate

Fe₃(PO₃)₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Fe²⁺ + PO₃^3-

iron(II) phosphite

Sn(CH₃COO)₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Sn⁴⁺ + CH₃COO^1-

tin(IV) acetate

Au₂SO₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Au⁺¹ + SO₄^2-

gold(I) sulphate, [aka gold(I) sulfate]

Co(ClO₂)₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Co³⁺ + ClO₂^-

cobalt(III) chlorite

Pd(MnO₄)₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Pd⁴⁺ + MnO₄^-1

palladium(IV) permanganate

Mn(NO₃)₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Mn²⁺ + NO₃^-1

manganese(II) nitrate

Practice Writing Formulas: Multivalent & Polyatomic

Write the chemical formula for the following compounds.

tin(II) bicarbonate Solution

SnCO₃^-

Sn(CO₃^-)₂

Sn₂CO₃^2-

SnCO₃^2-

Sn(HCO₃^-)

Sn(HCO₃^-)₂

Tin (Sn) has a 2+ charge, bicarbonate (HCO₃) has a 1- charge.

Use the criss-cross method to determine the subscripts on each ion.
Sn²⁺(HCO₃^1-)
= Sn₁(HCO₃^-)₂
= Sn(HCO₃^-)₂

platinum(IV) nitrite Solution

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Incorrect Attempts:

CHECK

Balance the charges of the ions. Look up the polyatomic ions. Multivalent ions show their charge in parentheses.

Pt⁴⁺ + NO₂^-1

Pt(NO₂)₄

iron(III) cyanide Solution

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Incorrect Attempts:

CHECK

Balance the charges of the ions. Look up the polyatomic ions. Multivalent ions show their charge in parentheses.

Fe³⁺ + CN^-

Fe(CN)₃

gold(I) acetate (C₂H₃O₂^-) Solution

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Incorrect Attempts:

CHECK

Balance the charges of the ions. Look up the polyatomic ions. Multivalent ions show their charge in parentheses.

Au⁺¹ + C₂H₃O₂^- or CH₃COO^-

AuC₂H₃O₂

tungsten(V) hydroxide Solution

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Incorrect Attempts:

CHECK

Balance the charges of the ions. Look up the polyatomic ions. Multivalent ions show their charge in parentheses.

W⁵⁺ + OH^-

W(OH)₅

vanadium(IV) bicarbonate Solution

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Incorrect Attempts:

CHECK

Balance the charges of the ions. Look up the polyatomic ions. Multivalent ions show their charge in parentheses.

V⁴⁺ + HCO₃^-1

V(HCO₃)₄

manganese(VI) carbonate Solution

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Incorrect Attempts:

CHECK

Balance the charges of the ions. Look up the polyatomic ions. Multivalent ions show their charge in parentheses. Make sure to reduce the subscripts by the greatest common factor.

Mn⁶⁺ + CO₃^2-

Mn₂(CO₃)₆

Mn(CO₃)₃

(After criss-cross, reduce fully from 2:6 to 1:3)

tin(IV) phosphate Solution

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Balance the charges of the ions. Look up the polyatomic ions. Multivalent ions show their charge in parentheses.

Sn⁴⁺ + PO₃^3-

Sn₃(PO₄)₄

Correcting Common Mistakes

Find what's wrong with the following names or formulas and correct them.

SnNO₃ [with tin(IV)] Solution

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Tin Sn⁴⁺, & nitrate, NO₃^-1...

Makes Sn(NO₃)₄

Sodium(II) bromide Solution

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Sodium is not multivalent so it does not use the stock roman numeral naming system. It can only be Na⁺¹.

The name is just sodium bromide.

Practice Naming Compounds: Multivalent & Polyatomic Oxyanions/Oxoanions, IUPAC

Modern, international IUPAC naming of compounds with polyatomic oxyanions, uses roman numerals with the oxyanion (as well as any multivalent cations). The only suffix used is 'ate'. Remember the roman numeral represents the oxidation number of the multivalent or non-oxide element. Name the following compounds, using full IUPAC notation.

E.g.) NO₃^1-
( # non-oxides )( oxidation state of non-oxide ) + ( # oxides )( oxidation state of oxide ) = polyatomic charge $\begin{align} (1)(N) + (3)(-2) & = -1 \\ \\ N - 6 & = -1 \\ \\ N & = -1 + 6 \\ \\ N & = +5 \end{align}$
∴ nitrate(V)

MgSO₄ Solution

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Mg²⁺ + SO₄^2-

magnesium sulphate(VI), [aka magnesium sulfate(VI)]

Na₂SO₂ Solution

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Na¹⁺ + SO₂^2-

sodium sulphate(II), [aka sodium sulfate(II)]

NH₄ClO Solution

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NH₄⁺ + ClO^-

ammonium chlorate(I)

CuCrO₄ Solution

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Cu²⁺ + CrO₄²⁺

copper(II) chromate(VI)

AlPO₃ Solution

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Al³⁺ + PO₃^3-

aluminum phosphate(III)

Fe₃(PO₄)₂ Solution

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The subscript '2' on (PO₄^3-)₂ does not contribute to the oxidation number of the phosphorus.

Fe₃²⁺ + PO₄^3-

iron(II) phosphate(V)

(Ti)₂(Cr₂O₇)₃ Solution

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'dichromate' is not used for the modern IUPAC naming of the polyatomic oxyanion.. use chromate(VI)

Ti³⁺ + Cr₂O₇^2-

titanium(III) chromate(VI)

Naming Hydrates

The name for the compound below is iron(II) sulphate hexahydrate. [note: sulphate is sometimes spelled sulfate]. Solution FeSO₄•6H₂O

True

False

For hydrates, you have to combine the 'Stock' and the 'Prefix' naming systems.

Calculations with Hydrates, Intro

0.48 mol of hydrated barium chloride is heated to its anhydrous salt, BaCl₂. If the mass of the hydrate is 117.2496 g, determine its molecular formula. Solution

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MM_BaCl₂ = 208.23 g/mol
MM_H₂O = 18.02 g/mol

You can calculate the molar mass given the mass and amount of moles. Let 'n' be the molar coefficient of H₂O in BaCl₂·nH₂O $\begin{align} \text{molar mass} & = \dfrac{\text{mass}}{\text{moles}} \\ \\ 208.23 + 18.02(n) & = \dfrac{117.2496\ g}{0.48\ mol} \\ \\ n & = 2 \end{align}$ Therefore there are 2 mols of H₂O: BaCl₂·2H₂O

Calculations with Hydrates

Epsom salts (magnesium sulphate heptahydrate, MgSO₄·7H₂O) are commonly used as bath salts. As an epsom salt is heated, to drive off the water, the mass continually decreases until the compound is anhydrous (lacking water). Under the right conditions, the salt, MgSO₄ does not decompose.

The anhydrous compound contains less moles of MgSO₄ than the fully hydrated compound. Solution

True

False

The moles of MgSO₄ are constant.

When the hydrate gives off water molecules, the amount of H₂O changes while the MgSO₄ stays constant because it cannot evaporate.

Find the amount of moles of water (H₂O) in 100 g of epsom salt, given the molecular weights of MgSO₄ is 120.3 g/mol, and 7H₂O is 126 g/mol. Solution

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mol

Hint Unavailable

Find moles of the whole epsom salt compound, MgSO₄·7H₂O $\begin{align} n & = \dfrac{mass}{molecular\ weight} \\ \\ & = \dfrac{100\ g}{120.3\ g/mol + 126\ g/mol} \\ \\ & = \dfrac{100\ \cancel{g}}{243.6\ \cancel{g}/mol} \\ \\ & = 0.406\ mol \\ \\ \end{align}$ Use molar ratios 7 H₂O to 1 MgSO₄·7H₂O $\begin{align} \dfrac{mol\ water}{molar\ coefficient\ water} & = \dfrac{mol\ epsom}{molar\ coefficient\ epsom} \\ \\ \dfrac{mol\ water}{7} & = \dfrac{0.406\ mol}{1} \\ \\ mol\ water & = \dfrac{0.406\ mol × 7}{1} \\ \\ & = 2.842\ mol\ water \\ \\ \end{align}$ [Note there is another method to get the answer using percent composition of water in the epsom, which is 51.157%, which is 51.157 g water, which is 2.84 mol water.]

Determine the mass of the anhydrous salt. Solution

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g

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Use molar ratios of 1 MgSO₄ to 1 MgSO₄·7H₂O $\begin{align} mol\ epsom & = 0.406\ mol \\ \\ mol\ magnesium\ sulphate & = 0.406\ mol \\ \\ \end{align}$ Convert mol to mass: $\begin{align} mass & = mol\ ×\ molecular\ weight \\ \\ & = 0.406\ \cancel{mol} × 120.3\ \tfrac{g}{\cancel{mol}} \\ \\ & = 48.8418\ g \\ \\ \end{align}$ [Note this could also be calculated using percent composition, not shown.]

ACIDS/BASES

Naming Acids: Rules

Match the example with the rule it falls under. Solution

Acidic Halogens

HF, HCl, HBr, HI

Polyatomic anions ending in -ate

Nitrate NO₃^-, Chlorate ClO₃^-, Sulfate/Sulphate SO₄^2-

Polyatomic anions ending in -ite

Nitrite NO₂^-, Chlorite ClO₂^-, Sulphite/Sulfite SO₃^2-

Replace the suffix with the new suffix -ous, then add the word acid last.

Add the Hydro prefix to the front, replace part of the end with the suffix ic, then add the word acid last.

Replace the suffix with the new suffix -ic, then add the word acid last.

None of the above.

E.g.) Hydrofluoric acid, Hydrochloric acid, Hydrobromic acid, Hydroiodic acid

Naming Acids: State Matters

Determine the circumstance where the names below are used for HBr. [2] Solution Hydrogen bromide,
Hydrobromic acid

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Use hydrogen bromide when the compound is a: solid Use hydrobromic acid when the compound is: dissolved in water (aqueous)

Naming Acids: Polyatomic Anions

Name the following acids using the mnemonic given to help you remember the charges and number of oxygens in the main polyatomic anions ending in -ate. (The vowel number is the charge, and the consonant number is the number of oxygens).

