Percentage of oxygen = 100 - (40 + 6.66) = 100 - 46.6 = 53.4%
Element |
% |
Atomic mass |
Relative ratio of atoms |
Simplest ratio |
C |
40 |
12 |
40/12 = 3.3 |
3.3/3.3 = 1 |
H |
6.66 |
1.008 |
6.61/1.008 = 6.5 |
6.5/3.3 = 2 |
O |
53.4 |
16 |
53.4/16 = 3.3 |
3.3/3.3 = 1 |
Hence, empirical formula = CH2O
Now, empirical formula mass
= 12 + (2 × 1) + 16 = 12 + 2 + 16 = 30″
Molecular mass = 2 x Vapour density = 2 × 30 = 60
n = Molecular mass / Empirical formula mass
= \(\frac{60}{30}\) = 2
Thus, Molecular formula = (Empirical formula) × n = (CH2O) × 2 = C2H4O2