empirical formula
EMPIRICAL FORMULA
An empirical formula is the simplest whole number ratio of elements in a compound.
Molecular formula Empirical formula
C
2
H
C
4
H
6
8
CH
3
CH
2
C
6
H
12
O
6
CH
2
O
The molecular formula is always a multiple of the empirical formula
The molecular formula is always a multiple of the empirical formula
Eg. Empirical formula is CH
2
, Mr = 42. Find molecular formula.
CH
2
= 12 + 1 + 1 = 14
42/14 = 3
Molecular formula is 3 x empirical formula:
CH
2
x 3
= C
3
H
6
Afl
For each of the following compounds the Mr and the empirical formula is shown. Work out the molecular formula of each compound.
i) empirical formula = CH ii) empirical formula = NH iii) empirical formula = CH
2
3
3
Mr = 30
Mr = 17
Mr = 98 iv) empirical formula = CH Mr = 78
Calculating empirical formula
What is the empirical formula if 12.0 g of calcium react with 2.8 g of nitrogen ?
mass
Ca
12.0
N
2.8
Calculating empirical formula
What is the empirical formula if 12.0 g of calcium react with 2.8 g of nitrogen ?
mass molar mass
Ca
12.0
40
N
2.8
14
Calculating empirical formula
What is the empirical formula if 12.0 g of calcium react with 2.8 g of nitrogen ?
mass molar mass number of moles
Ca
12.0
40
12.0/40
= 0.3
N
2.8
14
28/14
= 0.2
Calculating empirical formula
What is the empirical formula if 12.0 g of calcium react with 2.8 g of nitrogen ?
mass molar mass number of moles
Ca
12.0
40
12.0/40
= 0.3
N
2.8
14
28/14
= 0.2
divide by smallest = 0.3/0.2
= 0.2/0.2
Calculating empirical formula
What is the empirical formula if 12.0 g of calcium react with 2.8 g of nitrogen ?
mass molar mass number of moles
Ca
12.0
40
12.0/40
= 0.3
N
2.8
14
28/14
= 0.2
divide by smallest ratio
= 0.3/0.2
1.5
= 0.2/0.2
1.0
Calculating empirical formula
What is the empirical formula if 12.0 g of calcium react with 2.8 g of nitrogen ?
mass molar mass number of moles
Ca
12.0
40
12.0/40
= 0.3
N
2.8
14
28/14
= 0.2
divide by smallest ratio whole number ratio
= 0.3/0.2
1.5
3
= 0.2/0.2
1.0
2
Calculating empirical formula
What is the empirical formula if 12.0 g of calcium react with 2.8 g of nitrogen ?
mass molar mass number of moles
Ca
12.0
40
12.0/40
= 0.3
N
2.8
14
28/14
= 0.2
divide by smallest ratio whole number ratio
= 0.3/0.2
1.5
3
Ca
3
N
2
= 0.2/0.2
1.0
2
Now try these:
From the following data determine the empirical formulae of the compounds:
1) 4.0 g of sulfur reacts with the same mass of oxygen.
2) 6.5 g of potassium reacts with 1.3 g of oxygen.
3) 4.8 g of magnesium reacts with 14.2 g of chlorine.
4) 2.8 g of iron reacts with 1.2 g of oxygen.
5) 5.4 g of aluminium reacts with 48 g of bromine
ANSWERS
1) SO
2
2) K
2
0
3) MgCl
2
4) Fe
2
O
3
5) AlBr
3
Moles 6
–
Empirical & Molecular Formulas
C
3
H
6
CH
2
C
2
H
2
HO
70
56
34
C
3
H
8
C
6
H
6
C
6
H
6
O
6
a) b) c) d)
Ca 20%
S 0.5 g
Pb 90.7%
Br 80%
O 0.75 g
O 9.3%
K 1.82 g I 5.93 g
ANSWERS
a)CaBr
2 b)SO
3 c)Pb
3
O
4 d)KIO
3
O 2.24 g
3) 2.67 g of copper reacts with sulphur to form 4.01 g of copper sulphide. Work out the mass of sulphur that reacted with the copper and then work out the empirical formula of the copper sulphide.
(4)
4) Calculate the empirical and molecular formula of the following substances.
a) b) c)
Pb
P
C
59.4%
0.282 g
60.0%
Cl 40.6%
O 0.218 g
H 13.3%
M r
M r
349
220
O 26.7%
M r
60 (9)
