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zepelin [54]
3 years ago
8

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?
Chemistry
1 answer:
yan [13]3 years ago
6 0

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}

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