NICK the CAMEL ate CLAMS for SUPPER in PHOENIX

Mnemonic	Consonants, Vowels	Formula	Name (-ate's)
Nick	3, 1	NO₃^1-	Nitrate
Camel	3, 2	CO₃^2-	Carbonate
Clam	3, 1	ClO₃^1-	Chlorate
Supper	4, 2	SO₄^2-	Sulfate/Sulphate
Pheonix	4, 3	PO₄^3-	Phosphate

H₂SO₄ Solution

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Ions ending in -ate take the suffix -ic

sulfuric acid

H₂SO₃ Solution

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Ions ending in -ite take the suffix -ous

sulfurous acid

HNO₃ Solution

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Ions ending in -ate take the suffix -ic

nitric acid

HNO₂ Solution

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Ions ending in -ite take the suffix -ous

nitrous acid

HClO₃ Solution

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Ions ending in -ate take the suffix -ic

chloric acid

(Perchloric = ClO₄)

HClO₂ Solution

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Ions ending in -ite take the suffix -ous

chlorous acid

H₃PO₄ Solution

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Ions ending in -ate take the suffix -ic

phosphoric acid

H₃PO₃ Solution

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Ions ending in -ite take the suffix -ous

phosphorous acid

H₂CO₃ Solution

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Ions ending in -ate take the suffix -ic

carbonic acid

H₂CO₂ Solution

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Ions ending in -ite take the suffix -ous

carbonous acid

(Aside: common names are formic acid or methanoic acid)

Naming Bases

You can easily recognize bases that have hydroxide (OH^-) ions in them. Name the following bases. Remember the proper syntax of naming: lowercase and proper spacing --- for example, Ti(OH)₄ = titanium(IV) hydroxide.

KOH Solution

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potassium hydroxide

NH₄OH Solution

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ammonium hydroxide

Mg(OH)₂ Solution

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magnesium hydroxide

(Do not use prefix naming, like di, in bases)

Fe(OH)₃ Solution

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Multivalent ions, like Fe²⁺, Fe³⁺ use the stock naming system.

iron(III) hydroxide

(Don't use prefix naming — like mono, di, tri — in bases)

Pb(OH)₂ Solution

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Multivalent ions, like Pb²⁺, Pb⁴⁺ use the stock naming system.

lead(II) hydroxide

(Don't use prefix naming — like mono, di, tri — in bases)

Recognizing Bases in Grade 11

Which of the following is not a base? Solution

NH₃

NO₂

Ca(OH)₂

NaOH

Bases have different number of hydroxide (OH^-) ions in the formula.

The exception is ammonia, NH₃, which is the base form of ammonium, NH₄⁺.

NUCLEAR CHEM

Radioactive Decay

Uranium (U) decays into a Thorium (Th) isotope and what other kind of particle? Solution $^{238}_{92}\text{U} \, \xrightarrow{} \ ^{234}_{90}\text{Th} \, + \ ^{4}_{2}\text{He}$

Alpha particle

Beta negative particle

Beta positive particle

Gamma particle

Neutrino particle

$^{4}_{2}\text{He}$ is an alpha particle

Nuclear Reaction

The following reaction is nuclear fusion. Solution $1\ Neutron +\ ^{235}_{92}\text{U} \, \xrightarrow{} \ ^{92}_{36}\text{Kr} \, + \ ^{141}_{56}\text{Ba} + 3\ Neutrons$

True

False

The reaction is nuclear fission. Fission is the splitting of a large nucleus into smaller nuclei.

Flashcards: Matter and Bonding

How many neutrons in $^{59}_{28}\text{Ni}$

What is an energy level?

An orbit or shell corresponding to the period number, on which electrons are located.

What is the trend for increasing electronegativity on the periodic table?

Up a group, and right across a period

What is the trend for increasing electron affinity on the periodic table?

Up a group, and right across a period

What is the trend for increasing atomic radius on the periodic table?

Down a group, and left across a period

What is the trend for increasing ionization energy on the periodic table?

Up a group, and right across a period

How does the atomic radius of elemental fluorine (F) compare to its ion form (F^-)?

The fluorine ion has a larger radius because the extra electron causes more repulsion in the electron cloud and expands the size of the radius

How does the atomic radius of a cation compare to its neutral elemental form?

The atomic radius of the cation is always smaller than the neutral atom.

How are ionization energy, electronegativity, and electron affinity related?

Electronegativity is the attraction of the electrons by the element, which is the same trend for electron affinity. Ionization energy is the energy required to remove an electron, which is high for elements with high electronegativity and electron affinity

What is the difference between an ionic compound and a molecular compound?

Ionic compounds are a metal + a non-metal. Molecular compounds are two non-metals.

(another name for molecular compound is covalent compound)

What is the difference between an ionic compound and an alloy?

An alloy is composed of two or more metals while an ionic compound is composed of a metal + a non-metal. Alloys do not form intramolecular bonds like in ionic bonds.

What type of compounds are: SO₂ or NF₃?

Covalent/Molecular

What type of compound is Sodium Fluoride?

Ionic compound

What does a lewis dot diagram indicate?

The arrangement of bonding and non-bonding (lone pair) electrons in an atom or compound.

When drawing lewis structures, what is the next step to do if the octet on a central atom is incomplete?

Move peripheral lone pairs on adjacent atoms into bonding electrons that are shared with the central atom.

What are some of the main physical properties of molecular compounds?

Molecular compounds have a low melting point and are poor conductors of electricity when dissolved in water.

What are the three (four) intermolecular forces you have seen so far?

London, Dipole-dipole, van der Waals (London + Dipole-Dipole), and Hydrogen bonding

A molecule with a 0.6 electronegativity difference would be what type of polarity?

Polar covalent.

Name the molecule with the stock naming system:

Fe₂O₃

Iron (III) Oxide

Name the molecule with the prefix naming system:

Fe₂O₃

Diiron trioxide

Quantities in Chemical Reactions

Solutions and Solubility, Acid-Base

Solutions

Solutions are heterogeneous mixtures composed of a solute dissolved in a solvent. Solution

True

False

Solutions are homogeneous mixtures composed of a solute dissolved in a solvent. Homogeneous mixtures are uniform - consistent throughout.

Note that both solutes and solvents can consist of any combination of gas, liquid, or solid phases.

Solutions

An alloy is mixture of two or metal elements. The most common stainless steel alloy used is Stainless Steel 316, which contains a combination of the following elements: Fe, C, Cr, Ni, Mo, Mn, Si, and more. Stainless steel is a solution by definition. Solution

True

False

An alloy is a homogeneous mixture of elements, therefore it is a (solid) solution.

Ions in Solution

Electrolyte solutions contain aqueous ions that allow the conduction of electricity. Solution

True

False

Soluble ionic compounds form electrolytes that can propagate charge movement, conducting electricity.

Which of the following is an electrolyte solution? Solution

Chlorine and fluorine dissolved in tap water

Pure water

Sugar dissolved in water

Alcohol dissolved in water

None of the above

Only ionic compounds that dissolve in water form electrolyte solutions.

Hydrogen Bonding

A hydrogen bond is: Solution

A polar covalent bond

A nonpolar covalent bond

An ionic bond

A dipole-dipole bond

Not a bond, it is an attraction

Hydrogen 'bond' is a bit of a misnomer, as it isn't actually a bond by definition, it is more of an electrostatic 'attraction'.

Hydrogen bonding can occur between molecules that contain Hydrogen (H) covalently bonded to any of Nitrogen (N), Oxygen(O), or Fluorine (F). Furthermore, hydrogen bonding can occur between molecules with any of N-H, O-H, F-H bonds and the lone pairs on N, O, or F. Solution

True

False

True

Hydrogen Bonding

Which of the following pair of molecules cannot form hydrogen bonds? Solution

H₂O and CO₂

CN: and H₂O

NH₃ and H₂O

CF₄ and NH₃

CN: and CF₄

H₂O and CO₂
The O-H bond in H₂O can hydrogen bond to the lone pair of electrons on the oxygen in CO₂ (even when the oxygen in CO₂ is not bonded to hydrogen.)

CN: and H₂O
The lone pair on a nitrogen atom acts as a hydrogen bond acceptor to the O-H bond in H₂O.

NH₃ and H₂O
The N-H bond in NH₃ and the O-H bond in H₂O can clearly form strong hydrogen bonds.

CF₄ and NH₃
Surprisingly, the partial negative lone pair on fluorine in C-F: can hydrogen bond to the partial positive hydrogen in the H-N bond in NH₃.

CN: and CF₄
CN: and CF₄ can only form hydrogen bonds with other molecules containing either O-H, N-H, or F-H bonds.

Solubility of Compounds

Nonpolar compounds cannot dissolve in water because London dispersion forces are not as strong as dipole-dipole forces. Solution

True

False

Polar compounds dissolve in water because of the dipole-dipole interactions. Compounds that can hydrogen bond are even more soluble. A molecule with neither hydrogen bonds nor dipole-dipole will be forced out of water because water prefers to bond/interact with other water molecules in the presence of nonpolar solutes.

Solubility of Compounds

Which of the following statements regarding the solubility of a compound is incorrect? Solution

Covalent compounds dissolve in a solvent with a combination of one or more of dipole-dipole, London, or hydrogen bonding

Soluble ionic compounds dissolve as electrolytes with dissociated positive and negative ions attracted to the positive or negative dipole of the solvent

Nonpolar molecules dissolve in nonpolar solvents with London dispersion forces

A symmetrical molecule will tend to dissolve in nonpolar solvents rather than polar solvents

A diatomic molecule consisting of the same two atoms will dissolve in the universal solvent, water.

