MCQ
An organic compound contains $49.3\%$ carbon $6.84\%$ hydrogen and its vapour density is $73$. Molecular formula of the compound is
- A${C_3}{H_5}{O_2}$
- ✓${C_6}{H_{10}}{O_4}$
- C${C_3}{H_{10}}{O_2}$
- D${C_4}{H_{10}}{O_2}$
✓
Answer
Correct option: B.
${C_6}{H_{10}}{O_4}$
b
(b)
(b)
| Elements | No. of Moles | Simple ratio |
| $C = 12$ | $49.3/12 = 4.1$ | $4.1/2.7 = 13 \times 2 = 2.6 = 3$ |
| $H = 1$ | $6.84/1 = 6.84$ | $6.84/2.7 = 2.5 \times 2 = 5$ |
| $O = 16$ | $43.86/16 = 2.7$ | $2.7/2.7 = 1 \times 2 = 2$ |
Empirical formula = ${C_3}{H_5}{O_2}$
E.F. wt. = $12 \times 3 + 1 \times 5 + 16 \times 2 = 73$
Molecular wt = $V.D. \times 2 = 73 \times 2 = 146$
$n = \frac{{M.wt}}{{E.F.wt}} = \frac{{146}}{{73}} = 2$
Molecular formula $= (E.F)n = {({C_3}{H_5}{O_2})_2} = {C_6}{H_{10}}{O_4}$.
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