Question

Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of co2 and 1.37 mg of h2o. if the compound contains only carbon and hydrogen, what is its empirical formula?

Answer 1

Answer: The empirical formula for the given organic compound is $CH$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen 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 $CO_2=5.86mg=5.86\times 10^{-3}g$    (Conversion factor:  1 g = 1000 mg)

Mass of $H_2O=1.37mg=1.37\times 10^{-6}g$

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 $5.86\times 10^{-3}g$ of carbon dioxide, $\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g$ of carbon will be contained.

• For calculating the mass of hydrogen:

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

So, in $1.37\times 10^{-3}g$ of water, $\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g$ 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 =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles$

• 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 $0.133\times 10^{-3}moles$

For Carbon = $\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1$

For Hydrogen = $\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1$

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

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is $C_1H_{1}=CH$

Answer 2

The general equation for a combustion reaction would be expressed as:

CxHy + (x+y)O2 = xCO2 + y/2H2O

To determine the empirical formula of the substance, toluene, we calculate the moles of C and H in the product. The only source of carbon would be from carbon dioxide and for hydrogen, from water. We do as follows:

1.37 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H2O ) = 0.1521 mol H
5.86 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.1332 mol C

So we have,
C = 0.1332 C / 0.1332 = 1
H = 0.1521 H / 0.1332 = 1

Thus, the empirical formula woud be CH.