A diatomic molecule consisting of the same two atoms is nonpolar and would dissolve in a nonpolar solvent.

Percentage by Volume Concentration % V/V

Which of the following is the correct calculation of percentage by volume for 25mL of pure alcohol solute in 100mL of solution? Solution

$\dfrac{25\ mL + 100\ mL}{100\ mL} × 100\%$

$\dfrac{25\ mL}{100\ mL} × 100\%$

$\dfrac{25\ mL}{25\ mL + 100\ mL} × 100\%$

$\dfrac{100\ mL - 25\ mL}{100\ mL} × 100\%$

$\dfrac{25\ mL}{100\ mL - 25\ mL} × 100\%$

$\begin{align} \text{Percent by Volume} & = \dfrac{\text{Solute Volume}}{\text{Solution Volume}} × 100\% \\ \\ & = \dfrac{25\ mL}{100\ mL} × 100\% \\ \\ \end{align}$

Percentage by Weight Concentration % W/V

Calculate the percentage weight by volume of 5.0 g of pure hydrochloric acid in 0.25 L of solution. Solution

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% (W/V)

Hint Unavailable

Convert L --> mL, by multiplying by 1.000.
0.25 L = 250 mL $\begin{align} \text{Percent by Weight} & = \dfrac{\text{Solute Mass}}{\text{Solution Volume}} × 100\% \\ \\ & = \dfrac{5\ g}{250\ mL} × 100\% \\ \\ & = 2\% \ (W/V) \\ \\ \end{align}$

Percentage Weight by Weight Concentration % W/W

Calculate the percentage weight by weight of 25 g of pure sugar in 500g of solution. (The w/w concentration is probably used on your toothpaste). Solution

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% (W/W)

Hint Unavailable

$\begin{align} \text{Percent Weight by Weight} & = \dfrac{\text{Solute Mass}}{\text{Solution Mass}} × 100\% \\ \\ & = \dfrac{25\ g}{500\ g} × 100\% \\ \\ & = 5\% \ (W/W) \\ \\ \end{align}$

Summary of Concentrations (% V/V, % W/V, % W/W)

Determine the four INCORRECT combinations of units for calculating concentration (% V/V, % W/V, % W/W). Solution

g ÷ g

g ÷ L

g ÷ mL

g ÷ kg

kg ÷ mL

mL ÷ mL

kg ÷ kg

mL ÷ L

kg ÷ L

L ÷ L

The 6 correct combinations are:

g ÷ g

g ÷ mL

mL ÷ mL

kg ÷ kg

kg ÷ L

L ÷ L

The 4 incorrect combinations are:

g ÷ L

g ÷ kg

kg ÷ mL

mL ÷ L

The little guys go with the little guys: g, mL...
The big guys go with the big guys: kg, L...

Parts Per Million (PPM)

Which of the following would not equal 1.0 PPM? Solution

$\dfrac{1\ g\ \text{Solute}}{1,000,000\ g\ \text{Solution}}$

$\dfrac{1\ g\ \text{Solute}}{1,000,000\ mL\ \text{Solution}}$

$\dfrac{1\ g\ \text{Solute}}{1,000\ L\ \text{Solution}}$

$\dfrac{1\ mg\ \text{Solute}}{10\ L\ \text{Solution}}$

$\dfrac{1\ mg\ \text{Solute}}{1\ kg\ \text{Solution}}$

Parts Per Million (PPM) is a form of Percentage Weight by Weight Concentration % W/W.
Note that PPM is only used for dilute solutions in which 1g ≈ 1mL. $\begin{align} 1\ \text{PPM} & = \dfrac{1\ g\ \text{Solute}}{1,000,000\ g\ \text{Solution}} \\ \\ & = \dfrac{1\ g\ \text{Solute}}{1,000,000\ mL\ \text{Solution}} \\ \\ & = \dfrac{1\ g\ \text{Solute}}{1,000\ L\ \text{Solution}} \\ \\ & = \dfrac{1\ mg\ \text{Solute}}{1\ L\ \text{Solution}} \\ \\ & = \dfrac{1\ mg\ \text{Solute}}{1\ kg\ \text{Solution}} \\ \\ \end{align}$

Parts Per Million (PPM)

Calculate the mass of solute dissolved in a 5.0L solution that is 0.3 parts per million (ppm). Solution

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mg

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$\begin{align} 1\ \text{PPM} & = \dfrac{1\ mg\ \text{Solute}}{1\ L\ \text{Solution}} \\ \\ \\ \\ \\ \\ \\ \\ \text{PPM} & = \dfrac{mg\ \text{Solute}}{L\ \text{Solution}} \\ \\ 0.3\ \text{PPM} & = \dfrac{x\ mg\ \text{Solute}}{5.0\ L\ \text{Solution}} \\ \\ x\ mg\ \text{Solute} & = (0.3\ \text{PPM})(5.0\ L\ \text{Solution}) \\ \\ x & = 0.3 × 5.0 \\ \\ x & = 1.5\ mg \\ \\ \end{align}$

Parts Per Million (PPM)

The local lake water is considered safe to swim in when the level of E. coli bacteria is 100 ppm. If this lake water was put in a bathtub with a 150L volume, what number of E. coli bacteria would be in a bathtub? Solution

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bacteria

Hint Unavailable

In this case the number of solute organisms is used instead of the mass. $\begin{align} \text{PPM} & = \dfrac{n\ \text{Solute}}{L\ \text{Solution}} \\ \\ n\ \text{Solute} & = (\text{PPM})(L\ \text{Solution}) \\ \\ & = (100\ \text{PPM})(150\ L\ \text{Solution}) \\ \\ & = 15,000\ \text{bacteria} \\ \\ \end{align}$

Molarity

In solutions chemistry, the unit 'M' refers to Solution

mass, g

moles, mol

volume, L

concentration, mol/L

molar mass, g/mol

Molarity, mol/L

Calculating Molarity

1.0 mol/L of Na₂O molecules contains 2.0 mol/L of Na⁺ ions. Solution

True

False

One Na₂O molecule dissolves into 2 Na⁺ molecules and 1 O^2-. For every 1.0 mol/L of Na₂O molecules, there are 2.0 mol/L of Na⁺ ions and 1.0 mol/L of O^2- ions.

The concentration of chloride anions (Cl^-) in 1.0 mol of MgCl₂, dissolved in 1.0L of water is 2 mol/L Solution

True

False

MgCl₂ dissociates into 2 chloride (Cl^-) ions: $MgCl_2 \xrightarrow{} Mg^{2+} + 2Cl^{-}$ 1.0 mol of MgCl₂ dissolved in 1.0L of water = 1 mol/L MgCl₂.

Since there are every 1 MgCl₂ forms 2 Cl^- ions, then the molarity of chloride anions (Cl^-) = 2(1 mol/L)
= 2 mol/L

Calculating Molarity

Determine the molarity (molar concentration) of 1.0 mol NaCl dissolved in 250mL of water. Solution

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mol/L

Hint Unavailable

Although 1 mol NaCl dissociates into 2 mols of ions: Na⁺ + Cl^-, the molar concentration uses the mols of the solute, or compound instead of the total ion equivalent. $\begin{align} C & = \dfrac{n}{V} \\ \\ & = \dfrac{1.0\ \text{mol}}{0.250\ \text{L}} \\ \\ & = 4.0\ \text{mol/L} \\ \\ \end{align}$

Calculating Molarity

Determine the mass of potassium nitrate (KNO₃) required to make 1,250mL of a 0.50M solution. Solution

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g

Hint Unavailable

$\begin{align} C & = \dfrac{n}{V} \\ \\ n & = C × V \\ \\ & = 0.50 \text{mol/L} × 1.25 \text{L} \\ \\ & = 0.625 \text{mol} \\ \\ \end{align}$ $\begin{align} mass & = molar \ mass × moles \\ \\ m & = MM × n \\ \\ & = 101.1 \text{g/mol} × 0.625 \text{mol} \\ \\ & = 63.2\ \text{g} \\ \\ \end{align}$

Calculating Molarity

120mL of 4.5 mol/L potassium fluoride (KF) is added to 100mL of pure water. Determine the new molarity of the solution. Solution

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mol/L

Hint Unavailable

$\begin{align} n & = C × V \\ \\ & = 4.5 \text{mol/L} × 0.12 \text{L} \\ \\ & = 0.54 \text{mol} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ C_{new} & = \dfrac{n}{V_{total}} \\ \\ & = \dfrac{0.54 \text{mol}}{0.12 \text{L} + 0.10\text{L}} \\ \\ & = \dfrac{0.54 \text{mol}}{0.22 \text{L}} \\ \\ & = 2.45\ \text{mol/L} \\ \\ \end{align}$

Calculating Molarity

Determine the final concentration when 3.0 L of 0.30 M NaCl is mixed with 1.0 L of 1.5 M NaCl? Solution

C_f =

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mol/L

Hint Unavailable

$\begin{align} C & = \dfrac{n_{total}}{V_{total}} \\ \\ & = \dfrac{n_1 + n_2}{V_1 + V_2} \\ \\ & = \dfrac{C_1V_1 + C_2V_2}{V_1 + V_2} \quad\quad\quad\quad\quad\quad n_x = C_x × V_x \\ \\ & = \dfrac{(0.30\ M)(3.0\ L) + (1.5\ M)(1.0\ L)}{3.0\ L + 1.0\ L} \\ \\ & = \dfrac{0.9\ mol + 1.5\ mol}{4.0\ L} \\ \\ & = \dfrac{2.4\ mol}{4.0\ L} \\ \\ & = 0.6 \tfrac{mol}{L} \\ \\ \end{align}$

0.4 mol of NaCl is mixed with 0.3 mol of AlCl₃ in 2.0L of distilled water. Calculate the final concentration of chloride ions in solution [Cl^-]. Solution

C_f =

Hint Clear Info

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M

Hint Unavailable

NaCl and AlCl₃ dissociate (break apart) into cations and anions in solution.

