5(i)
The compound ‘A’ has the following percentage composition by mass, carbon 26.7%, oxygen 71.1%, hydrogen 2.2%. Determine the empirical formula of ‘A’. If the relative molecular mass of ‘A’ is 90, what is the molecular formula of ‘A’?
[At. wt. C=12, H=1, O=16]
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| Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
|---|---|---|---|---|
| Carbon | 26.7 | 12 | $\frac{26.7}{12}$ = 2.225 | $\frac{2.225}{2.2}$ = 1 |
| Hydrogen | 2.2 | 1 | $\frac{2.2}{1}$ = 2.2 | $\frac{2.2}{2.2}$ = 1 |
| Oxygen | 71.1 | 16 | $\frac{71.1}{16}$ = 4.44 | $\frac{4.44}{2.2}$ = 2 |
Simplest ratio C : H : O = 1 : 1 : 2
Hence, empirical formula is CHO2.
Empirical formula weight = 12 + 1 + 2(16) = 45
n = $\frac{\text{Relative molecular mass}}{\text{Empirical formula weight}}$ = $\frac{90}{45}$ = 2
Molecular formula = 2 × (CHO2) = C2H2O4