1. Mass Relationships in Chemical Reactions
Chapter 3

2. Atomic mass is the mass of an atom in atomic mass units (amu)
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H = amu
16O = amu
3.1

3. Average atomic mass of lithium:
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
(7.42% x 6.015) + (92.58% x 7.016)
100
= amu
3.1

4. Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly grams of 12C
1 mol = NA = x 1023
Avogadro’s number (NA)
3.2

5. Molar mass is the mass of 1 mole of in grams marbles atoms
eggs
shoes
Molar mass is the mass of 1 mole of in grams
marbles
atoms
1 mole 12C atoms = x 1023 atoms = g
1 12C atom = amu
1 mole 12C atoms = g 12C
1 mole lithium atoms = g of Li
For any element
atomic mass (amu) = molar mass (grams)
3.2

6. One Mole of: S C Hg Cu Fe
3.2

7. 1 g = 6.022 x 1023 amu = molar mass in g/mol M NA = Avogadro’s number
1 amu = 1.66 x g M
= molar mass in g/mol
NA = Avogadro’s number
3.2

8. Do You Understand Molar Mass?
How many atoms are in g of potassium (K) ?
1 mol K = g K
1 mol K = x 1023 atoms K
1 mol K
39.10 g K x x
6.022 x 1023 atoms K
1 mol K =
0.551 g K
8.49 x 1021 atoms K
3.2

9. molecular mass (amu) = molar mass (grams)
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
SO2 1S
32.07 amu 2O
+ 2 x amu
SO2
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = amu
1 mole SO2 = g SO2
3.3

10. Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = x 1023 atoms H
1 mol C3H8O
60 g C3H8O x
8 mol H atoms
1 mol C3H8O x
6.022 x 1023 H atoms
1 mol H atoms x =
72.5 g C3H8O
5.82 x 1024 atoms H
3.3

11. Heavy
Light
KE = 1/2 x m x v2
v = (2 x KE/m)1/2
F = q x v x B
3.4

12. Percent composition of an element in a compound =
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
C2H6O
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% % % = 100.0%
3.5

13. Types of Formulas Empirical Formula
The formula of a compound that expresses the smallest whole number ratio of the atoms present.
Ionic formula are always empirical formula
Molecular Formula
The formula that states the actual number of each kind of atom found in one molecule of the compound.

14. To obtain an Empirical Formula
1. Determine the mass in grams of each element present, if necessary.
2. Calculate the number of moles of each element.
3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.
If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers
* Be careful! Do not round off numbers prematurely

15. A sample of a brown gas, a major air pollutant, is found to contain 2
A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance.
require mole ratios so convert grams to moles
moles of N = 2.34g of N = moles of N
14.01 g/mole
moles of O = g = moles of O
16.00 g/mole
Formula:

16. Calculation of the Molecular Formula
A compound has an empirical formula of NO2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?
empirical formula mass: (16.00) = g/mol
n = molar mass = g/mol emp. f. mass g/mol
n = 2
N2O4

17. Empirical Formula from % Composition
A substance has the following composition by mass: % Na ; % B ; % H
What is the empirical formula of the substance?
Consider a sample size of 100 grams
This will contain: grams of Na, grams of B, and grams H
Determine the number of moles of each
Determine the simplest whole number ratio

18. g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
g C
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
g H
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
3.6

19. Mass Changes in Chemical Reactions
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
Convert moles of sought quantity into desired units
3.8

20. Other units Molarity Gases Moles solute / L solution
22.4 L = 1 mole of ANY GAS at STP

21. Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
1 mol CH3OH
32.0 g CH3OH x
4 mol H2O
2 mol CH3OH x
18.0 g H2O
1 mol H2O x =
209 g CH3OH
235 g H2O
3.8

22. Limiting Reagents
6 red left over
6 green used up
3.9

23. Method 1 Pick A Product Try ALL the reactants
The lowest answer will be the correct answer
The reactant that gives the lowest answer will be the limiting reactant

24. Limiting Reactant: Method 1
10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?
2 Al + 3 Cl2  2 AlCl3
Start with Al:
Now Cl2:
10.0 g Al 1 mol Al mol AlCl g AlCl3
27.0 g Al mol Al mol AlCl3
= 49.4g AlCl3
35.0g Cl mol Cl mol AlCl g AlCl3
71.0 g Cl mol Cl mol AlCl3
= 43.9g AlCl3

25. Solving for Multiple Products
Once you determine the LR, you should only start with it!
A + B  X + Y + Z
A  X
B  X
To find Y and Z
B  Y
B  Z
Let’s say B is the LR!
There is no need to use A to find Y and Z
It will give you the wrong answer – a lot of extra work for nothing

26. Method 2 Convert one of the reactants to the other REACTANT
See if there is enough reactant “A” to use up the other reactants
If there is less than the GIVEN amount, it is the limiting reactant
Then, you can find the desired species

27. Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed OR
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
1 mol Al
27.0 g Al x
1 mol Fe2O3
2 mol Al x
160. g Fe2O3
1 mol Fe2O3 x =
124 g Al
367 g Fe2O3
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9

28. Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
g Al2O3
2Al + Fe2O Al2O3 + 2Fe
1 mol Al
27.0 g Al x
1 mol Al2O3
2 mol Al x
102. g Al2O3
1 mol Al2O3 x =
124 g Al
234 g Al2O3
3.9

29. Finding Excess Practice
10.0g of aluminum reacts with 35.0 grams of chlorine gas 2 Al + 3 Cl2  2 AlCl3
We found that chlorine is the limiting reactant, and 43.8 g of aluminum chloride are produced.
35.0 g Cl2 1 mol Cl2 2 mol Al g Al
71 g Cl mol Cl mol Al
= 8.8 g Al USED!
10.0 g Al – 8.8 g Al = 1.2 g Al EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

30. Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
Theoretical Yield
x 100
3.10

31. Chemistry In Action: Chemical Fertilizers
Plants need: N, P, K, Ca, S, & Mg
3H2 (g) + N2 (g) NH3 (g)
NH3 (aq) + HNO3 (aq) NH4NO3 (aq)
2Ca5(PO4)3F (s) + 7H2SO4 (aq)
3Ca(H2PO4)2 (aq) + 7CaSO4 (aq) + 2HF (g)
fluorapatite