Name: Law of Multiple Proportions
Uploaded: 2019-07-19
Duration: 6 min 43 s
Description: The Law of multiple proportions is applicable only when two elements combine to form two or more compounds. Watch this video to know how this law is applicable.

Question

Butyric acid contains only

C, H

and

O

. A

4.24 mg

sample of butyric acid is completely burnt. It gives

8.45 mg

of

{CO}_{2}

and

3.46 mg

of

H_{2} O

. The molecular mass of butyric acid was determined by experiment to be

88 amu

. What is molecular formula?

Accepted Answer

Given,

$4.24 mg$ butyric acid gives $8.45 mg$ ${CO}_{2}$ and ${CO}_{2}$ $H_{2} O$ .

Molecular mass of ${CO}_{2} = 12 + (2 \times 16) = 12 + 32 = 44 g {mol}^{- 1}$ .

$44 g$ of ${CO}_{2}$ contains $12 g C$ .

$∴ 44 mg$ of ${CO}_{2}$ contains $12 mg C$ .

Thus, $1 mg$ of ${CO}_{2}$ contains $\frac{12}{44} \times 1 = 0.273 mg C$ .

$C, H$ of $4.24 mg$ contains $0.273 \times 8.45 = 2.306 mg C$ .

It is present in $4.24 mg$ compound. Therefore, the percentage of carbon in the given compound will be as follows,

% of $C = \frac{2.306}{4.24} \times 100 = 54.4 % .$

Molecular mass of $H_{2} O = (2 \times 1) + 16 = 18 g {mol}^{- 1}$

$18 g$ of $H_{2} O$ contains $2 g H$ .

$∴ 18 mg$ of $H_{2} O$ contains $2 mg H$ .

Thus, $3.46 mg$ of $H_{2} O$ contains $\frac{2}{18} \times 3.46 = 0.384 mg H$ .

Therefore, the percentage of hydrogen in the given compound will be as follows,

% of $H = \frac{0.384}{4.24} \times 100 = 9.1 %$ .

The percentage of oxygen,

$% O = 100 - [% C + % H] % O = 100 - [54.4 + 9.1] % O = 100 - 63.5 = 36.5$

$C = \frac{54.4}{12} = 4.53 \Rightarrow \frac{4.53}{2.28} = 1.99 \Rightarrow 2 . H = \frac{9.1}{1} = 9.1 \Rightarrow \frac{9.1}{2.28} = 3.99 \Rightarrow 4 . O = \frac{36.5}{16} = 2.28 \Rightarrow \frac{2.28}{2.28} = 1 \Rightarrow 1 . C : H : O = 2 : 4 : 1 .$

Empirical formula $= C_{2} H_{4} O_{1}$
Empirical formula mass $C_{2} H_{4} O_{1} = (2 \times 12) + (4 \times 1) + 16 = 24 + 4 + 16 = 44 u .$

Molecular mass of butyric acid $= 88 u$ .

$n = \frac{Molecular mass}{Empirical formula mass} n = \frac{88}{44} = 2 .$

Therefore, the molecular formula,

$M . F = n \times E . F M . F = 2 \times (C_{2} H_{4} O) M . F = C_{4} H_{8} O_{2}$ .

Butyric acid contains only $C, H$ and $O$ . A $4.24 mg$ sample of butyric acid is completely burnt. It gives $8.45 mg$ of ${CO}_{2}$ and $3.46 mg$ of $H_{2} O$ . The molecular mass of butyric acid was determined by experiment to be $88 amu$ . What is molecular formula?

Important Questions on Some Basic Concepts of Chemistry

Learn and Prepare for any exam you want !

Butyric acid contains only C,Hand O. A 4.24 mg sample of butyric acid is completely burnt. It gives 8.45 mg of CO2 and 3.46 mg of H2O. The molecular mass of butyric acid was determined by experiment to be 88 amu. What is molecular formula?

Important Questions on Some Basic Concepts of Chemistry

Learn and Prepare for any exam you want !

Butyric acid contains only $C, H$ and $O$ . A $4.24 mg$ sample of butyric acid is completely burnt. It gives $8.45 mg$ of ${CO}_{2}$ and $3.46 mg$ of $H_{2} O$ . The molecular mass of butyric acid was determined by experiment to be $88 amu$ . What is molecular formula?