1. Percent Composition and Chemical Formulas
Chemical Quantities
Unit 7
Chapter 10, Section 10.3
Percent Composition and Chemical Formulas

2. Objectives When you complete this presentation, you will be able to …
describe how to calculate the percent by mass of an element in a compound.
interpret an empirical formula.
distinguish between empirical and molecular formulas.

3. Introduction
Chemical formulas tell us about the number of atoms in a compound.
In general, there are two kinds of chemical formulas.
In molecular formulas, the total number of atoms in the compound is used.
In empirical formulas, the lowest whole number ratio of atoms in the compound is used.

4. Molecular Formulas
In molecular formulas, the total number of atoms in the compound is used.
For example, benzene has 6 carbon atoms and 6 hydrogen atoms in each molecule.
Therefore, its molecular formula is C6H6.

5. Molecular Formulas
In molecular formulas, the total number of atoms in the compound is used.
For example, acetic acid has 2 carbon atoms, 4 hydrogen atoms, and 2 oxygen atoms in each molecule.
Therefore, its molecular formula is C2H4O2.

6. Molecular Formulas
In molecular formulas, the total number of atoms in the compound is used.
For example, propane gas has 3 carbon atoms and 8 hydrogen atoms in each molecule.
Therefore, its molecular formula is C3H8.

7. Empirical Formulas
In empirical formulas, the lowest whole number ratio of atoms in the compound are used.
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).

8. Empirical Formulas
In empirical formulas, the lowest whole number ratio of atoms in the compound are used.
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).

9. Empirical Formulas
In empirical formulas, the lowest whole number ratio of atoms in the compound are used.
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).

10. Percent Composition
If we know the mass of a compound and the mass of one or more of the elements in the compound, then we can find the percent composition of those atoms in the compound.
% mass of an element = × 100%
mass of atoms
mass of compound

11. Percent Composition For example,
1.716 g of C, g of H, and g of O combine together to form g of a compound.
% mass C = × 100%
mass of C
mass of compound
= × 100%
1.716 g
4.579 g
= 37.48%
% mass H = × 100%
mass of H
mass of compound
= × 100%
0.577 g
4.579 g
= 12.60%
% mass O = × 100%
mass of O
mass of compound
= × 100%
2.286g
4.579 g
= 49.92%

12. Sample Problem 10.9
When a 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 = g
mass of O = 5.40 g
mass of Mg = g – 5.40 g = 8.20 g
Unknown: %Mg = ?%
%O = ?%
% Mg = × 100%
mass of Mg
mass of compound
= × 100%
8.20 g
13.60 g
= 60.3%
% O = × 100%
mass of O
mass of compound
= × 100%
5.40 g
13.60 g
= 39.7%

13. Percent Composition
If we know the chemical formula of a compound, then we can find the percent composition of each of the atoms in the compound.
We use the molar mass of the compound and the average atomic masses of the atoms in the compound.
percent composition = × 100%
atomic mass of atoms
molar mass of compound

14. Percent Composition For example:
the percent composition of benzene, C6H6, is:
%C = × 100%
atomic mass of C atoms
molar mass of C6H6
= × 100%
6(12.01 g/mol)
78.12 g/ mol
= %
%H = × 100%
atomic mass of H atoms
molar mass of C6H6
= × 100%
6(1.01 g/mol)
78.12 g/ mol
= 7.76%

15. Percent Composition For example:
the percent composition of acetic acid, C2H3O2, is:
%C = × 100%
atomic mass of C atoms
molar mass of C2H4O2
= × 100%
2(12.01 g/mol)
60.06 g/ mol
= %
%H = × 100%
atomic mass of H atoms
molar mass of C2H4O2
= × 100%
4(1.01 g/mol)
60.06 g/ mol
= 6.73%
%O = × 100%
atomic mass of O atoms
molar mass of C2H4O2
= × 100%
2(16.00 g/mol)
60.06 g/ mol
= %

16. 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/mol
mass of C = 3 × 12.0 g/mol = 36.0 g/mol
mass of H = 8 × 1.0 g/mol = 8.0 g/mol
Unknown: %C = ?%
%H = ?%
% Mg = × 100%
mass of C
molar mass of C3H8
= × 100%
36.0 g/mol
44.0 g/mol
= 81.8%
% O = × 100%
mass of H
molar mass of C3H8
= × 100%
8.0 g/mol
44.0 g/mol
= 18%

17. Percent Composition
Practice Problems: find the percent composition of …
C6H14
NaCl
KNO3
CuSO4
FeCO3
%C = [(6×12.01)/(86.20)]×100% = 83.60%
%H = [(14×1.01)/(86.20)]×100% = 16.40%
%Na = [(22.99)/(58.44)]×100% = 39.34%
%Cl = [(35.45)/(58.44)]×100% = 60.66%
%K = [(39.10)/(101.11)]×100% = 38.67%
%N = [(14.01)/(101.11)]×100% = 13.86%
%O = [(3×16.00)/(101.11)]×100% = 47.47%
%Cu = [(63.55)/(159.61)]×100% = 39.82%
%S = [(32.06)/(159.61)]×100% = 20.09%
%O = [(4×16.00)/(159.61)]×100% = 40.10%
%Fe = [(55.85)/(115.86)]×100% = 48.20%
%C = [(12.01)/(115.86)]×100% = 10.37%
%O = [(3×16.00)/(115.86)]×100% = 41.43%

