Kiel Benedick M.

Gianan

12 – Canary

Chapter 4 – Atoms in Reaction

Self-Check

1. What is the difference between relative atomic mass and average atomic mass of an element?
Relative and average atomic mass both describe properties of an element related to its
different isotopes. However, relative atomic mass is a standardized number that's assumed to
be correct under most circumstances, while average atomic mass is only true for a specific
sample.
2. What does mole signify?
Mole, also spelled mol, in chemistry, a standard scientific unit for measuring large
quantities of very small entities such as atoms, molecules, or other specified particles. The
mole designates an extremely large number of units, 6.02214076 × 1023.
3. What are the differences among the four basic types of chemical reactions?
Representation of four basic chemical reactions types: synthesis, decomposition, single
replacement and double replacement.
4. What is the difference between oxidation and reduction? Oxidizing and reducing agents?
Oxidation occurs when a reactant loses electrons during the reaction. Reduction occurs
when a reactant gains electron during the reaction. This often occurs when metals are reacted
with acid.
An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction
while a reducing agent is typically in one of its lower possible oxidation states, and is known
as the electron donor. A reducing agent is oxidized, because it loses electrons in the redox
reaction.
5. How does a balanced chemical equation explain the law of conservation of mass?
Matter cannot be created or destroyed in chemical reactions. This is the law of
conservation of mass. In every chemical reaction, the same mass of matter must end up in the
products as started in the reactants. Balanced chemical equations show that mass is conserved
in chemical reactions.

CHEMISTRY 1 – CHAPTER 4 15
Practice Exercise

Practice Exercise 4.2

1. Find the molar mass of diatomic oxygen.

The atomic mass of an element is multiplied by the number of atoms in the molecule, and
then the masses of all the elements in the molecule are added. The molecule O 2 is diatomic.
A single oxygen atom has a mass of 16 atomic mass units. Therefore, the mass of
O2 molecule = 2 × 16 = 32 amu.
2. Calculate the molar mass of acetaminophen (C 8H9NO₂), an over-the-counter pain reliever
also known as paracetamol.
(Carbon) x 8 + (Hydrogen) x 8 + (Nitrogen) + (Oxygen) x 2
12.0107 x 8 + 1.00794 x 9 + 14.0067 + 15.9994 x 2
= 151.16256 g/mol

Practice Exercise 4.3

1. How many moles of carbon dioxide (CO2) does 250 grams of the substance have? How many
molecules of CO2 are there? The molar mass of CO2 is 44.01 g/mol.
Using Avogrado’s number 6.022×1023 for molecules per mole to calculate the number of
molecules from the number of moles.
Moles of carbon dioxide (CO2) = 250 g
= 5.68 mol
Molecules of carbon dioxide (CO2) = 250 g CO2 6.022×1023 molecules
44.01 g/mol 1 mol

= 3.420 × 1024 molecules

2. Sodium bicarbonate (NaHCO3) is commercially known as baking soda. How many grams
and atoms of oxygen are there in 15 grams of sodium bicarbonate? The molar mass of
NaHCO3 is 84.01 g/mol.
Given:
Molecular mass of Baking Soda is = 84 g/mol
84 g/mol of Baking Soda = 48 g/mol Oxygen
CHEMISTRY 1 – CHAPTER 4 16
 15 g Baking Soda contains = 48 × 15 g Oxygen = 8.571 g/mol Oxygen.
84
Then.
32 g/mol Oxygen = 6.022×1023 atoms
∴ 8.571 g/mol Oxygen contains = (6.022×1023) × 8.571 = 1.613 × 1023 atom of Oxygen.
32
3. What is the mass of 4.0x1023 formula units of potassium chloride (KCI)?
KCl = 4.0 × 10²³ formula units × 1 mol × 10²³ formula units × 74.55 g
6.022 1 mol
KCl = 50 g

Practice Exercise 4.4

Tartaric acid (C4H6O6) is naturally found in fruits like grapes and banana. It is used in the
food industry as a leavening agent that gives a sour taste. Determine the percent composition of
each element in the compound.

Solution:

First, find the formula mass of C4H6O6.

Formula of C4H6O6 = (atomic mass of C × 4) + (atomic mass of H × 6) + (atomic mass of O × 6)

= (12.0107 amu × 4) + (1.00784 amu × 6) + (15.999 amu × 6)

= (48.0428 amu) + (6.0470 amu) + (95.994 amu)

= 150.0838 amu

Applying the general formula for percent composition for each constituent element,

% Carbon (C) = 48.0428 amu × 100 = 32.011%

150.0838 amu

% Hydrogen (H) = 6.0470 amu × 100 = 4.029%

150.0838 amu
CHEMISTRY 1 – CHAPTER 4 17
% Oxygen (O) = 95.994 amu × 100 = 63.960%

150.0838 amu

100%

Practice Exercise 4.5

What are the empirical and molecular formulas of a compound that is 40.27% potassium,
26.78% chromium, and 32.95% oxygen? The molar mass of the compound is 194.2 g/mol.