NaCl has 1 [Cl^-] per molecule.
= 1(0.4 mol) of Cl^- per molecule = 0.4 mol

AlCl₃ has 3 [Cl^-] per molecule.
= 3(0.3 mol) of Cl^- per molecule = 0.9 mol $\begin{align} \text{total concentration of Cl} & = \dfrac{\text{total moles Cl}}{\text{total volume}} \\ \\ & = \dfrac{1(0.4 mol) + 3(0.3 mol)}{2.0L} \\ \\ & = \dfrac{0.4 mol + 0.9 mol}{2.0L} \\ \\ & = 0.65\ M \\ \\ \end{align}$

Dilution

Dilutions involve the addition of pure water to a solution.

Diluting a 2.0 mol/L solution of NaCl with distilled water will change the amount of moles of NaCl. Solution

True

False

Dilutions with pure water do not change the solvent amount (moles). Moles are constant.

However, the concentration does change.

25mL of a 4mol/L solution is used to prepare a 2.5mol/L solution. Determine the final volume in milliliters. Solution

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mL

Hint Unavailable

$\begin{align} mols_1 & = mols_2 \\ \\ C_1 V_1 & = C_2 V_2 \\ \\ (4 \text{mol/L})(0.025 \text{L}) & = (2.5 \text{mol/L})(V_2) \\ \\ V_2 & = \dfrac{(4 \text{mol/L})(0.025 \text{L})}{2.5 \text{mol/L}} \\ \\ & = 0.04 \text{L} \\ \\ & = 40\ \text{mL} \\ \\ \end{align}$

Dilution

45mL of stock solution is taken from a 2.0L supply and used to make 100mL of 0.5mol/L solution. Determine the concentration of the stock solution. Solution

Hint Clear Info

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mol/L

Hint Unavailable

Givens:
V₁ = 45mL = 0.045L
V₂ = 100mL = 0.1L
C₂ = 0.5mol/L
Find C₁ $\begin{align} C_1V_1 & = C_2V_2 \\ \\ C_1 & = \dfrac{C_2V_2}{V_1} \\ \\ & = \dfrac{(0.5)(0.1)}{0.045} \\ \\ & = 1.1\ mol/L \\ \\ \end{align}$

Gas Solubility

Gases cannot dissolve in liquids. Solution

True

False

Many gases dissolve in water, for instance the carbon dioxide in soda. Also swimming pools contain a small amount of chlorine gas.

Solubility Curve

Describe the slope of NH_{3 (g)} and AgNO_{3 (aq)} on a solubility curve of concentration (g/100 mL) over temperature (˚C). Solution
NH₃ is a gas at room temperature. The solubility of gases decreases with increasing temperature; therefore the slope is negative.

AgNO₃ is a solid, ionic compound that is soluble is water. The solubility of ionic compounds increases with increasing temperature; therefore the slope is positive.

Solubility

Which of the following statements is incorrect? Solution

Polar liquids dissociate in a polar solvent to form a homogenous solution

Oil is immiscible in a polar solvent like water

Increasing temperature increases the solubility of ionic compounds

Decreasing the temperature of a solvent allows more gas to dissolve in solution

Solid compounds are not considered miscible.

Polar liquids cannot dissociate, but they can dissolve. Solid ionic compounds dissociate, or break apart in solution.

Solubility

Determine the solubility of each ionic compound in water, as soluble, or insoluble.

Order (Ranking)	Cation	Anion	Solubility	Exceptions
1 (Highest)	Li⁺, Na⁺, K⁺, NH₄⁺	NO₃^-, ClO₃^-, CH₃COO^-	Soluble	Insoluble: Ca(ClO₃)₂
2	Ag⁺, Hg²⁺, Pb⁺	OH^-, PO₄^3-, CO₃^2-, O₂^-, S^2-,	*Insoluble*	Soluble: BaO, Ba(OH)₂, and Group 2 Sulfides
3		Cl^-, Br^-, I^-	Soluble
4	Ca²⁺, Sr²⁺, Ba²⁺		*Insoluble*
5 (Lowest)	Mg²⁺, Al³⁺, Fe²⁺, Fe³⁺, Cu²⁺, Zn²⁺	SO₄^2-	Soluble

Potassium Hydroxide, KOH Solution

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Hint Unavailable

Soluble

Due to potassium: KOH

Calcium Carbonate, CaCO₃ Solution

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Hint Unavailable

Insoluble

Due to Carbonate: CaCO₃

Calcium Nitrate, Ca(NO₃)₂ Solution

Hint Clear Info

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Hint Unavailable

Soluble

Due to Nitrate: Ca(NO₃)₂

Barium Chloride, BaCl₂ Solution

Hint Clear Info

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Hint Unavailable

Soluble

Due to Chloride: BaCl₂

Silver Chloride, AgCl Solution

Hint Clear Info

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Hint Unavailable

Insoluble

Due to Silver: AgCl

Mercury(II) Sulfate/Sulphate, HgSO₄ Solution

Hint Clear Info

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Hint Unavailable

Insoluble

Due to Mercury: HgSO₄

Magnesium Oxide, MgO Solution

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Hint Unavailable

Insoluble

Due to Oxide: MgO

Calcium Sulfate/Sulphate, CaSO₄ Solution

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Hint Unavailable

Insoluble

Due to Calcium, CaSO₄

Lead(I) Nitrate, PbNO₃ Solution

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Hint Unavailable

Soluble

Due to Nitrate: PbNO₃

Lead, Pb, is a very insoluble substance, however when combined with Nitrate, NO₃, which is the most soluble anion, the compound can dissociate/dissolve in water.

Strontium Acetate, Sr(CH₃COO)₂ Solution

Hint Clear Info

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Hint Unavailable

Soluble

Due to Acetate: Sr(CH₃COO)₂

Iron(II) Phosphate, Fe₃(PO₄)₂ Solution

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Hint Unavailable

Insoluble

Due to Phosphate: Fe₃(PO₄)₂

* Barium Oxide, BaO Solution

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Hint Unavailable

Soluble

*Exception.
Ba²⁺ and O₂² are both insoluble oxides, except when combined, the compound is soluble!

Lead(I) Chloride, PbCl Solution

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Hint Unavailable

Insoluble

Due to Lead: PbCl

* Beryllium Sulfide, BeS Solution

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Hint Unavailable

Soluble

*Exception.
Sulfides, S^2- are insoluble, except when combined with the group 2 cations, alkaline earth metals.

Solubility

Hardwater contains Ca²⁺ ions dissolved in solution. Ions are taken out of solution by precipitation into solids. Which of the following compounds would be best at lowering the calcium ion concentration in water. Solution

NaClO₃^-

NaNO₃

NaCl

NaSO₄

AgCl

NaSO₄ would precipitate the Ca²⁺ because calcium is more reactive and will displace the sodium ions.

The sulfate/sulphate becomes insoluble when bound to calcium - this precipitates the calcium out of solution, lowering its concentration. None of the other compounds would precipitate calcium.

Net Ionic Equation

Determine the net ionic equation for the reaction of potassium carbonate and aluminum nitrate. [Side note: the product of the reaction is not common in nature, meaning it usually doesn't form spontaneously, naturally.] Solution $K_2CO_{3(aq)} + Al(NO_3)_{3(aq)} \xrightarrow{}$
FIRST: Separate the ions of all soluble compounds, leaving the solid, liquid, and gaseous compounds.
$\begin{align} K_2CO_{3(aq)} + Al(NO_3)_{3(aq)} & \xrightarrow{} KNO_{3(aq)} + Al_2(CO_3)_{3(s)} \\ \\ 3K_2CO_{3(aq)} + 2Al(NO_3)_{3(aq)} & \xrightarrow{} 6KNO_{3(aq)} + Al_2(CO_3)_{3(s)} \\ \\ 6K^+_{(aq)} + 3CO_{3(aq)}^{2-} + 2Al^{3+}_{(aq)} + 6NO_{3(aq)} & \xrightarrow{} 6K^+_{(aq)} + 6NO_{3(aq)} + Al_2(CO_3)_{3(s)} \\ \\ \end{align}$ SECOND: Cancel ions on the left and right side, leaving the net ionic equation.
$\begin{align} \cancel{6K^+_{(aq)}} + 3CO_{3(aq)}^{2-} + 2Al^{3+}_{(aq)} + \cancel{6NO_{3(aq)}} & \xrightarrow{} \cancel{6K^+_{(aq)}} + \cancel{6NO_{3(aq)}} + Al_2(CO_3)_{3(s)} \\ \\ 3CO_{3(aq)}^{2-} + 2Al^{3+}_{(aq)} & \xrightarrow{} Al_2(CO_3)_{3(s)} \\ \\ \end{align}$

Solubility of Relative Unknowns

Four test tubes are filled with the following four ions: Na⁺, Ca²⁺, SO₄²⁺, and OH^-. However, it is not known which ion is in which test tube. Four different mixtures are prepared from the test tubes, and the observations are recorded in the table below. Using your understanding of solubility, determine which ion is in each test tube. Solution

Mixtures Test Tube 2 Test Tube 4

Test Tube 1: Soluble Soluble

Test Tube 3: Insoluble Insoluble

Test Tube 1: Na⁺

Test Tube 2: OH^-

Test Tube 3: Ca²⁺

Test Tube 4: SO₄²⁺

(Note that it could be switched where Test Tube 2 is SO₄²⁺ and Test Tube 4 is OH^-.)

