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

Empirical formula -
`C_(7)H_(8)`

Step-by-step 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* (1molCO_(2))/(44gCO_(2))* (1mol\,C)/(1mol\,CO_(2))= 1`

`0.00151* (1molH_(2)O)/(44gH_(2)O)* (2mol\,H)/(1mol\,H_(2)O)= 1.4`

1:1.4

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

`1:1.4* 7= 7:8`

Therefore, empirical formula -
`C_(7)H_(8)`