Solution for empirical formula:

First, suppose that each compound has a mass of 100g.

K = 40.27 g

Cr = 26.78 g

O = 32.95 g

Next, determine the molecular weight of each element.

K = 40.27 g × 1 mol = 1.030 mol

39.10 g

Cr = 26.78 g × 1 mol = 0.5150 mol

52 g

O = 32.95 g × 1 mol = 2.059 mol

16 g

Next, divide the number of moles of each element by the least number of moles.

K = 1.030 mol = 2
0.5150 mol

CHEMISTRY 1 – CHAPTER 4 18
Cr = 0.5150 mol = 1

0.5150 mol

O = 2.059 mol = 4
0.5150 mol

Lastly, write the empirical formula.

Potassium chromate = K2CrO4

Solution for molecular formula:

First, represent a molecular formula.

= (K2CrO4)n

Then, calculate the empirical mass.

= empirical mass = (39.10 g/mol × 2) + (52.00 g/mol × 1) + (16.00 g/mol × 4)

= empirical mass = 194.2 g/mol

Then, divide the molar mass by the empirical mass.

N = molar mass N = 1

Empirical mass

N = 194.2 g/mol.

194.2 g/mol.

Lastly, multiply the subscripts by the value of n to obtain the molecular formula.

Molecular formula = (K2CrO4)1

= K2CrO4

CHEMISTRY 1 – CHAPTER 4 19
Practice Exercise 4.6

What is the oxidation number of each of the elements in each reactant and product in the
equation? Identify the oxidizing and reducing agents.

0 -2 +5 -2 +3 +5 -2 0
2A1(s) + 3Ni(NO3)2(aq) → 2Al(NO3)3(aq) + 3Ni(s)
Aluminum is oxidized

= Al ⟶ Al3+ + 3e –

Nickel is reduced

= Ni + 2e− ⟶ Ni(s) + 2NO3 –

Al is the reducing agent.

Ni is the oxidizing agent.

Practice Exercise 4.7

1. The combustion of glucose (C6H12O6) occurs in the presence of oxygen, producing carbon
dioxide and water.
C6H12O6 (g)+ O2 (g) ⟶ CO2 (g) + H2O (g)
6 C atoms : 1 C atoms
12 H atoms : 2 H atoms
6 O atoms : 3 O atoms

C6H12O6 (g) + O2 (g) ⟶ 6CO2 (g) + H2O (g)

6 C atoms : 6 C atoms
12 H atoms : 2 H atoms
6 O atoms : 13 O atoms

C6H12O6 (g) + O2 (g) ⟶ 6CO2 (g) + 6H2O (g)

6 C atoms : 6 C atoms

CHEMISTRY 1 – CHAPTER 4 20
 12 H atoms : 12 H atoms
6 O atoms : 18 O atoms

C6H12O6 (g) + 6O2 (g) ⟶ 6CO2 (g) + H2O (g)

6 C atoms : 6 C atoms
12 H atoms : 12 H atoms
18 O atoms : 18 O atoms

2. The reaction between calcium carbide (kalburo, CaC2) and water forms acetylene (C₂H₂)
and calcium hydroxide [Ca(OH)2].

CaC2 (g) + H2O (g) ⟶ C2H2 (g) + Ca(OH)2 (g)

1 Ca atoms : 1 Ca atoms
2 C atoms : 2 C atoms
2 H atoms : 4 H atoms
1 O atoms : 2 O atoms

CaC2 (g) + 2H2O (g) ⟶ C2H2 (g) + Ca(OH)2 (g)

1 Ca atoms : 1 Ca atoms
2 C atoms : 2 C atoms
4 H atoms : 4 H atoms
2 O atoms : 2 O atoms

Love of Lab – A Balancing Act (p. 102A)

A. Write and balance the chemical equation for each of these reactions.

a. Solid calcium carbonate is synthesized from calcium oxide powder and carbon dioxide gas.

Ca(OH)2 + CO2 = CaCO3 + H2O.

CHEMISTRY 1 – CHAPTER 4 21
b. The reaction between zinc metal and hydrochloric acid forms solid zinc chloride and
hydrogen gas.

Zn + 2HCl = ZnCl2 + H2.

c. The reaction between the aqueous solutions of silver nitrate and potassium chromate forms
silver chromate precipitate and aqueous potassium nitrate.

K2Cr2O7 + 2AgNO3 → Ag2Cr2O7 + 2KNO3

Test Yourself A

1. B
2. B
3. D
4. C
5. C
6. D
7. D
8. D
9. C
10. C
11. A
12. B

CHEMISTRY 1 – CHAPTER 4 22