Part 1: Formula Mass Calculations
A formula mass is the sum of the atomic masses of all of the atoms that are found in the
formula.
To calculate a formula mass:
 Find the atomic mass of each element in the formula
 Multiply each atomic mass by the number of times the atom appears in the
formula
 Find the sum
Example
·
First calculate the formula mass for Na2SO4 . Find the atomic mass of each
element from the periodic table. Multiply it by the number of times it appears in
the formula and add up the total
2 Sodium atoms Na 2 x
= 45.98
22.99
1 Sulfur atom
S
1x
= 32.06
32.06
4 Oxygen atoms O
4x
= 64.00
16.00
Formula mass = 142.04
Use the example above as a model and calculate the formula mass (Molar Mass) for each
of the following:
1.
NaCl
2.
KNO3
3.
ZnCO3
4.
BaSO4
5.
(NH4)2SO4
6.
Ca(OH)2
7.
Ti2O3
8.
Sr(NO3)2
9.
PbCl2
10.
Cu(NO3)2
11.
CH3CH2OH
12.
Al2(SO4)3
Part II Percent Composition
According to the law of definite proportions, compounds, contain definite proportions of
each element by mass. The sum of all of the atomic masses of elements in a formula is
called the formula mass. If it is expressed in grams, then it is called a gram formula
mass or molar mass. If it represents the sum of all of the masses of all of the elements in
a molecule then it is called a molecular mass. To find the percentage of each element in a
compound it is necessary to compare the total mass of each element with the formula
mass.
Example Calculate the mass of each element in potassium carbonate, K2CO3 .
·
·
First calculate the formula mass for K2CO3 . Find the atomic mass of each
element from the periodic table. Multiply it by the number of times it appears in
the formula and add up the total
2 Potassium
K
2x
= 78.20
atoms
39.10
1 carbon atom
C
1x
= 12.01
12.01
3 Oxygen atoms O
3x
= 48.00
16.00
138.21
To find the percent of each element divide the part of the formula mass that
pertains to that element with the total formula mass
78.20 X
Percent of
K
56.58
138.21 100 = %
Potassium
=
12.01 X
Percent
C
8.69
138.21 100 = %
of Carbon
=
48.00 X
Percent of
O
34.73
138.21 100 = %
Oxygen
=
Percent Composition Calculations
Using the example above, Calculate the percentage composition of each element in the
formulas shown below. Show your work and circle your answers
Magnesium carbonate MgCO 3
1.
2.
Sulfuric Acid H2SO4
3.
Sodium Nitrate NaNO3
4.
Iron (III) Phosphate FePO4
5.
Ammonium sulfate (NH4)2SO4
6.
Zinc Hydroxide, Zn(OH) 2
7.
Sodium hydroxide, NaOH
8.
Potassium Chloride, KCl
9.
Potassium chlorate, KClO3
10. Silver Nitrate, AgNO3
11. Calcium phosphate, Ca3(PO4)2
12.
Butane, C4H10
Empirical and Molecular Formulas
Empirical formula determination
The empirical formula is the simpliest ratio of the numbers of atoms of each element that
make a compound. To find the empirical formula of a compound:
 Divide the amount of each element (either in mass or percentage) by its atomic
mass. This calculation gives you moles of atoms for each element that appears in
the formula
 Convert the results to small whole number ratios. Often the ratios are obvious. If
they are not divide all of the other quotients by the smallest quotient
Example: Analysis of a certain compound showed that 39.348 grams of it contained
0.883 grams of hydrogen, 10.497 grams of Carbon, and 27.968 grams
of Oxygen. Calculate the empirical formula of the compound.
·
First divide the amount by the atomic mass to get the number of moles
of each kind of atom in the formula
·
Hydrogen
H =
0.883 g
1.01 g mol-1
= 0.874 mol
Carbon
C =
10.497 g
12.01 g mol-1
= 0.874 mol
Oxygen
O =
27.968 g
16.00 g mol-1
= 1.748 mol
Analysis of the ratio s shows that the first tow are identical and that the
third is twice the other two. Therefore the ratio of H to C to O is 1 to 1 to
2. The empirical formula is HCO2
Molecular formula Determination
To calculate the molecular formula from the empirical formula it is necessary to know
the molecular (molar) mass.
 Add up the atomic masses in the empirical formula to get the factor Divide this
number into the molecular formula mass. If the number does not divide evenly
you probably have a mistake in the empirical formula or its formula mass
 Multiply each subscript in the empirical formula by the factor to get the molecular
formula
Example: Suppose the molecular mass of the above compound HCO2 is 90.0. Calculate
the molecular formula.


The empirical formula mass of
1 H @ 1.0
=
1 C @ 12.0 =
2 O @ 16.0 =
is
1.0
12.0
32.0
45.0
Note that 45 is exactly half of the molecular mass of 90. So the formula mass
of HCO2 is exactly half of the molecular mass. Hence the molecular formula is
double that of the empirical formula or H2C2O4.
Part III Empirical and Molecular Formula Calculations
1. A certain compound contains 4.0 g of calcium and 7.1 g of chlorine. Is relative
molecular mass is 111. Find its empirical and molecular formulas.
2. A certain compound has 25.9% nitrogen and 74.1% oxygen. Its relative
molecular mass is 108. Find its empirical and molecular formula
3. A certain compound was found to contain 54.0 g of carbon and 10.5 grams
of hydrogen Its relative molecular mass is 86.0. Find the empirical and the
molecular formulas.
4. A certain compound was found to contain 26.4 g of carbon, 4.4 grams
of hydrogen and 35.2 grams of oxygen. Its relative molecular mass is 60.0. Find
the empirical and the molecular formula.
5. A certain compound was found to contain 78.2 % Boron and 21.8 %
hydrogen. Its relative molecular mass is 27.7. Find the empirical and the
molecular formula.
6.
A certain compound contains 27.3%Carbon, 4.5 % hydrogen, 36.4% oxygen,
and 31.8% nitrogen. Its relative molecular mass is 176.0. Find its empirical and
molecular formulas.