Questions: A 9.112 gram sample of an organic compound containing C, H, and O is analyzed by combustion analysis and 11.56 grams of CO2 and 3.156 grams of H2O are produced. In a separate experiment, the molar mass is found to be 104.1 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 =

Solution Steps

First, calculate the moles of carbon in the compound. The mass of CO\(_2\) produced is 11.56 grams. The molar mass of CO\(_2\) is 44.01 g/mol.

\[ \text{Moles of } \text{CO}_2 = \frac{11.56 \, \text{g}}{44.01 \, \text{g/mol}} = 0.2626 \, \text{mol} \]

Since each mole of CO\(_2\) contains one mole of carbon, the moles of carbon are also 0.2626 mol.

Next, calculate the moles of hydrogen. The mass of H\(_2\)O produced is 3.156 grams. The molar mass of H\(_2\)O is 18.02 g/mol.

\[ \text{Moles of } \text{H}_2\text{O} = \frac{3.156 \, \text{g}}{18.02 \, \text{g/mol}} = 0.1752 \, \text{mol} \]

Each mole of H\(_2\)O contains two moles of hydrogen, so the moles of hydrogen are:

\[ 0.1752 \, \text{mol} \times 2 = 0.3504 \, \text{mol} \]

The total mass of the compound is 9.112 grams. The mass of carbon and hydrogen in the compound can be calculated as follows:

• Mass of carbon: \(0.2626 \, \text{mol} \times 12.01 \, \text{g/mol} = 3.153 \, \text{g}\)
• Mass of hydrogen: \(0.3504 \, \text{mol} \times 1.008 \, \text{g/mol} = 0.3532 \, \text{g}\)

The mass of oxygen in the compound is:

\[ 9.112 \, \text{g} - (3.153 \, \text{g} + 0.3532 \, \text{g}) = 5.6058 \, \text{g} \]

The molar mass of oxygen is 16.00 g/mol, so the moles of oxygen are:

\[ \frac{5.6058 \, \text{g}}{16.00 \, \text{g/mol}} = 0.3504 \, \text{mol} \]

The mole ratio of C:H:O is 0.2626:0.3504:0.3504. To find the simplest whole number ratio, divide each by the smallest number of moles, 0.2626:

• C: \(\frac{0.2626}{0.2626} = 1\)
• H: \(\frac{0.3504}{0.2626} \approx 1.333\)
• O: \(\frac{0.3504}{0.2626} \approx 1.333\)

To convert 1.333 to a whole number, multiply all ratios by 3:

• C: \(1 \times 3 = 3\)
• H: \(1.333 \times 3 = 4\)
• O: \(1.333 \times 3 = 4\)

Thus, the empirical formula is C\(_3\)H\(_4\)O\(_4\).

The empirical formula mass of C\(_3\)H\(_4\)O\(_4\) is:

\[ 3(12.01) + 4(1.008) + 4(16.00) = 104.1 \, \text{g/mol} \]

Since the empirical formula mass is equal to the given molar mass, the molecular formula is the same as the empirical formula: C\(_3\)H\(_4\)O\(_4\).

Final Answer