Question

1.50 g of a certain compound x, known to be made of carbon and hydrogen, and to have a molecular molar mass of 78. g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 5.08 g water 1.04 g use this information to find the molecular formula of x.

Explanation:

Step1: Calculate moles of carbon

Molar mass of \(CO_2 = 44\space g/mol\).
Moles of \(CO_2=\frac{5.08\space g}{44\space g/mol}= 0.1155\space mol\).
Since 1 mole \(CO_2\) has 1 mole C, moles of C \(= 0.1155\space mol\).
Mass of C \(=0.1155\space mol\times12\space g/mol = 1.386\space g\).

Step2: Calculate moles of hydrogen

Molar mass of \(H_2O=18\space g/mol\).
Moles of \(H_2O=\frac{1.04\space g}{18\space g/mol}=0.0578\space mol\).
Since 1 mole \(H_2O\) has 2 moles H, moles of H \(=2\times0.0578 = 0.1156\space mol\).
Mass of H \(=0.1156\space mol\times1\space g/mol=0.1156\space g\).

Step3: Check ratio of C and H

Mass of C + H \(=1.386 + 0.1156=1.5016\approx1.50\space g\) (matches mass of compound).
Ratio of moles C : H \(=0.1155:0.1156\approx1:1\).
Empirical formula is \(CH\) (empirical formula mass \(=13\space g/mol\)).

Step4: Determine molecular formula

Molecular molar mass \(=78\space g/mol\).
\(n=\frac{78\space g/mol}{13\space g/mol}=6\).
Molecular formula \(=(CH)_6 = C_6H_6\).

Answer:

\(C_6H_6\)