Empirical Formula
• The simplest ratio of atoms
• For example, the empirical formula of
H2O2 (hydrogen peroxide) is HO (1
hydrogen atom for every oxygen atom)
• What is the empirical formula of glucose,
C6H12O6? Sucrose, C12H22O11?
How to calculate EF
Q: Calculate the EF of a compound containing
49.5% C, 5.2% H, 28.8% N and 16.5 % O by
mass.
1.Arrange the elements in % or mass
49.5g C, 5.2g H, 28.8g N, 16.5g O
2.Divide each by the molar mass of the element to get
moles
C49.5g
H5.2g
N28.8g
O16.5g
12gmol-1 1 gmol-1 14gmol-1 16gmol-1
C 4.125 H 5.20 N
mol
mol
2.056
mol
O 1.031
mol
4. Divide everything by the SMALLEST
NUMBER of moles
4.125
5.2
2.056
1.031
1.031
1.031
1.031
1.031
=
4
:
5
:
2
:
1
5. Write the formula with the simplest whole
number ratio
C4H5N2O ANSWER
Now try these! Calculate the EF:
1. Nutrasweet is 57.14% C, 6.16% H,
9.52% N and 27.18% O.
2. A compound is 72.2% Mg and 27.8% N
3. An oxide of nitrogen contains 42.05 g N
and 95.95 g of O.
4. Mercury forms a compound that is 73.9%
Hg and 26.1% Cl.
5. Vitamin C contains 40.92% C, 4.58% H
and 54.50% O.
Now try these - ANSWERS
1.
2.
3.
4.
5.
C14H18N2O5
Mg3N2
NO2
HgCl2
C 3H 4O 3
From EF to molecular formula
• The molecular formula is the ACTUAL
NUMBER OF EACH ATOM in the compound
HO
Multiplication
factor
x2
H2O2 (hydrogen peroxide)
CH2O
CH2O
X6
x2
C6H12O6 (glucose)
CH3COOH (ethanoic acid)
EF
MF
How to work out the MF
1. Calculate the molar mass for the empirical
formula.
Example: for a compound where the EF is
HO and actual molar mass is 34 gmol-1
Molar mass of EF = 1 + 16 = 17 gmol-1
2. Divide the molar mass MF ÷ EF
Actual MF ÷ EF 34 ÷ 17 gmol-1 = 2
3. Multiply the formula by the factor calculated in
part 2.
HO x 2 = H2O2
Questions – Molecular formula
1. A compound has an empirical formula of
CH2Cl and a molar mass of 99 gmol-1.
Calculate the molecular formula.
2. A compound has 75.46% C, 4.43% H
and 20.1% O and a molar mass of 318
gmol-1. Calculate the molecular formula.
Answers – Molecular formula
1. The EF is CH2Cl. The molar mass for
this is 12 + 2 x 1 + 35.5 = 49.5. 99 ÷
49.5 = 2, therefore the molecular formula
is TWICE that of the EF – C2H4Cl2.
2. The EF is C10H7O2. The molar mass for
this is 12 x 10 + 7 x 1 + 2 x 16 = 159.
318 ÷ 159 = 2, therefore the molecular
formula is TWICE that of the EF –
C20H14O4.