Question

One gram of a certain compound has $0.2729$ g of carbon and $0.7271$ g of oxygen. Molecular weight of the compound is $88$. The formula of the compound is:

• A. $CO$
• B. $C{O}_2$
• C. $C_3{O}_2$
• D. $C_2{O}_4$

Answer 1

Answer: (D)

$C_2{O}_4$

The ratio of the number of gram atoms between $C$ and $O$ is,

Moles of carbon $=\frac{0.2729}{12}$

Moles of oxygen $=\frac{0.7271}{16}$

The ratio of moles of $C:O=\frac{0.2729}{12}:\frac{0.7271}{16}$

$=0.0224:0.0454$

$=1:2$

$\therefore$ The empirical formula is $C{O}_2$ and empirical weight is $44$ g.

$Molecularweight=(Empiricalformula{)}_n$

$n=\frac{\text{Molecular weight}}{\text{Empirical weight}}=\frac{88}{44}=2$

So, the molecular formula is $C_2{O}_4$.