Question

$60gm$ of an organic compound on analysis is found to have $C=24g,H=4g$ and $O=32g$. The empirical formula of compound is:

• A. $C{H}_{2}O$
• B. $CHO$
• C. $C_2{H}_{2}O$
• D. $C_2{H}_2{O}_2$

Answer 1

Answer: (A)

$C{H}_{2}O$

As the mass of organic compound as well as its constituents elements are given, now first to find that how many atoms are there of the particular type of element, by dividing the mass by their molar mass.

Atoms of $C=\frac{24}{12}=2$

Atoms of $H=\frac{4}{1}=4$

Atoms of $O=\frac{32}{16}=2$

Now, secondly, we take the ratio of all atoms of all type present.

So, $2:4:2=1:2:1$

So, empirical formula must be $C{H}_{2}O$.

Hence, the correct option is $\text{A}$