Q. A compound $(60\, g )$ on analysis produce carbon, hydrogen and oxygen $24\, g,\, 4\, g$ and $32\, g$ respectively. The empirical formula is :
Rajasthan PMTRajasthan PMT 2005
Solution:
$\because 60\, g$ of compound contains $24\, g$ of carbon
$\therefore \%$ of carbon $=\frac{24 \times 100}{60}=40 \%$
Similarly, percentage of hydrogen
$=\frac{4 \times 100}{60}=6.66 \%$
$\therefore $ percentage of oxygen
$=100-(40+6.66)$
$=53.34 \%$
Element $t$
Percentage
Atomic $c$ mass
Relative number of atoms
Ratio
$C$
$40\%$
$12$
$40/12=3.33$
$3.33/3.3=3=1$
$H$
$6.66\%$
$1$
$6.66/1=6.66$
$6.66/3.33=2$
$O$
$53.34\%$
$16$
$53.34/16=3.33$
$3.33/3.33=1$
Therefore, empirical formula $= CH _{2} O$
| Element $t$ | Percentage | Atomic $c$ mass | Relative number of atoms | Ratio |
|---|---|---|---|---|
| $C$ | $40\%$ | $12$ | $40/12=3.33$ | $3.33/3.3=3=1$ |
| $H$ | $6.66\%$ | $1$ | $6.66/1=6.66$ | $6.66/3.33=2$ |
| $O$ | $53.34\%$ | $16$ | $53.34/16=3.33$ | $3.33/3.33=1$ |