1. PERCENT COMPOSITION,
EMPIRICAL &
MOLECULAR FORMULAS
1

2. Percent Composition
oNew food labels are required to
describe the ingredients using
percents of the daily reccom-
mended allowance
• These numbers tell
what part of the total #
of calories can be ob-
tained from a product
• AKA percent composition
2

3. Percent Composition
o To get the information found on
food labels the chemists had to
know what fraction of the whole
was each component
• Component/total and then
multiply by 100
• There are a couple of
procedures used to calculate
percent compositions
3

4. Calculating PC given formula
What percentage of
Hydrogen and Oxygen is
in Water (H2O)?
Assume you have 1 mole of
water, and calculate its molar
mass
(2•1.008g) + (1•15.994g) =
18.01g
4

5. Calculating PC given formula
oThere are 2 mols of H atoms for
every 1 mol of Water molecules
oHow much do 2 mols of H atoms
weigh?
H: (2•1.008g)= 2.016g H
o Percent of H in Water?
2.016g H
X 100%= 11.2%
18.01 g H2O
5

6. Calculating PC given formula
oThere is 1 mol of O atoms for
every 1 mol of Water molecules
oHow much does 1 mol of O
atoms weigh?
O: (1•15.994g)= 15.994g O
o Percent of O in Water?
15.994 O
X 100%= 88.8%
18.01 g H2O
6

7. Percent Composition
o Another method of calculating
the percent composition is by
experimental analysis.
• the overall mass of the sample
is measured.
• then the sample is decomp-
osed or separated into its
component elements
7

8. Percent Composition
o The masses of the component
elements are then determined
and the percent composition is
calculated as before
• by dividing the mass of each
element by the total mass of
the sample
• then multiplying by 100
8

9. Calculating PC given sample
Find the percent composition
of a compnd that contains
1.94g of carbon, 0.48g of
Hydrogen, and 2.58g of
Sulfur in a 5.0g sample of the
compnd.
9

10. Calculating PC given sample
o Calculate the percents for
each element much like you
would calculate the percents
for anything.
C: 1.94g/5.0g X 100% = 38.8%
H: 0.48g/5.0g X 100% = 9.6%
S: 2.58g/5.0g X 100% = 51.6%
10

11. Empirical Formulas
o Once the percent compositions
are determined then they can
be used to calculate a simple
chem formula for the compnd
• key is to convert the percents
by mass into amounts in moles
• Then, compare the moles using
ratios to determine coefficients
11

12. Calculating Empirical Formulas
What is the empirical formula
of a compound that is 80%C
and 20%H by mass
oSince we have been given per-
cents rather than masses we
need to make an assumption.
• Let’s suppose we have a total
sample that weighs 100 g.
12

13. Calculating Empirical Formulas
o This allows us to say that if we
had a 100 grams of sample,
• 80 g is Carbon
• 20 g is Hydrogen
o Now that we have a set of
masses we need to convert
them to moles
• Divide by the molar masses
from the Periodic Table
13

14. Calculating Empirical Formulas
1 mole C
80g C = 6.7mol C
12 g C
1 mole H
20g H = 20 mol H
1gH
• Now calculate the simplest ratio
of each by dividing both values
by the smallest value
14

15. Calculating Empirical Formulas
Divide each mole value by the
smaller of the two values:
C: 6.7/6.7=1
H: 20/6.7 = 2.98  3
Ratio is 1 C’s for every 3 H’s;
so the formula is =
CH3
15

16. Calculating Empirical Formulas
Determine the empirical
formula of a compound
containing 25.9g of N and
74.1g of O.
Notice we have masses
this time not percents,
we can convert masses
directly to moles
16

17. Calculating Empirical Formulas
1 mol N
25.9g N = 1.85 mol N
14 g N
1.85 mol
1 mol O
74.1g O = 4.63 mol O
16 g O
1.85 mol
17

18. Calculating Empirical Formulas
Is the final answer N1O2.5?
Of course not!
We need a whole number
ratio…
Each part of the ratio is multiplied
by a number that converts the
fraction to a whole number
N2(1)O2(2.5)= N2O5
18

19. Molecular Formulas
o The empirical formula indicates
the simplest ratio of the atoms
in the compnd
• However, it does not tell you
the actual numbers of atoms in
each molecule of the compnd
• For instance, glucose has the
molecular formula of C6H12O6
• Empirical form would be CH2O
19

20. Molecular Formulas
o The empirical formula of CH2O,
could be several compnds.
• C2H4O2 or C3H6O3 or C100H200O100
o It’s more important to know the
exact numbers of atoms
involved
The numbers of atoms define
the properties of the compnd
20

21. Molecular Formulas
o The molecular formula is
always a whole-number
multiple of the emp. formula
o In order to calculate the
molecular formula you must
have 2 pieces of information
• Empirical formula
• Molar mass of the unknown
compound (must be given)
21

22. Calculating Molecular Formulas
Find the molecular formula of a
compound that contains
56.36 g of O and 54.6 g of P.
If the molar mass of the
compound is 189.5 g/mol.
1) Find the Empirical Formula
2) Find the MM of the Emp. Form.
3) Find the ratio of the 2 molar
masses (Mol MM/Emp MM)
22

23. 1)Find the Empirical Formula
56.36g O 1 mol O
= 3.5 mol O
16 g O 1.8 mol
1 mol P
54.6g P = 1.8 mol P
31g P
1.8 mol
Empirical formula: P1O2
23

24. 2) Find the MM of the Emp Form.
MM of PO2: (1•31g P) + (2•16g O)
= 63g/mol
3)Find the ratio of the 2 molar
masses (mol MM/emp MM)
GIVEN 189.5 g/mol
= 3.00
CALCULATED 63 g/mol
24

25. Calculating Molecular Formulas
o So the Molecular formula is
3 times heavier than the
Empirical formula
• Therefore, the molecular
formula has 3 times more
atoms than the emp. formula
P3(1)O2(3)= P3O6
25