Honors Chemistry
In Class Notes
Percent Composition and Chemical Formulas
1
Objectives

When you complete this presentation, you will be able to:
o describe how to calculate the percent by mass of an element in a compound.
o interpret an empirical formula.
o distinguish between empirical and molecular formulas.
Introduction

Chemical formulas tell us about the number of atoms in a compound.

In general, there are two kinds of chemical formulas.
o In ________________ formulas, the ________________ ________________ of atoms in the
compound is used.
o In ________________ formulas, the ________________ ________________ ________________
________________ of atoms in the compound is used.

In molecular formulas, the total number of atoms in the compound is used.
o For example, benzene has 6 carbon atoms and 6 hydrogen atoms in each
molecule.

Therefore, its molecular formula is C6H6.
o For example, acetic acid has 2 carbon atoms, 4 hydrogen atoms, and 2 oxygen
atoms in each molecule.

Therefore, its molecular formula is C2H4O2.
o For example, propane gas has 3 carbon atoms and 8 hydrogen atoms in each
molecule.


Therefore, its molecular formula is C3H8.
In empirical formulas, the lowest whole number ratio of atoms in the compound
are used.
o For example, benzene has 6 carbon atoms and 6 hydrogen atoms in each
molecule.

Therefore, its molecular formula is C6H6 and its empirical formula is
CH (we divide all subscripts by 6).
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The Mole: A Measurement of Matter
o For example, acetic acid has 2 carbon atoms, 6 hydrogen atoms, and 2 oxygen
atoms in each molecule.

Therefore, its molecular formula is C2H4O2. and its empirical formula
is CH2O (we divide all subscripts by 2).
o For example, propane gas has 3 carbon atoms and 8 hydrogen atoms in each
molecule.

Therefore, its molecular formula is C3H8 and its empirical formula is
also C3H8 (there is no common divisor for all subscripts).
Percent Composition

If we know the ________________ of a compound and the ________________ of one or more
of the elements in the compound, then we can find the ________________ ________________
of those atoms in the compound.
% mass of an element =

mass of atoms
 100 %
mass of compound
For example,
o 1.716 g of C, 0.577 g of H, and 2.286 g of O combine together to form 4.579 g
of a compound.
Sample Problem 10.9
When a 13.60 g sample of a compound containing only magnesium and oxygen is
decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this
compound?
Known:
mass of compound = 13.60 g
Unknown: % Mg = ?%
mass of O = 5.40 g
% O = ?%
mass of Mg = 13.60 g – 5.40 g = 8.20 g
Honors Chemistry

In Class Notes
Percent Composition and Chemical Formulas
If we know the ________________ formula of a compound, then we can find the percent
composition of each of the atoms in the compound.

We use the ________________ ________________ of the compound and the ________________
________________ ________________ of the atoms in the compound.
% composition =

atomic mass of atoms
 100 %
molar mass of compound
For example:
o the percent composition of benzene, C6H6, is:
o the percent composition of acetic acid, C2H3O2, is:
Sample Problem 10.10
Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained
from petroleum. Calculate the percent composition of propane
Known:
molar mass of C3H8 = 44.0 g
Unknown: % C = ?%
mass of C = 3  12.01 g/mol = 36.03 g/mol
% H = ?%
mass of H = 8  1.01 g/mol = 8.08 g/mol
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The Mole: A Measurement of Matter
Practice Problems: find the percent composition of …
1. C6H14
2. NaCl
3. KNO3
4. CuSO4
5. FeCO3
Empirical Formulas

In ________________ ________________, the lowest whole number ratios of atoms in a
compound is used.
o Benzene, C6H6, has an empirical formula of ________________.

A ratio of 6/6 = 1/1.
o Acetic acid, C2H4O2, has an empirical formula of ________________.

A ratio of 2/4/2 = 1/2/1
o Propane, C3H8, has an empirical formula of ________________.


