Q. An organic compound with $ C=40 \% $ and $ H=6.7\% $ will have the empirical formula
Chhattisgarh PMTChhattisgarh PMT 2005
Solution:
Elements $\%$
No.of moles
Simple Ratio
$C =40 \%$
$40 / 12=3.33$
$3.33 / 3.33=1$
$H =6.7 \%$
$6.7 / 1=6.7$
$6.7 / 3.33=2$
$O =53.3 \%$
$53.3 / 16=3.33$
$3.33 / 3.33=1$
Thus, emipirical formula $= CH _{2} O$
| Elements $\%$ | No.of moles | Simple Ratio |
|---|---|---|
| $C =40 \%$ | $40 / 12=3.33$ | $3.33 / 3.33=1$ |
| $H =6.7 \%$ | $6.7 / 1=6.7$ | $6.7 / 3.33=2$ |
| $O =53.3 \%$ | $53.3 / 16=3.33$ | $3.33 / 3.33=1$ |