Molarity Calculation with Volume

Determine the volume of 0.75 M KCl needed to react completely with 100.0 mL of 0.50M NaNO₃. Solution

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mL

Hint Unavailable

$\begin{array}{lcl} & 1KCl_{(aq)} \ \ + & 1NaNO_{3(aq)} \xrightarrow{} KNO_{3(aq)} + NaCl_{(aq)} \\ \\ &\!\! 0.75\ M & 0.50\ M \\ &\!\! V = x\ mL & 100.0\ mL \end{array}$ $\begin{align} \dfrac{n_{KCl}}{1} & = \dfrac{n_{NaNO_3}}{1} \\ \\ C_{KCl} × V_{KCl} & = C_{NaNO_3} × V_{NaNO_3} \quad\quad\quad\quad n = C × V \\ \\ V_{KCl} & = \dfrac{C_{NaNO_3} × V_{NaNO_3}}{C_{KCl}} \\ \\ & = \dfrac{0.50\ M × 0.1\ L}{0.75\ M} \\ \\ & = 0.067\ L \\ \\ & = 67\ mL \\ \\ \end{align}$

Molarity Calculation with Volume

If 0.50 L of 1.500 mol/L Na₂O is mixed with excess AgClO₃, what mass of precipitate will form?
$\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Svante Arrhenius 1884

Arrhenius defined acids and bases as Solution

Acids are proton (H⁺) donors, and bases are hydroxide (OH^-) acceptors

Acids are proton (H⁺) donors, and bases are proton (H⁺) acceptors

Acids are proton (H⁺) donors, and bases are hydroxide (OH^-) donors

Acids are electron (e^-) acceptors, and bases are electron (e^-) donors

Acids are electron (e^-) donors, and bases are electron (e^-) acceptors

According to Arrhenius,
Acids are proton (H⁺) donors, and bases are hydroxide (OH^-) donors

(Arrhenius' definitions become outdated when Brønsted and Lowry offered their definitions, and later Lewis...)

Johannes Brønsted and Thomas Lowry 1923

Brønsted and Lowry defined acids and bases as Solution

Acids are proton (H⁺) donors, and bases are hydroxide (OH^-) acceptors

Acids are proton (H⁺) donors, and bases are proton (H⁺) acceptors

Acids are proton (H⁺) donors, and bases are hydroxide (OH^-) donors

Acids are electron (e^-) acceptors, and bases are electron (e^-) donors

Acids are electron (e^-) donors, and bases are electron (e^-) acceptors

According to Brønsted and Lowry,
Acids are proton (H⁺) donors, and bases are proton (H⁺) acceptors

(Brønsted and Lowry's definition was more general than Arrhenius' and encompassed more circumstances)

Brønsted Amphoteric

A water molecule can act as both a Brønsted acid and a Brønsted base. Solution

True

False

Water can donate a proton to become OH^-. Water can also accept a proton to become H₃O⁺. This depends on what else is there with the water. For example...

Arrhenius, Brønsted-Lowry

Ammonia (NH₃) violates which definition of a base? Solution

Arrhenius

Brønsted-Lowry

Lewis

None of the above

Ammonia does not contain hydroxide (OH^-) ions.

For example in pure water, Ammonia, NH₃ is a Brønsted-Lowry base because it is a proton acceptor, but is not a hydroxide donor according to Arrhenius' definition. (Arrhenius defined bases as hydroxide donors). $NH_3 + H_2O \longleftrightarrow NH_4^{+} + OH^{-}$ As you can see the base, ammonia is a proton acceptor (Brønsted-Lowry base) but does not fall under the Arrhenius category of a base (hydroxide donor).

Brønsted-Lowry Base Strength

Which of the following Brønsted bases is the strongest? Solution

F^-

Cl^-

Br^-

I^-

At^-

A Brønsted base is a proton acceptor. Fluoride ions are more electronegative and have a smaller ionic radius.
Furthermore, the conjugate acid HF is the weakest acid. Remember that the conjugate base of a weak acid is a strong base.

Acid Base Neutralization

Hydrochloric acid neutralizes sodium hydroxide base in the following reaction. What are the products of this reaction? Solution $HCl_{(aq)} + NaOH_{(aq)} \longleftrightarrow \ ?$

H₂ + NaOCl

Na + Cl + H₂O

Na + Cl₂ + H₂O

NaCl + H₂ + O₂

NaCl + H₂O

Acid + Base ↔ Salt + Water... $HCl_{(aq)} + NaOH_{(aq)} \longleftrightarrow \ NaCl + H_2O$

Identifying Acids and Bases

Which of the following mixtures would result in a neutralization reaction? Solution

Mixture Compound 1 Compound 2

A HCl HNO₃

B K₂SO₄ H₂O

C OH^- H₂CO₃

D CH₃COO^- NaOH

A

B

C

D

None of the above

An acid + base results in a neutralization reaction... This would be mixture C.

Acid = H₂CO₃
Base = OH^-

Acids and Bases

Which of the following statements about acids and bases is false?

Acids turn blue litmus red

Bases turn red litmus blue

Strong acids are strong electrolytes and conduct much more electricity than weak acids

A solution is acidic when [H₃O⁺] < [OH^-]

Strong acids and bases dissociated almost completely while weaker ones dissociate much less

pH Units

What are the units used for pH?

mol

L

mol/L

M

pH does not have units

Calculating pH

Which sample has an incorrect pH calculation in the following table? Solution

Sample [H⁺_(aq)] pH

A 1.0 × 10^—13 13

B 1.0 × 10^—2 2

C 1.0 × 10^—8 8

D 1.0 × 10⁰ 1

E 1.0 × 10^—5 5

A

B

C

D

E

[H⁺_(aq)] = 1.0 × 10^-pH

[1.0 × 10⁰] has a pH = 0

Calculating pH

Calculate the pH of a solution with [H⁺_(aq)] = 2.8 × 10^-3 mol/L.

pH = (2.8 × 10^-3)²

pH = log(2.8 × 10^-3)

pH = $log\left(\dfrac{1}{2.8 × 10^{-3}}\right)$

pH = -log(2.8 × 10^-3)

pH = $\dfrac{1}{2.8 × 10^{-3}}$

Which of the following quantities of H⁺ would make an acidic 5.0 L solution? Solution

1.0 × 10^-7 mol

9.0 × 10^-8 mol

9.0 × 10^-7 mol

5.0 × 10^-7 mol

1.0 × 10^-8 mol

Note: you must divide the moles by the 5.0L volume to obtain the concentration in mol/L.

The acidic solution has a pH less than 7 and a [H⁺] > 1.0 × 10^-7 mol/L. $\begin{align} pH & = -log\left( [H_3O^+] \right) \\ \\ pH & = -log\left( [H^+] \right) \\ \\ & = -log\left[ \dfrac{n\ H^+}{V} \right] \\ \\ & = -log\left[ \dfrac{9.0 × 10^{-7}}{5.0\ L} \right] \\ \\ & = -log\left[ 1.8 × 10^{-7} \right] \\ \\ & = 6.7 \\ \\ \end{align}$

Calculating pH

Calculate the pH of a solution of 0.150 mol/L of H₂SO₄, a strong acid. Solution

pH =

Hint Clear Info

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Hint Unavailable

H₂SO₄ contains 2 protons per molecule.

So, 2 H₃O⁺ molecules will be made from one molecule of H₂SO₄. $\begin{align} pH & = -log\left( [H_3O^+] \right) \\ \\ & = -log\left( 2 × 0.150\ mol/L \right) \\ \\ & = -log\left( 0.30\ mol/L \right) \\ \\ & = 0.52 \\ \\ \end{align}$

Conjugate Pair of Acids and Bases

Which of the following is a conjugate acid-base pair? Solution $H_2CO_3 + NaOH \xrightarrow{} NaHCO_3 + H_2O$

H₂CO₃ and H₂O

NaOH and NaHCO₃

H₂CO₃ and NaOH

H₂CO₃ and NaHCO₃

NaHCO₃ and H₂O

The carbonic acid H₂CO₃ dissociates into the bicarbonate anion, which forms a salt with the available sodium ions, as sodium bicarbonate: NaHCO₃

Conjugate Pair of Acids and Bases

Which of the following is a conjugate acid-base pair? $NH_{3(aq)} + H_2O_{(l)} \xrightarrow{} NH^{4+}_{(aq)} + OH^-_{(aq)}$

Acid Base Titrations

The following table shows the ratio of strong acid and strong base required to neutralize. Determine which pair has an incorrect ratio. Solution

Acid Base

2 H₃PO₄ 3 Mg(OH)₂

1 H₂SO₃ 2 NaOH

1 HCl 1 KOH

1 H₃O⁺ 3 OH^-

2 HNO₂ 1 Ba(OH)₂

2 H₃PO₄ and 3 Mg(OH)₂

1 H₂SO₃ and 2 NaOH

1 HCl and 1 KOH

1 H₃O⁺ and 3 OH^-

HNO₂ and 1 Ba(OH)₂

1 H₃O⁺ and 3 OH^- is the incorrect ratio. Instead, the correct ratio should be:
1 H₃O⁺ and 1 OH^- $1H_3O^+ + 1OH^- \xrightarrow{} 2H_2O$

Titrations and Indicators

Universal indicators are used to test a wide range of pH.

Creative Commons Credit: CC BY-SA 3.0, Kiyok, 2008

The point at which the indicator starts to change color is the: Solution

Midpoint

Titrant

Endpoint

Equivalence point

Inflection point

As the pH of solution changes, the indicator will change color at the endpoint when the specific pH for that indicator is reached.

(Note that many metallic ions make different color solutions and this is not an endpoint).

What is the equivalence point in a titration? Solution

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Hint Unavailable

The equivalence point is the point at which the amount of moles of the acid equals the moles of the base to completely neutralize the other.

Some titrations with polyprotic acids or polyhydroxic bases.

