Question

Combustion analysis of toluene, a common organic solvent, gives 7.03 mg of CO2 and 1.64 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 compound is CH

Step-by-step 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, hydrogen and oxygen respectively.

We are given:

Mass of
`CO_2=7.03mg=0.00703g`

Mass of
`H_2O=1.64mg=0.00164g`

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

So, in 0.00703 g of carbon dioxide,
`(12)/(44)* 0.00703=0.00192g` of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.00164 g of water,
`(2)/(18)* 0.00164=0.000182g` 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}}=(0.00192g)/(12g/mole)=0.00016moles`

Moles of Hydrogen =
`\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=(0.000182g)/(1g/mole)=0.000182moles`

• 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 0.00016 moles.

For Carbon =
`(0.00016)/(0.00016)=1`

For Hydrogen =
`(0.000182)/(0.00016)=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`