Question

Combustion analysis of toluene, a common organic solvent, gives 6.45 mg of CO2 and 1.51 mg of H2O. If the compound contains only carbon and hydrogen, what is its empirical formula?

Answer 1

Answer:

Empirical formula - $C_{7}H_{8}$

Explanation:

From the given,

Mass of $CO_{2}$ = 6.45 mg= 0.00645 g

Molar mass of carbon = 12 g/mol

Molar mass of $CO_{2}$ = 44 g/mol

Given mass of $H_{2}O$ = 1.51 mg= 0.00151 g

Molar mass of water = 18 g/mol

Molar mass of hydrogen = 1.0 g/mol

$0.00645\times \frac{1molCO_{2}}{44gCO_{2}}\times \frac{1mol\,C}{1mol\,CO_{2}}= 1$

$0.00151\times \frac{1molH_{2}O}{44gH_{2}O}\times \frac{2mol\,H}{1mol\,H_{2}O}= 1.4$

1:1.4

Those two values are multiplied by two.(molar ratio -7)

$1:1.4\times 7= 7:8$

Therefore, empirical formula - $C_{7}H_{8}$