Q. A compound contains $54.55 \%$ carbon, $9.09 \%$ hydrogen and $36.36 \%$ oxygen . The empirical formula of this compound is
Some Basic Concepts of Chemistry
Solution:
Element
$\%$
At. weight
$\%$ Atomic weight
Simplest molar ratio
$C$
$54.55$
$12$
$\frac{54.55}{12}=4.54$
$\frac{4.54}{2.275}=2$
$H$
$9.09$
$1$
$\frac{9.09}{1}=9.09$
$\frac{9.09}{2.27}=4$
$O$
$36.36$
$16$
$\frac{36.36}{16}=2.27$
$\frac{2.27}{2.27}=1$
$\therefore$ Empirical formula is $C _{2} H _{4} O$.
| Element | $\%$ | At. weight | $\%$ Atomic weight | Simplest molar ratio |
|---|---|---|---|---|
| $C$ | $54.55$ | $12$ | $\frac{54.55}{12}=4.54$ | $\frac{4.54}{2.275}=2$ |
| $H$ | $9.09$ | $1$ | $\frac{9.09}{1}=9.09$ | $\frac{9.09}{2.27}=4$ |
| $O$ | $36.36$ | $16$ | $\frac{36.36}{16}=2.27$ | $\frac{2.27}{2.27}=1$ |