Question

Complete combustion of a sample of a hydrocarbon gives 0.66g of carbon dioxide, 0.36 g of water. The empirical formula of the compound is

Answer 1

Answer:The empirical formulae of the hydrocarbon is C3H4

Explanation:

Answer 2

The empirical formula of the compound is $CH_3$.

Explanation:

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

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 0.66 g

Mass of $H_2O$=0.36 g

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 0.66 g of carbon dioxide, =$\frac{12}{44}\times 0.66=0.18g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.36 g of water, =$\frac{2}{18}\times 0.36=0.04g$ of hydrogen will be contained.

Mass of C = 0.18  g

Mass of H = 0.04 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{0.18g}{12g/mole}=0.015moles$

Moles of H=$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.04g}{1g/mole}=0.04moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =$\frac{0.015}{0.015}=1$

For H =$\frac{0.04}{0.015}=3$

The ratio of C : H = 1: 3

Hence the empirical formula is $CH_3$.

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