6(iii)
An organic compound ‘X’ contains carbon, oxygen and hydrogen only. The percentage of carbon and hydrogen are 47.4% and 10.5% respectively. The relative molecular mass of ‘X’ is 76. Find the empirical formula and the molecular formula of ‘X’.
[Atomic weight: C = 12, O = 16, H = 1]
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| Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
|---|---|---|---|---|
| Carbon | 47.4 | 12 | $\frac{47.4 }{12}$ = 3.95 | $\frac{3.95}{2.63 }$ = $\frac{3 }{2}$ |
| Hydrogen | 10.5 | 1 | $\frac{10.5}{1}$ = 10.5 | $\frac{10.5}{2.63 }$ = 4 |
| Oxygen | 42.1 | 16 | $\frac{42.1}{16}$ = 2.63 | $\frac{2.63}{2.63 }$ = 1 |
Simplest ratio of whole numbers = C : H : O = $\frac{3}{2}$ : 4 : 1 = 3 : 8 : 2
Hence, empirical formula is C3H8O2
Empirical formula weight = (3 x 12) + (8 x 1) + (2 x 16)= 36 + 8 + 32 = 76
Empirical formula weight = Relative molecular mass = 76
So, molecular formula = C3H8O2