Ch 7.3
Using Chemical Formulas
The Mass of a Mole of an Element
• Remember: The atomic mass of an element
(a single atom) is expressed in atomic
mass units (amu).
• Molar Mass: is the atomic mass of an
element expressed in grams/mole (g/mol).
• Carbon = 12.01 g/mol
• Hydrogen = 1.01 g/mol
• When dealing with molar mass, round off to
two decimals. 12.011g/mol -> 12.01g/mol
The Mass of a Mole of a Compound
• You calculate the mass of a molecule by
adding up the molar masses of the atoms
making up the molecules.
• Example: H2O
– H = 1.01 g x 2 atoms = 2.02 g/mol
– O = 16.00 g x 1 atom = 16.00 g/mol
• Molar Mass of H2O = 2.02 g/mol + 16.00 g/mol =
18.02 g/mol
• This applies to both molecular and ionic
compounds
• Find the molar mass of PCl3
– P = 30.97 g x 1 atom = 30.97 g/mol
– Cl = 35.45 g x 3 atoms = 106.35 g/mol
– PCl3 = 30.97 g + 106.35 g = 137.32 g/mol
• What is the molar mass of Sodium
Hydrogen Carbonate (NaHCO3) ?
– Na = 22.99 g x 1 atom = 22.99 g/mol
– H = 1.01 g x 1 atom = 1.01 g/mol
– C = 12.01 g x 1 atom = 12.01 g/mol
– O = 16.00 g x 3 atoms = 48.00 g/mol
– NaHCO3 = 22.99 + 1.01 + 12.01 + 48.00
= 84.01 g/mol
Converting Moles to Mass
• You can use the molar mass of an element
or compound to convert between the mass of
a substance and the moles of a substance.
• Mass (g) = # of moles x mass (g)
1 mole
Example: If molar mass of NaCl is 58.44 g/mol,
what is the mass of 3.00 mol NaCl?
Mass of NaCl = 3.00 mol x 58.44g = 175 g NaCl
1 mol
Example 2: Moles to Mass
• What is the mass of 9.45 mol of Aluminum
Oxide (Al2O3)?
• Find molar mass of Al2O3
= 101.96 g/mol
• Mass = 9.45 mol Al2O3 x 101.96 g Al2O3
1 mol Al2O3
= 964 g Al2O3
Converting Mass to Moles
• You can invert the conversion factor to find
moles when given the mass.
• Moles = mass (g) x 1 mole
mass (g)
Example: If molar mass of Na2SO4 142.05
g/mol, how many moles is 10.0 g of Na2SO4?
Moles of Na2SO4 = 10.0 g x 1 mol =
142.05 g
= 0.0704 mol Na2SO4
Example 2: Mass to Moles
• How many moles are in 75.0 g of Dinitrogen
Trioxide? N2O3
• Find molar mass of N2O3
= 76.02 g/mol
• Moles = 75.0 g N2O3 x 1 mole = 0.987 mol
N 2O 3
76.02 g
Percent Composition
• Percent Composition: the relative amount
of the elements in a compound.
• Also known as the percent by mass
• It can be calculated in two ways:
– Using Mass Data
– Using the Chemical Formula
% mass of element= mass of element x100%
mass of compound
Example
• When a 13.60 g sample of a compound containing
Mg and O is decomposed, 5.40 g O is obtained.
What is the % composition of this compound?
Mass of compound: 13.60 g
Mass of oxygen: 5.40 g O
Mass of magnesium: 13.60 g - 5.40 g = 8.20 g Mg
% Mg = 8.20 g Mg x 100% = 60.3%
13.60 g
% O = 5.40 g O x 100% = 39.7%
13.60 g
• Find the percent composition of Cu2S.
• Find mass of Cu and S
– Cu = 63.55 x 2 = 127.10 g
– S = 32.07 g
• Find mass of Cu2S
– 127.10 g + 32.07 g = 159.17 g
% Composition
– Cu = 127.10 g x 100% = 79.85%
159.17 g
– S = 32.07 g x 100% = 20.15%
159.17 g
Homework
• 7.3 pg 253 #30-33
Ch 7.4
Determining Chemical Formulas
Empirical Formulas
• Empirical Formula: shows the smallest
whole-number ratio of the atoms of the
elements in a compound.
• Example:
– The Empirical Formula for Hydrogen Peroxide
(H2O2) is HO with a 1:1 ratio.
– The Empirical Formula for Carbon Dioxide (CO2)
is CO2 with a 1:2 ratio.
Determining the Empirical
Formula of a Compound
• A compound is found to contain 25.9%
Nitrogen and 74.1% Oxygen. What is the
Empirical Formula of the compound?
• 25.9 g N x 1 mol N = 1.85 mol N
14.01 g N
• 74.1 g O x 1 mol O = 4.63 mol O
16.00 g O
• N1.85O4.63 = N1O2.5 = N2O5
Molecular Formulas
• Molecular Formula: tells the actual number
of each kind of atom present in a molecule
of a compound
• Example:
– The Molecular Formula for Hydrogen Peroxide is
H2O2.
– The Molecular Formula for Carbon Dioxide is CO2
• It is possible to find the Molecular Formula
using the Empirical Formula if you know the
molar mass of the compound.
Finding the Molecular Formula
• Calculate the molecular formula of a
compound whose molar mass is 60.0 g/mol
and empirical formula is CH4N
• Step 1: Find the empirical formula molar mass
– 12.01 + (4 x 1.01) + 14.01 = 30.06 g/mol
• Step 2: Divide molar mass by EF molar mass
– 60.0 g/mol = 1.99  2
30.06 g/mol
• Step 3: Multiply empirical formula by 2
– CH4N x 2 = C2H8N2
Homework
• 7.4 pg 253 #36-38