18. Empirical Formulas
In empirical formulas, the lowest whole number ratios of atoms in a compound is used.
Benzene, C6H6, has an empirical formula of CH.
A ratio of 6/6 = 1/1.
Acetic acid, C2H4O2, has an empirical formula of CH2O.
A ratio of 2/4/2 = 1/2/1
Propane, C3H8, has an empirical formula of C3H8.
A ratio of 3/8 =3/8.

19. Empirical Formulas
If we know the percent composition of a compound, then we can find 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.
Second, we find the number of mols of each atom.
Third, we find the lowest whole number ratio of mols.

20. Empirical Formulas
For example, we have a compound with 52.2% C, 13.1% H, and 34.7% 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

21. Empirical Formulas
For example, we have a compound with 52.2% C, 13.1% H, and 34.7% 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

22. Empirical Formulas
For example, we have a compound with 52.2% C, 13.1% H, and 34.7% 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
This means that there is a ratio of 2:6:1 for C:H:O
The empirical formula is C2H6O

23. Empirical Formulas
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% 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

24. Empirical Formulas
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% 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 = mol S
nO = mO/MO = 36.7 g/16.0 g/mol = 2.29 mol O

25. Empirical Formulas
For example, we have a compound with 44.9% K, 18.4% S, and 36.7% 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
This means that there is a ratio of 2:1:4 for K:S:O
The empirical formula is K2SO4

26. 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.
mN = 25.9 g
mO = 74.1 g

27. 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?
Next, we find the number of mols of each atom.
nN = mN/MN = 25.9 g/14.0 g/mol = 1.85 mol
nO = mO/MO = 74.1 g/16.0 g/mol = 4.63 mol
Finally, we find the lowest whole number ratio of mols.
nO/nN = 4.63 mol/1.85 mol = 2.5/1 = 5/2
This means that there is a ratio of 5:2 for O:N
The empirical formula is N2O5

28. Empirical Formulas H2O CH CH2O CH2Cl NH5O
Practice problems: find the empirical formulas of compounds that have the following percent compositions.
11.21% H and 88.79% O
92.24% C and 7.76% H
39.99% C, 6.73% H, and 53.28% O
24.27% C, 4.08%H, and 71.65% Cl
39.96% N, 14.40% H, and 45.64% O
H2O CH
CH2O
CH2Cl
NH5O

29. Molecular Formulas
In molecular formulas, the total number of atoms in the compound is used.
Benzene has a molecular formula of C6H6.
Acetic acid has a molecular formula of C2H4O2.
Propane has a molecular formula of C3H8.

30. Molecular Formulas
If we know the empirical formula of a compound, then we can find the molecular formula of the compound, if we know the molar mass of the compound.
First, we determine the empirical mass of the compound.
Second, we divide the molar mass by the empirical mass to get our multiplier.
Third, we multiply the subscripts of the empirical formula by the multiplier to get the molecular formula.

31. Molecular Formulas
For example, we have a compound with a molecular formula of CH2O and a molar mass of g/mol. What is the molecular formula of the compound?
First, we determine the empirical mass of the compound.
EM = (1×12.01) + (2×1.01) + (1×16.00) = 30.03

32. Molecular Formulas
For example, we have a compound with a molecular formula of CH2O and a molar mass of g/mol. What is the molecular formula of the compound?
Next, we divide the molar mass by the empirical mass to find the multiplier.
M/EM = (180.16)/(30.03) = ≈ 6

33. Molecular Formulas
For example, we have a compound with a molecular formula of CH2O and a molar mass of g/mol. What is the molecular formula of the compound?
Finally, we multiply the subscripts of the empirical formula by the multiplier to get the molecular formula.
CH2O ⇒ C6H12O6

34. Sample Problem 10.12
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N.
First, we determine the empirical mass.
EM = (1×12.0) + (4×1.0) + (1×14.0) = 30.0
Next, divide molar mass by empirical mass.
M/EM = (60.0)/(30.0) = 2
Finally, multiply to get the molecular formula.
CH4N ⇒ C2H8N2

35. Molecular Formulas C2H4O2 C3H6 Na2C2O4 C2H4Cl2 NH5O
Practice problems: find the molecular formulas of compounds that have the following empirical formulas and molar masses.
CH2O; M = 60.0 g/mol
CH2; M = 42.1 g/mol
NaCO2; M = g/mol
CH2Cl; M = g/mol
NH5O; M = g/mol
C2H4O2
C3H6
Na2C2O4
C2H4Cl2
NH5O