Mass Relationships in
Chemical Reactions
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
3.1
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
3.1
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
3.2
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.02 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
3.2
1 amu = 1.66 x 10-24 g or 1 g = 6.02 x 1023 amu
M = molar mass in g/mol
NA = Avogadro’s number
3.2
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
3.2
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.1 amu
+ 2 x 16.0amu
64.1 amu
3.3
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
3.3
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
C2H6O
3.5
Types of Formulas
• Empirical Formula
The formula of a compound that
expresses the
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
of the compound.
To obtain an Empirical Formula
1. Determine the mass in grams of each
element present, if necessary.
2. Calculate the number of
element.
of each
3. Divide each by the smallest number of
moles to obtain the
whole
number ratio.
4. If whole numbers are not obtained* in
step 3), multiply through by the smallest
number that will give all whole numbers
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.
Empirical Formula from % Composition
A substance has the following composition by
mass: 60.80 % Na ; 28.60 % B ; 10.60 % H
What is the empirical formula of the substance?
Sample Exercise 3.13 Calculating an Empirical Formula
Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula
of ascorbic acid?
Solution
Sample Exercise 3.13 Calculating an Empirical Formula
Continued
Practice Exercise
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?
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
gC
g H2O
mol H2O
mol H
gH
g of O = g of sample – (g of C + g of H)
Empirical formula
Divide by smallest subscript
Empirical formula
3.6
Mass Changes in Chemical Reactions
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
3.8
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
3.8
Limiting Reagents
3.9
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
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:
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
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
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
3.9
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
2Al + Fe2O3
mol Al2O3
g Al2O3
Al2O3 + 2Fe
3.9
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.
3.10