Name: Empirical Formula and Molecular Formula (II)
Uploaded: 2019-07-19
Duration: 8 min 7 s
Description: The video describes one of the most important applications of the mole concept, finding the empirical formula of any compound. The method is explained with s...

Question

0.21 g

of an organic compound containing

C, H, O

and

N

gave upon combustion

0.462 g

of

{CO}_{2}

and

0.1215 g

of

H_{2} O

. The ammonia produced on distillation of

0.104 g

of this compound with

NaOH

, required

15 mL

of

\frac{N}{20} H_{2} {SO}_{4}

for neutralisation. Find the empirical formula of the compound.

Accepted Answer

Let the compound be $C_{x} H_{y} O_{z} N_{p}$ .

$0.462 g$

According to the principle of atomic conservation,

$Weight of C in 0.462 g of {CO}_{2} = Weight of C in 0.21 g of organic compound \frac{12}{44} \times 0.462 g = Weight of C in 0.21 g of organic compound % of C = \frac{\frac{12}{44} \times 0.462 g}{0.21} \times 100 = 60 %$

$Weight of H in 0.1215 g of H_{2} O = Weight of H in 0.21 of organic compound % of H = \frac{\frac{2}{18} \times 0.1215}{0.21} \times 100 = 6.43 %$

$Organic compound + NaOH \overset{Distillation}{\to} {NH}_{3} 0.104 {NH}_{3} + H_{2} {SO}_{4} \to {({NH}_{4})}_{2} {SO}_{4} 15 ml, \frac{1}{20} N Molar equivalence of {NH}_{3} = mmoles of {NH}_{3} \times 1 = 15 \times \frac{1}{20} \frac{Weight of {NH}_{3}}{17} \times 1000 = \frac{3}{4} Weight of {NH}_{3} = \frac{3 \times 17}{4000} g So, weight of N present in the above weight of ammonia = \frac{3 \times 17}{4000} \times \frac{14}{17} = \frac{42}{4000} % of N = \frac{1.05 \times 10^{- 2}}{0.104} \times 100 = 10 % of O = 100 - (% C + % H + % N) = 100 - (60 + 6.4 + 10) = 23.6 %$

Element $%$ by weight $\frac{% by weight}{Atomic weight} = x$ $Mole ratio$

Carbon $60$ $\frac{60}{12} = 5$ $\frac{5}{0.72} = 7$

Hydrogen $6.4$ $\frac{6.4}{1} = 6.4$ $\frac{6.4}{0.72} = 9$

Nitrogen

Oxygen

$10$

$23.6$

$\frac{10}{14} = 0.72$

$\frac{23.6}{16} = 1.47$

$\frac{0.72}{0.72} = 1$

$\frac{1.47}{0.72} = 2$

So, the empirical formula will be $C_{7} H_{9} {NO}_{2}$ .

Important Questions on Empirical, Molecular and Structural Formulae

Learn and Prepare for any exam you want !

0.21 g of an organic compound containing C,H,O and N gave upon combustion 0.462 g of CO2 and 0.1215 g of H2O. The ammonia produced on distillation of 0.104 g of this compound with NaOH, required 15 mL of N20 H2SO4 for neutralisation. Find the empirical formula of the compound.

Important Questions on Empirical, Molecular and Structural Formulae

Learn and Prepare for any exam you want !