Question

When 3.915 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 12.74 grams of CO2 and 3.913 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 54.09 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

Answer: The empirical formula and the molecular formula of the hydrocarbon is $C_2H_3$ and $C_4H_6$ respectivley.

Explanation:

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

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y'  are the subscripts of Carbon, hydrogen

We are given:

Mass of $CO_2$ = 12.74 g

Mass of $H_2O$= 3.913 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 12.74 g of carbon dioxide, =$\frac{12}{44}\times 12.74=3.474g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 3.913 g of water, =$\frac{2}{18}\times 3.913=0.435g$ of hydrogen will be contained.  

Mass of C = 3.474 g

Mass of H = 0.435 g

Step 1 : convert given masses into moles.

Moles of C =$\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.474g}{12g/mole}=0.289moles$

Moles of H=$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.435g}{1g/mole}=0.435moles$

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

For C =$\frac{0.289}{0.289}=1$

For H =$\frac{0.435}{0.289}=1.5$

The ratio of C : H = 1: 1.5

The whole number ratio will be = 2: 3

Hence the empirical formula is $C_2H_3$.

The empirical weight of $C_2H_3$ = 2(12.01)+3(1.008)= 27.04 g.

The molecular weight = 54.09 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{54.09}{27.04}=2$

The molecular formula will be=$2\times C_2H_3=C_4H_6$