Acid Base Titrations

What volume of 2.0 M H₃PO₄ will neutralize 500 mL of 1.5 M NaOH? Write the neutralization equation. Solution

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L

Hint Unavailable

Neutralization occurs when moles of acid = moles of base: $\begin{align} n_{acid} & = n_{base} \\ \\ (c_a)(V_a)(x_a) & = (c_b)(V_b)(x_b) \quad\quad\quad\quad\quad \text{x is the number of protons or hydroxides} \\ \\ (V_a) & = \dfrac{(c_b)(V_b)(x_b)}{(c_a)(x_a)} \\ \\ & = \dfrac{(1.5\ M)(0.5\ L)(1\ OH)}{(2.0\ M)(3\ H)} \\ \\ & = \dfrac{0.75}{6} \\ \\ & = 0.125\ L \\ \\ \end{align}$

Acid Base Titrations Lab

A chemistry student has titrated a strong base using a strong acid. A 0.5M hydrochloric acid (HCl) was slowly dripped one drop at a time from a buret into a beaker containing an unknown concentration of 50mL of Mg(OH)₂. The volume of acid that changed the colour of the indicator in the unknown sample was recorded in the table. Find the unknown concentration of base. Solution

Trial: 1 2 3

Initial buret reading (mL): 42.0 30.4 19.0

Final buret reading (mL): 30.4 19.0 6.9

Volume of HCl added: 11.6 11.4 12.1

Hint Clear Info

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M

Hint Unavailable

$\begin{align} \text{Average Volume HCL} & = \dfrac{11.6 + 11.4 + 12.1}{3} \\ \\ & = 11.7\ mL \\ \\ \\ \\ \\ \\ n_{acid} & = n_{base} \\ \\ (c_a)(V_a)(x_a) & = (c_b)(V_b)(x_b) \quad\quad\quad\quad\quad \text{x is the number of protons or hydroxides} \\ \\ (c_b) & = \dfrac{(c_a)(V_a)(x_a)}{(V_b)(x_b)} \\ \\ & = \dfrac{(0.5\ M)(0.0117\ L)(1\ H)}{(0.05\ L)(2\ OH)} \\ \\ & = 0.06\ M \\ \\ \end{align}$

Titration Curves

The following titration curve represents Solution

A weak acid added to a strong base

A strong acid added to a strong base

A strong acid added to a weak base

A stronger base added to a strong acid

None of the above

At the equivalence point, the moles of acid is equal to the moles of base. The equivalence point of the titration is > 7 as you can see on the graph. An acidic equivalence point indicates a stronger acid, a basic equivalence point indicates a stronger base, and a neutral equivalence point indicates equal strength of acid and base.

Solutions Units

1 mmol equals Solution

1000 mol/L

1 mol/L

1 mol

1000 mol

1 × 10^-3 mol

Milli, 'm', represents 1/1000th

pH and pOH

There is a constant relationship between pH and pOH.

The following equations always apply to acids and bases. Solution 1 × 10^-14 = [OH^-][H⁺]
14 = pH + pOH

True

False

True. These equations are used often, see below.

Determine the pH of a 0.5 M solution of NaOH Solution

pH =

Hint Clear Info

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Hint Unavailable

Determine pOH based on the concentration [OH^-] = 0.5 M $\begin{align} pOH & = -log(0.5) \\ \\ pOH & = 0.3 \\ \\ \end{align}$ pH + pOH = 14 $\begin{align} pH & = 14 - pOH \\ \\ pH & = 14 - 0.3 \\ \\ pH & = 13.7 \\ \\ \end{align}$

Concentration Combinations

A 1.1 L of a 0.6 M solution of H₃O⁺ is added to a 2.5 M stock solution of H₃O⁺. If the total concentration is 1.5 M, determine the volume of the stock solution. Solution

V =

Hint Clear Info

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L

Hint Unavailable

Let the volume of the stock solution be V_s $\begin{align} concentration & = \dfrac{moles}{volume} \\ \\ concentration & = \dfrac{n_1 + n_s}{V_1 + V_s} \\ \\ concentration & = \dfrac{C_1V_1 + C_2V_s}{V_1 + V_s} \\ \\ 1.5 & = \dfrac{(0.6)(1.1) + 2.25 V_s}{1.1 + V_s} \\ \\ 1.5 & = \dfrac{0.66 + 2.25 V_s}{1.1 + V_s} \\ \\ 1.5(1.1 + V_s) & = 0.66 + 2.25 V_s \\ \\ 1.65 + 1.5V_s & = 0.66 + 2.25 V_s \\ \\ 1.5V_s - 2.25 V_s & = 0.66 - 1.65 \\ \\ -0.75 V_s & = -0.99 \\ \\ V_s & = \dfrac{-0.99}{-0.75} \\ \\ V_s & = 1.32\ L \\ \\ \end{align}$

Acids

Calculate the pH assuming the acid completely dissociates into PO₄^3-. Solution

pH =

Hint Clear Info

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Hint Unavailable

H₂PO₄ → 3H⁺ + PO₄^3- Convert mass to mol $\begin{align} n & = \dfrac{m}{MM} \\ \\ n & = \dfrac{20.0}{98} \\ \\ & = 0.20\ mol \\ \\ \end{align}$ Stoichiometry: 1H₂PO₄ → 3H⁺ $\begin{align} & = 0.20\ mol × 3 \\ \\ & = 0.60\ mol\ H^+ \\ \\ \end{align}$ Convert mol H⁺ to concentration first, [H₃O⁺] $\begin{align} C & = \dfrac{n}{V} \\ \\ C & = \dfrac{0.60\ mol}{2\ L} \\ \\ & = 0.3\ M \\ \\ \end{align}$ Calculate pH based on concentration, [H₃O⁺] $\begin{align} pH & = -log[H_3O^+] \\ \\ & = -log[0.3] \\ \\ & = 0.5 \\ \\ \end{align}$

Acid Base Reaction pH

80mL of a 2.5M solution of sulfuric acid is added to 110mL of a 1.1mol/L solution of sodium hydroxide.

Determine the total number of moles of protons (H⁺) and hydroxides (OH^-) in solution before neutralization occurs. Solution

Hint Clear Info

mol H⁺ =
mol OH^- =

Incorrect Attempts:

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Hint Unavailable

$\begin{array}{lcl} H_2SO_4 \quad\quad + & & 2NaOH \quad \xrightarrow{} \quad 2H_2O \quad + \quad Na_2SO_4 \\ \\ \\ n = CV & & n = CV \\ = (2.5 mol/L)(0.08L) & & = (1.1 mol/L)(0.11L) \\ = 0.2 mol & & = 0.121 mol \\ = 0.2 mol × 2H^+ & & = 0.121 mol × 1OH^- \\ = 0.4 mol & & = 0.121 mol \\ \end{array}$ There are total 0.4 mol of protons (H⁺), and 0.121 mol of hydroxides (OH^-) in solution, before neutralization.

Determine the limiting reagent and the excess amount of moles of protons or hydroxides after neutralization. Solution

Hint Clear Info

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mol

Hint Unavailable

The NaOH is limiting because the lower amount of total moles.

Neutralization = 1 H⁺ + 1 OH^- = 1 H₂O $\begin{align} Net\ amount & = \text{Excess} - \text{Limiting} \\ \\ & = 0.4 mol - 0.121 mol \\ \\ & = 0.279\ mol\ of\ H^+ \\ \\ \end{align}$

Calculate the pH Solution

pH =

Hint Clear Info

Incorrect Attempts:

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Hint Unavailable

first need to calculate the concentration of [H⁺]

(Use total Volume) $\begin{align} C & = \dfrac{n}{V} \\ \\ C & = \dfrac{0.279}{0.08 + 0.11} \\ \\ & = 1.46\ M \\ \\ \end{align}$ then calculate pH $\begin{align} pH & = -log[H^+] \\ \\ pH & = -log[1.46] \\ \\ pH & = -0.16 \\ \\ \end{align}$

Flashcards: Solutions and Solubility

What is the difference between a homogeneous and heterogeneous mixture?

Homogeneous mixtures are uniform/consistent throughout and the components are indistinguishable. The components of heterogeneous mixtures can be distinguished visually.

Explain why a metal alloy is or is not a solution

An alloy is a uniform mixture of two or more metals. It is possible to have solid solutions, liquid solutions, and gas solutions.

What is an electrolyte?

An electrolyte is a solution composed of dissolved (aqueous) ions that allow the solution to conduct electricity.

Explain how the saturation point of a solution can be changed.

By changing the temperature of the solution. (For a solid, increasing the temperature increases the saturation point while decreasing the temperature decreases the saturation point. The opposite is true for gases.)

Explain why sugar water is or is not an electrolyte solution.

Sugar is not an electrolyte solution because it is not ions dissolved in a solvent — electrolytes are made only of aqueous ions. Sugar is a covalent/molecular compound and does not dissolve as ions in solution.

What are the aqueous ions formed when sulfuric acid (H₂SO₄) dissolves?

H⁺ and SO₄^2-

What is the difference between dissolving and dissociation?

Dissociation is any chemical splitting of a molecule into its smaller parts, while dissolving is the chemical splitting of a molecule due to the presence of a liquid such as water.

Correct the following statement:

Hydrogen bonding can occur whenever there are hydrogen bonds in a molecule.

Hydrogen bonding can occur when the hydrogen is bonded to any one of three highly electronegative atoms: nitrogen, oxygen, or fluorine. The recipient of the hydrogen bond must have a partial negative dipole (like the oxygen in OH), or a negative charge like a lone pair of electrons.

What is parts per million (PPM)?

This is a density, calculated in several ways as either:
grams particles ÷ million grams solute
grams particles ÷ million mL solute
grams particles ÷ thousand L solute
...

What is molarity (M)?

The amount of solute per in a certain volume of solution.

mole/Liter

What is the concentration of chloride ions in a 3M solution of MgCl₂?

How is volume calculated, given concentration and moles?

$V = \dfrac{n}{C}$

How does temperature affect the solubility of a gas in water?

The relationship is inverse. For example, the solubility of gases decreases with increasing temperature.

Predict the solubility of any cation bonded to nitrate (NO₃^-)

Soluble

(nitrate is one of the most soluble ions known: any ionic compound containing nitrate is soluble in water)

State the net ionic equation of
$K_2CO_{3(aq)} + Al(NO_3)_{3(aq)} \xrightarrow{} KNO_{3(aq)} + Al_2(CO_3)_{3(s)}$

$3CO_{3(aq)}^{2-} + 2Al^{3+}_{(aq)} \xrightarrow{} Al_2(CO_3)_{3(s)}$

(net ionic equations only include the ions that contribute to making the solid product)

What is a spectator ion?

