Question

A hydrocarbon is burnt completely to give 3.447g of

Answer 1

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, iron and oxygen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y'  are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=3.447g$

Mass of $H_2O=1.647g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 3.447 g of carbon dioxide, $\frac{12}{44}\times 3.447=0.940g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 1.647 g of water, $\frac{2}{18}\times 1.647=0.183g$ of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.940g}{12g/mole}=0.0783moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.183g}{1g/mole}=0.183moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0783 moles.

For Carbon = $\frac{0.0783}{0.0783}=1$

For Hydrogen = $\frac{0.183}{0.0783}=2.34\approx 2$

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 2

Hence, the empirical formula for the given compound is $CH_2$