Question

a 9.545 gram sample of an organic compound containing only c, h, and o is analyzed by combustion analysis and 19.07 g co₂ and 7.808 g h₂o are produced.
in a separate experiment, the molar mass is found to be 88.11 g/mol. determine the empirical formula and the molecular formula of the organic compound.
enter the elements in the order c, h, o
empirical formula =
molecular formula =

Explanation:

Step1: Calculate moles of C from \(CO_2\)

Molar mass of \(CO_2 = 44.01\space g/mol\).
Moles of \(CO_2=\frac{19.07\space g}{44.01\space g/mol}\approx0.4333\space mol\).
Since 1 mole \(CO_2\) has 1 mole C, moles of \(C = 0.4333\space mol\).
Mass of \(C=0.4333\space mol\times12.01\space g/mol\approx5.204\space g\).

Step2: Calculate moles of H from \(H_2O\)

Molar mass of \(H_2O = 18.02\space g/mol\).
Moles of \(H_2O=\frac{7.808\space g}{18.02\space g/mol}\approx0.4333\space mol\).
Since 1 mole \(H_2O\) has 2 moles H, moles of \(H = 2\times0.4333\space mol = 0.8666\space mol\).
Mass of \(H=0.8666\space mol\times1.008\space g/mol\approx0.874\space g\).

Step3: Calculate mass and moles of O

Mass of \(O=9.545-(5.204 + 0.874)=3.467\space g\).
Molar mass of \(O = 16.00\space g/mol\).
Moles of \(O=\frac{3.467\space g}{16.00\space g/mol}\approx0.2167\space mol\).

Step4: Determine mole ratios (for empirical formula)

Divide moles by the smallest number (\(n_O = 0.2167\)):
\(n_C=\frac{0.4333}{0.2167}\approx2\), \(n_H=\frac{0.8666}{0.2167}\approx4\), \(n_O=\frac{0.2167}{0.2167}=1\).
Empirical formula is \(C_2H_4O\).

Step5: Calculate molecular formula

Molar mass of empirical formula \(C_2H_4O=(2\times12.01)+(4\times1.008)+16.00 = 44.06\space g/mol\).
\(n=\frac{88.11\space g/mol}{44.06\space g/mol}\approx2\).
Molecular formula is \((C_2H_4O)_2 = C_4H_8O_2\).

Answer:

empirical formula = \(C_2H_4O\)
molecular formula = \(C_4H_8O_2\)