Spectator ions are ions that are present on both sides of the ion equation and do not contribute to make the solid product.

Gases and Atmospheric Chemistry

Hydrocarbons and Energy

	Ag⁺	Cu²⁺	Fe³⁺	Ti⁴⁺
Cl^-
O^2-
PO₄^3-		( )		( )

Reactivity Ranking	Element
1 (Most Reactive)	Lithium, Li
2	Potassium, K
3	Barium, Ba
4	Calcium, Ca
5	Sodium, Na
6	Magnesium, Mg
7	Aluminum, Al
8	Zinc, Zn
9	Chromium, Cr
10	Iron, Fe
11	Cadmium, Cd
12	Cobalt, Co
13	Nickel, Ni
14	Tin, Sn
15	Lead, Pb
16	Hydrogen, H
17	Copper, Cu
18	Mercury, Hg
19	Silver, Ag
20	Platinum, Pt
21 (Least Reactive)	Gold, Au

Isotope	Relative Abundance
²⁸Si	92.2 %
²⁹Si	4.7 %
³⁰Si	3.1 %

	2 NH₃	1 CH₄
Mole Ratio	2	1
Molecules	2	1
Molecules in 1 mole		6.02 × 10²³

	2 NH₃	1 CH₄
Mole Ratio	2	1
Molecules	2	1
Molecules in 1 mole	1.2 × 10²⁴	6.02 × 10²³

Element	Electronegativity
O	3.44
H	2.2
C	2.55
Al	1.61
Cl	3.16
Li	0.98
S	2.58
Na	0.93
F	3.98

Compound	Electronegativity Difference (∆EN)
H₂O	1.4
CH₄
AlCl₃
Li₂S
NaF

ClO₄^–	perchlorate
ClO₃^–	chlorate
ClO₂^–	chlorite
ClO^–	hypochlorite

BrO₄^–	perbromate
BrO₃^–	bromate
BrO₂^–	bromite
BrO^–	hypobromite

Mass	Mols	Ratio	Subscript
1.9 g Oxygen	$= \dfrac{1.9\ g}{16.0\ g/mol} = 0.1188\ mol$	$= \dfrac{0.1188}{0.1188} = 1$	= 1
3.0 g Magnesium	$= \dfrac{3.0\ g}{24.31\ g/mol} = 0.1234\ mol$	$= \dfrac{0.1234}{0.1188} = 1.04$	≈ 1

Component	Mass (g)	Moles (mol)	Amount (#)
Ba	56.22	0.409379	0.409379/0.409379 = 1
Cl	29.02496	2(0.409379)	2(0.409379)/0.409379 ≈ 2
H₂O	14.755	0.8188	0.8188/0.409379 ≈ 2

Mixtures	Test Tube 2	Test Tube 4
Test Tube 1:	Soluble	Soluble
Test Tube 3:	Insoluble	Insoluble

Mixture	Compound 1	Compound 2
A	HCl	HNO₃
B	K₂SO₄	H₂O
C	OH^-	H₂CO₃
D	CH₃COO^-	NaOH

Sample	[H⁺_(aq)]	pH
A	1.0 × 10^—13	13
B	1.0 × 10^—2	2
C	1.0 × 10^—8	8
D	1.0 × 10⁰	1
E	1.0 × 10^—5	5

Acid	Base
2 H₃PO₄	3 Mg(OH)₂
1 H₂SO₃	2 NaOH
1 HCl	1 KOH
1 H₃O⁺	3 OH^-
2 HNO₂	1 Ba(OH)₂

Trial:	1	2	3
Initial buret reading (mL):	42.0	30.4	19.0
Final buret reading (mL):	30.4	19.0	6.9
Volume of HCl added:	11.6	11.4	12.1

Pressure:	101,325 Pa
Volume:	18,500 mL
Temperature:	10 ˚C

Reactant 1	Reactant 2	Product	Ratio
H_{2 (g)}	S_(g)	H₂S_(g)	1:1:1
H_{2 (g)}	Cl_{2 (g)}	2HCl_(g)	1:1:2
2H_{2 (g)}	O_{2 (g)}	2H₂O_(g)	2:2:2
N_{2 (g)}	2O_{2 (g)}	2NO_{2 (g)}	1:2:2
N_{2 (g)}	3H_{2 (g)}	2NH_{3 (g)}	1:3:2

Hydrocarbon	Boiling Point (˚C)
CH₄	-164
C₃H₆	-42
C₄H₁₀	-1
C₅H₁₂	36.1
C₈H₁₈	125

Heat Event	Heat (J)
Heat ice from -8˚C to 0˚C	x Joules
Convert ice from solid to liquid at 0˚C	y Joules
Heat water from 0˚C to 30˚C	z Joules

FAVORITES

OPTIONS

CLASS OPTIONS

GRADE

11

Chemistry SCH3U

TABLE OF CONTENTS

Matter, Chemical Bonding, and Naming

Background: Elements

Which of the following is not a metal? Solution Sodium Gallium Silicon Tin Magnesium Silicon (Si) is a non-metal

Background: Atomic Numbers

Background: Substances

Bohr Nuclear Model

The following three terms are equivalent: Orbit, Shell, Energy Level True False

Electron Transitions

In an atom, energy is released in the form of light in certain wavelengths when: Electrons transition or fall to a lower energy level Electrons leave the ground state Electrons change direction of travel on an energy level Electrons transition to a higher energy level Protons and neutrons collide

Electrons on Shells (Energy Levels)

The maximum number of electrons on the third energy level is: Solution Hint Clear Info Incorrect Attempts: CHECK electrons Hint Unavailable The maximum number of electrons on an energy level equals 2(n)2. 2(3)2 = 18. 2 electrons in the first, 8 in the second, and actually 18 in the third.

Periodic Trends

Which element will have the largest atomic radius? Solution Nitrogen, N Phosphorus, P Lithium, Li Rubidium, Rb Tin, Sn Atomic radius increases down a group, and from right-to-left across a period. The lower-left elements have the largest atomic radius (size).

Periodic Trends: Atomic Radius of Anions

Periodic Trends: Atomic Radius of Cations

The atomic radius of a cation is always smaller than the original neutral element. Solution True False Because removing an electron decreases some of the electron-electron repulsions in a neutral element causing the cation to be smaller than its neutral atom.

Periodic Trends: Ionization & Electron Affinity

Ionization energy is defined as the energy required to add an electron to an atom or ion. Electron affinity is the energy change that occurs when an electron is added to an atom. Solution True False Ionization energy is defined as the energy required to remove an electron to an atom or ion.

Periodic Trends

Periodic Trends

Which element has the highest electronegativity? Solution Zinc, Zn Mercury, Mg Potassium, K Bromine, Br Cesium, Cs Electronegativity is the ability of an atom to attract electrons when bonded. Electronegativity increases left-to-right across a period, and bottom-to-top up a group.

An element with a low electron affinity will most likely also have High ionization energy High polarity High electronegativity Large atomic radius Low melting point

Background: Compounds

Which of the following would form a molecular compound? Solution Metal + Metal Metal + Nonmetal Nonmetal + Nonmetal Crystal Lattice Ions None of the above Molecular compounds form by combining two nonmetals. Ionic compounds are formed by combining a metal + a nonmetal.

Background: Ionic Versus Covalent

An ionic bond forms with atoms that Solution Share electrons Share electrons and neutrons Transfer electrons Transfer protons None of the above Ionic bonds form when electrons are transferred from the positive cation to the negative anion.

Electron Dot Diagrams with Ionic Compounds

Forming Covalent Compounds

Identifying Covalent Compounds

Which of the following is a covalent compound? Solution Video KF MgCl2 NH3 Na+OH- NaNO3 Covalent is two non-metals...

Lewis Structures: Intro

A Lewis structure for an element shows an arrangement of dots around the symbol of the element. These dots indicate the: Nuclear charge Atomic Number Total number of electrons Number of electrons in the outer, valence shell Charge on the element

Lewis Structures of Covalent Compounds

Carbon would form the central atom in the molecule below. Solution CH2BrF True False

Lewis Structures of Covalent Compounds

Draw the Lewis Structures of the following molecules using the rules below.

CH4 Solution

H2O Solution Video

Br2 Solution

NH3 Solution Video

N2H2 Solution

OH- Solution

CO2 Solution Video

Physical Properties of (Covalent) Molecular Compounds

Molecular compounds have a high melting point and are poor conductors of electricity when dissolved in water. Solution True False Molecular compounds have a low melting point and are poor conductors of electricity when dissolved in water.Ionic compounds have a high melting and boiling point.

Molecular Forces

Which of the following is not an intermolecular force? Solution London Dispersion Van der Waals Dipole-Dipole Covalent Hydrogen Bonding Covalent is an intra-molecular bond...

PERIODIC TRENDS

Electronegativity

Rank the elements in terms of increasing electronegativity (from bottom to top). Solution H, Br, C, F, P, Mg, K, I Mg Br F I K H P C Electronegativity increases up and to-the-right on the periodic table.

Neon has a higher electronegativity than Fluorine. Solution True False Rule notwithstanding, noble gases have extremely low electronegativity because of their full valence octet.

Electronegativity Difference (∆EN)

Polarity Based on Electronegativity

Determine the polarity of the following molecules using the following guidelines. (* Indicates an exception to the guidelines).

