Question

When 6.597 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.74 grams of CO2 and 5.935 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 40.06 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

first calculate % composition

% of carbon

molar mass of CO₂ = 44gm/mole and molar mass of C =12gm/mole that mean in 44 gmof CO₂ contain 12 gm of carbon.then 21.74 gm CO₂ contain 21.7412/44 = 5.929 gm of carbon

6.597 gm of sample = 100% then 5.929 gm = 5.929100/6.597 = 89.874%

% of carbon = 90.01%

% of hydrogen

molar mass of H₂O = 18gm/mole and molar mass of H₂ = 2gm/mole that mean 18 gm H₂O contain 2 gm H₂ then 5.935 gm  H₂O contain 5.9352/18 = 0.6594 gm hydrogen

6.597 gm of sample = 100% then 0.6594 gm = 0.6594100/6.597 = 9.99%

% of hydrogen = 9.99%

calculate atomic retio by dividing % composition by atomic mass.

mass retio of carbon = 90.01/12 = 7.50

mass retio of hydrogen = 9.99/1 = 9.99

now calculate simplest retio by dividing atomic retio by smallest atomic retio retio

simplest retio of carbon = 7.50/7.50 = 1

simplest retio of hydrogen = 9.99/7.50 = 1.33

convert fractional number into whole number by multiplying by 3

for carbon = 13 = 3

for hydrogen = 1.333 = 4

Empirical formula of compound = C₃H₄

Empirical formula weight = (12.013) + (1.008 4) = 40.06

Empirical formula weight = molar mass therefore

Empirical formula = molacular formula = C₃H₄