Question

Vitamin C $(M=176)$ is a compound of $C,H,O$ found in many natural sources, especially citrus fruits. When a $1.0\text{g}$ sample of vitamin C is placed in a combustion chamber and burned, the following data is obtained.
$\text{Mass of}C{O}_2\text{absorber after combustion}=83.85\text{g}$
$\text{Mass of}C{O}_2\text{absorber before combustion}=82.35\text{g}$
$\text{Mass of}{H}_{2}O\text{absorber after combustion}=37.96\text{g}$
$\text{Mass of}{H}_{2}O\text{absorber before combustion}=37.55\text{g}$
What is the empirical formula of Vitamin C?

• A. $C{H}_{2}O$
• B. $CHO$
• C. $C_3{H}_4{O}_3$
• D. $C_2{H}_{2}O$

Answer 1

The correct option is C $C_3{H}_4{O}_3$

Mass of $C{O}_2$ produced =$83.85-82.35=1.5\text{g}$
$\therefore \text{Percentage of C}=\frac{12}{44}\times \frac{1.5}{1.0}\times 100=40.9\%$
Mass of $H_{2}O$ produced =$37.96-37.55=0.41\text{g}$
$\therefore \text{Percentage of H}=\frac{2}{18}\times \frac{0.41}{1.0}\times 100=4.56\%$
Hence, percentage of oxygen= $100-(40.9+4.56)=54.54\%$
Now,
$N_C:N_H:N_O=\frac{40.9}{12}:\frac{4.56}{1}:\frac{54.54}{16}$
$\Rightarrow 3:4:3$
Hence, Empirical formula of Vitamin C is $C_3{H}_4{O}_3$.