Question

1. a 15.00 - g sample of an unknown hydrocarbon is analyzed by combustion analysis. the sample produced 50.70 grams of carbon dioxide and 10.42 grams of water. find the empirical formula of the unknown.

Explanation:

Step1: Calculate moles of carbon

The mass of carbon in \(CO_2\) is calculated. The molar mass of \(CO_2\) is \(M_{CO_2}=44\space g/mol\), and the molar mass of \(C\) is \(M_C = 12\space g/mol\).
The mass of \(C\) in \(50.70\space g\) of \(CO_2\) is \(m_C=\frac{12}{44}\times50.70\space g\approx13.83\space g\).
The moles of \(C\), \(n_C=\frac{m_C}{M_C}=\frac{13.83\space g}{12\space g/mol}\approx1.15\space mol\).

Step2: Calculate moles of hydrogen

The mass of hydrogen in \(H_2O\) is calculated. The molar mass of \(H_2O\) is \(M_{H_2O}=18\space g/mol\), and the molar mass of \(H\) is \(M_H = 1\space g/mol\).
The mass of \(H\) in \(10.42\space g\) of \(H_2O\) is \(m_H=\frac{2}{18}\times10.42\space g\approx1.16\space g\).
The moles of \(H\), \(n_H=\frac{m_H}{M_H}=\frac{1.16\space g}{1\space g/mol}=1.16\space mol\).

Step3: Find the ratio of moles

Divide the number of moles of each element by the smaller number of moles. Here, \(n_C\approx1.15\space mol\) and \(n_H = 1.16\space mol\).
\(\frac{n_C}{n_C}\approx1\), \(\frac{n_H}{n_C}\approx1\) (since \(1.16\div1.15\approx1\)).

Answer:

\(CH\)