1. Percent Composition Review
The percent composition of elements in any compound can be determined using the comparing ratio equation:
n x molar mass of element
molar mass of compound
x 100
n is the number of moles of the element in 1 mole of the compound
%C =
2 x ( g)
46.07 g
x 100 = 52.14%
C2H6O
%H =
6 x (1.008 g)
46.07 g
x 100 = 13.13%
%O =
1 x (15.99 g)
46.07 g
x 100 = 34.73%
52.14% % % = 100.0%

2. Percent Composition Review
Penicillin, the first of a now large number of antibiotic was discovered accidentally by the Scottish bacteriologist Alexander Aeming in 1928, but he was never able to isolate it as a pure component. This and similar antibiotics have saved millions of lives that might have been lost due to infections. Penicillin-F has the formula C14 H20 N2 SO4. Compute the mass percent of each element.
%C =
14 x ( g) g
x 100 = 53.8 %
%S =
1 x ( g) g
x 100 = 52.14%
%H =
20 x (1.008 g) g
x 100 = 6.45%
10.26%
%O =
4 x ( g) g
x 100 = 13.13%
%N =
2 x ( g) g
x 100 = 8.98%
20.49%

3. Determining the Formula of a Compound
empirical formula
– smallest whole number multiple of the components
molecular formula = (empirical formula)n
[n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH
If the molar mass for any compound is known one can get the chemical or molecular formula by the following:
a. calculate empirical mass = ∑ atomic mass
b. find the multiplication factor = MM/EM
c. molecular formula = empirical formula x factor
Note: if MM = EM
molecular formula = empirical formula

4. Divide through by the smallest number of mols
Determining the Formula of a Compound
Caffeine a stimulant found in coffee, tea, and chocolate, contain % C, 5.159% H, % N, and % O by mass has a molar mass g/mol. Determine the molecular formula of caffeine.
Assume 100g of caffeine
mC = g nc = 49.48/ = 4.12 mol C
mH = g nH =5.159/ = mol H
mN = g nN = 28.87/ = 2.06 mol N
mO = g nO = 16.49/ = 1.03 mol O 4
1.03 5
1.03 2
1.03 1
1.03
Divide through by the smallest number of mols

5. empirical formula ---C4 H5 N2 O
Empirical mass =
(4)(12.011) + (5)(1.008) + (2)(14.007) + (1)(15.999) =
MM/EM = 194.2/ = 2
molecular formula ---(C4 H5 N2 O)2
molecular formula ---C8 H10 N4 O2

6. Empirical Formulas from Percent Composition
We can use percent composition data to calculate empirical formulas.
Assume that you have 100 grams of sample.
Benzene (molar mass g/mol) is 92.2% carbon and 7.83% hydrogen, what is the empirical formula.
If we assume 100 grams of sample, we have g carbon and 7.83 g hydrogen.

7. Empirical Formulas from Percent Composition
Calculate the moles of each element:
1 mol C
g C
92.2 g C ×
= 7.68 mol C
1 mol H
1.008 g H
7.83 g H ×
= 7.77 mol H
Divide by the smallest number to get the formula.
7.68 =
7.77 1 1

8. Molecular Formulas
The empirical formula for benzene is CH. This represents the ratio of C to H atoms of benzene.
The actual molecular formula is some multiple of the empirical formula, (CH)n.
Benzene has a molar mass of g/mol. Find n to find the molecular formula. =
78.00 g/mol
g/mol n
n = 6 so the molecular formula is C6H6.

9. Review
The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule.
The molecular formula of benzene is C6H6
The empirical formula of benzene is CH.
The molecular formula of octane is C8H18
The empirical formula of octane is C4H9.

10. Conclusions
The percent composition of a substance is the mass percent of each element in that substance.
The empirical formula of a substance is the simplest whole number ratio of the elements in the formula.
The molecular formula is a multiple of the empirical formula.