H2 Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Non-polar Covalent \begin{align} \Delta \text{EN} & = 2.20 - 2.20 \\ \\ & = 0 \\ \\ \end{align}

CH4 Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Non-polar Covalent \begin{align} \Delta \text{EN} & = 2.55 - 2.20 \\ \\ & = 0.35 \\ \\ \end{align}

LiF Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Ionic \begin{align} \Delta \text{EN} & = 3.98 - 0.98 \\ \\ & = 3.00 \\ \\ \end{align}

H2O Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable Polar Covalent \begin{align} \Delta \text{EN} & = 3.44 - 2.20 \\ \\ & = 1.24 \\ \\ \end{align}

MgI2 Solution Hint Clear Info Incorrect Attempts: CHECK Hint Unavailable *Ionic* \begin{align} \Delta \text{EN} & = 2.66 - 1.31 \\ \\ & = 1.35 \\ \\ \end{align} This molecule is ionic because it is made of a metal plus a non-metal, and it can dissolve into Mg2+ and 2I- in water as an electrolyte.

Exceptions to Polarity Based on Electronegativity

Polarity

Physical Properties, Boiling and Melting Point Based on Polarity

Using the polarity of the following molecules as a guide, order the molecules from low to high boiling points (from bottom to top). Solution CH4 H2O CO2 LiF Non-polar covalent < Polar covalent < Ionic. Rank from low ∆EN to high ∆EN: 0.35 (CH4), 0.89 (CO2), 1.24 (H2O), 3.00 (LiF)

Ionic Compounds

Which of the following ions would fit into this space indicated by 'Z' below? Solution MgZ2 Na2+ Li+ Ca2+ O2- Br- Br-

Chemical Formulas

NAMING

When to Use All the Different Naming Systems

Compounds use different naming systems and it is important to know when to use each one.

Naming Molecular/Covalent

Prefix Naming

Order the prefixes, with lowest at the top, by dragging them around. Solution deca tri penta di hepta octa tetra hexa nona mono 1) Mono2) di3) tri4) tetra5) penta6) hexa7) hepta8) octa9) nona10) deca

Naming Covalent Molecules

Which of the following is not a metal? Solution

Sodium

Gallium

Silicon

Tin

Magnesium

Silicon (Si) is a non-metal

The following three terms are equivalent: Orbit, Shell, Energy Level

True

False

The maximum number of electrons on the third energy level is: Solution

Hint Clear Info

Incorrect Attempts:

CHECK

electrons

Hint Unavailable

The maximum number of electrons on an energy level equals 2(n)².

2(3)² = 18.

2 electrons in the first, 8 in the second, and actually 18 in the third.

Which element will have the largest atomic radius? Solution

Nitrogen, N

Phosphorus, P

Lithium, Li

Rubidium, Rb

Tin, Sn

Atomic radius increases down a group, and from right-to-left across a period. The lower-left elements have the largest atomic radius (size).

The atomic radius of a cation is always smaller than the original neutral element. Solution

True

False

Because removing an electron decreases some of the electron-electron repulsions in a neutral element causing the cation to be smaller than its neutral atom.

Ionization energy is defined as the energy required to add an electron to an atom or ion. Electron affinity is the energy change that occurs when an electron is added to an atom. Solution

True

False

Ionization energy is defined as the energy required to remove an electron to an atom or ion.

Which element has the highest electronegativity? Solution

Zinc, Zn

Mercury, Mg

Potassium, K

Bromine, Br

Cesium, Cs

Electronegativity is the ability of an atom to attract electrons when bonded.

Electronegativity increases left-to-right across a period, and bottom-to-top up a group.

An element with a low electron affinity will most likely also have

High ionization energy

High polarity

High electronegativity

Large atomic radius

Low melting point

Which of the following would form a molecular compound? Solution

Metal + Metal

Metal + Nonmetal

Nonmetal + Nonmetal

Crystal Lattice Ions

None of the above

Molecular compounds form by combining two nonmetals. Ionic compounds are formed by combining a metal + a nonmetal.

An ionic bond forms with atoms that Solution

Share electrons

Share electrons and neutrons

Transfer electrons

Transfer protons

None of the above

Ionic bonds form when electrons are transferred from the positive cation to the negative anion.

Which of the following is a covalent compound? Solution Video

KF

MgCl₂

NH₃

Na⁺OH^-

NaNO₃

Covalent is two non-metals...

A Lewis structure for an element shows an arrangement of dots around the symbol of the element. These dots indicate the:

Nuclear charge

Atomic Number

Total number of electrons

Number of electrons in the outer, valence shell

Charge on the element

Carbon would form the central atom in the molecule below. Solution CH₂BrF

True

False

CH₄ Solution

H₂O Solution Video

Br₂ Solution

NH₃ Solution Video

N₂H₂ Solution

OH^- Solution

CO₂ Solution Video

Molecular compounds have a high melting point and are poor conductors of electricity when dissolved in water. Solution

True

False

Molecular compounds have a low melting point and are poor conductors of electricity when dissolved in water.

Ionic compounds have a high melting and boiling point.

Which of the following is not an intermolecular force? Solution

London Dispersion

Van der Waals

Dipole-Dipole

Covalent

Hydrogen Bonding

Covalent is an intra-molecular bond...

Rank the elements in terms of increasing electronegativity (from bottom to top). Solution H, Br, C, F, P, Mg, K, I

Mg

Br

F

I

K

H

P

C

Electronegativity increases up and to-the-right on the periodic table.

Neon has a higher electronegativity than Fluorine. Solution

True

False

Rule notwithstanding, noble gases have extremely low electronegativity because of their full valence octet.

H₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Non-polar Covalent

$\begin{align} \Delta \text{EN} & = 2.20 - 2.20 \\ \\ & = 0 \\ \\ \end{align}$

CH₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Non-polar Covalent

$\begin{align} \Delta \text{EN} & = 2.55 - 2.20 \\ \\ & = 0.35 \\ \\ \end{align}$

LiF Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Ionic

$\begin{align} \Delta \text{EN} & = 3.98 - 0.98 \\ \\ & = 3.00 \\ \\ \end{align}$

H₂O Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Polar Covalent

$\begin{align} \Delta \text{EN} & = 3.44 - 2.20 \\ \\ & = 1.24 \\ \\ \end{align}$

Which of the following ions would fit into this space indicated by 'Z' below? Solution MgZ₂

Na²⁺

Li⁺

Ca²⁺

O^2-

Br^-

Br^-

Order the prefixes, with lowest at the top, by dragging them around. Solution

deca

tri

penta

di

hepta

octa

tetra

hexa

nona

mono

1) Mono
2) di
3) tri
4) tetra
5) penta
6) hexa
7) hepta
8) octa
9) nona
10) deca

CO Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

Determine the number of each element, and don't forget to add the suffix.

carbon monoxide

(not monocarbon monoxide - because mono never goes on the 1^st word.)

NO Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Determine the number of each element, and don't forget to add the suffix.

nitrogen monoxide

(Remember to only use mono on the second word, never the first).

N₂O Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Determine the number of each element, and don't forget to add the suffix.

dinitrogen monoxide

(drop the 'o' on the mono suffix)

B₂H₆ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Determine the number of each element, and don't forget to add the suffix.

diboron hexahydride

(Still add the suffix -ide to the end of hydrogen = hydride)

CH₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Determine the number of each element, and don't forget to add the suffix.

carbon tetrahydride

PCl₅ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Determine the number of each element, and don't forget to add the suffix.

phosphorus pentachloride

CO₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Determine the number of each element, and don't forget to add the suffix.

carbon dioxide

N₂F₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Determine the number of each element, and don't forget to add the suffix.

dinitrogen tetrafluoride

P₂O₅ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Determine the number of each element, and don't forget to add the suffix.

diphosphorus pentoxide

(drop the 'a' on the penta suffix)

Dihydrogen monosulfide Solution

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¹
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Incorrect Attempts:

CHECK

The prefix in-front of each element indicates the amount.

H₂S

Nitrogen trihydride Solution

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¹
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Incorrect Attempts:

CHECK

The prefix in-front of each element indicates the amount.

NH₃

Diphosphorus trisulfide Solution

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Incorrect Attempts:

CHECK

The prefix in-front of each element indicates the amount.

P₂S₃

Silicon tetrachloride Solution

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¹
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Incorrect Attempts:

CHECK

The prefix in-front of each element indicates the amount.

SiCl₄

Nitrogen triiodide Solution

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Incorrect Attempts:

CHECK

The prefix in-front of each element indicates the amount.

NI₃

Dinitrogen trioxide Solution

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¹
²
³
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Incorrect Attempts:

CHECK

The prefix in-front of each element indicates the amount.

N₂O₃

Sulfur trioxide Solution

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¹
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³
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Incorrect Attempts:

CHECK

The prefix in-front of each element indicates the amount.

SO₃

Trioxygen Solution

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¹
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Incorrect Attempts:

CHECK

The prefix in-front of each element indicates the amount.

O₃

(common name: ozone)

Dihydrogen sulfide Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Where should prefixes be used?

The last name always has a prefix, even if it's just one.
The correct name for H₂S: dihydrogen monosulfide

C₁F₄ Solution

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³
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Incorrect Attempts:

CHECK

What subscripts should not be used?

Don't write subscripts of 1.
The correct formula is just: CF₄

MgO Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Mg²⁺O^2-

Magnesium is not multivalent, so no roman numerals are required. Don't use prefix naming with ionic compounds (metal plus non-metal). magnesium oxide

AlCl₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Al³⁺Cl^1-

Aluminum is not multivalent, so no roman numerals are required. aluminum chloride

InF₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

indium fluoride

Na₂O Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

sodium oxide

CaSe Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

calcium selenide

BaI₂ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

barium iodide

ScF₃ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

scandium fluoride

ZrBr₄ Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

zirconium bromide

Gallium oxide Solution

⁰
¹
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Incorrect Attempts:

CHECK

Start with the charges on the ions, and then use the criss-cross method to balance the charges

Ga₂O₃

Lithium iodide Solution

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Incorrect Attempts:

CHECK

Start with the charges on the ions, and then use the criss-cross method to balance the charges

LiI

Germanium oxide Solution

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Incorrect Attempts:

CHECK

Start with the charges on the ions, and then use the criss-cross method to balance the charges

GeO₂