A ratio of 3/8 =3/8.
If we know the percent composition of a compound, then we can find the empirical
formula of the compound.
o First, we assume that we have 100 g of the compound and find the mass of
each atom in the compound.
o Second, we find the number of mols of each atom.
o Third, we find the lowest whole number ratio of mols.
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Honors Chemistry

In Class Notes
Percent Composition and Chemical Formulas
For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O.
o First, we assume that we have 100 g of the compound and find the mass of
each atom in the compound.

52.2 g of C

13.1 g of H

34.7 g of O
o Next, we we find the number of mols of each atom.

nC = mC/MC = 52.2 g/12.0 g/mol = 4.35 mol C

nH = mH/MH = 13.1 g/1.01 g/mol = 13.0 mol H

nO = mO/MO = 34.7 g/16.0 g/mol = 2.17 mol O
o Finally, we find the lowest whole number ratio of mols.

nC/nO = 4.35 mol/2.17 mol = 2/1

nH/nO = 13.0 mol/2.17 mol = 6/1
o This means that there is a ratio of 2:6:1 for C:H:O


The empirical formula is C2H6O
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.
o First, we assume that we have 100 g of the compound and find the mass of
each atom in the compound.

44.9 g of K

18.4 g of S

36.7 g of O
o Next, we we find the number of mols of each atom.

nK = mK/MK = 44.9 g/39.1 g/mol = 1.15 mol K

nS = mS/MS = 18.4 g/32.1 g/mol = 0.573 mol S

nO = mO/MO = 36.7 g/16.0 g/mol = 2.29 mol O
o Finally, we find the lowest whole number ratio of mols.

nK/nS = 1.15 mol/0.573 mol = 2/1

nO/nS = 2.29 mol/0.573 mol = 4/1
o This means that there is a ratio of 2:1:4 for K:S:O

The empirical formula is K2SO4
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The Mole: A Measurement of Matter
Sample Problem 10.11
A compound is analyzed and found to contain25.9% nitrogen and 74.1% oxygen. What is
the empirical formula of the compound?
 First, we assume that we have 100 g of the compound and find the mass of each
atom in the compound.

Next, we find the number of mols of each atom.

Finally, we find the lowest whole number ratio of mols.
Practice problems: find the empirical formulas of compounds that have the following
percent compositions.
1. 11.21% H and 88.79% O
2. 92.24% C and 7.76% H
3. 39.99% C, 6.73% H, and 53.28% O
4. 24.27% C, 4.08%H, and 71.65% Cl
5. 39.96% N, 14.40% H, and 45.64% O
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Honors Chemistry
In Class Notes
Percent Composition and Chemical Formulas
7
Molecular Formulas

In ________________ ________________, the total number of atoms in the compound is used.
o Benzene has a molecular formula of ________________.
o Acetic acid has a molecular formula of ________________.
o Propane has a molecular formula of ________________.

If we know the ________________ ________________ of a compound, then we can find the
molecular formula of the compound, if we know the ________________ ________________ of
the compound.
o First, we determine the empirical mass of the compound.
o Second, we divide the molar mass by the empirical mass to get our multiplier.
o Third, we multiply the subscripts of the empirical formula by the multiplier
to get the molecular formula.

For example, we have a compound with a molecular formula of CH2O and a molar
mass of 180.16 g/mol. What is the molecular formula of the compound?
o First, we determine the empirical mass of the compound.
EM = (1×12.01) + (2×1.01) + (1×16.00) = 30.03
o Next, we divide the molar mass by the empirical mass to find the multiplier.
M/EM = (180.16)/(30.03) = 5.999333999 ≈ 6
o Finally, we multiply the subscripts of the empirical formula by the multiplier
to get the molecular formula.
CH2O ⇒ C6H12O6
Sample Problem 10.12
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and
empirical formula is CH4N.
The Mole: A Measurement of Matter
Practice problems: find the molecular formulas of compounds that have the following
empirical formulas and molar masses.
1. CH2O; M = 60.0 g/mol
2. CH2; M = 42.1 g/mol
3. NaCO2; M = 134.0 g/mol
4. CH2Cl; M = 98.96 g/mol
5. NH5O; M = 35.06 g/mol
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