Question

A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 \(\mathrm{mg}\) of the compound yields 16.01 \(\mathrm{mg}\) \(\mathrm{CO}_{2}\) and 4.37 \(\mathrm{mg} \mathrm{H}_{2} \mathrm{O}\) . The molar mass of the compound is 176.1 \(\mathrm{g} / \mathrm{mol} .\) What are the empirical and molecular formulas of the compound?

Short answer

The empirical formula of the compound is \(\mathrm{CH}_{2}\mathrm{O}\), and the molecular formula is \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\).

Step-by-step solution

Determine the moles of CO2 and H2O produced

Use the masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) produced to calculate the moles of each. Molar mass of \(\mathrm{CO}_{2} = 12.01 + (2 \times 16.00) = 44.01 \: \mathrm{g}/\mathrm{mol}\) Molar mass of \(\mathrm{H}_{2}\mathrm{O} = (2 \times 1.01) + 16.00 = 18.02 \: \mathrm{g}/\mathrm{mol}\) Moles of \(\mathrm{CO}_{2} = \frac{16.01 \: \mathrm{mg}}{44.01 \: \mathrm{g/mol}} = 0.3635 \: \mathrm{mol}\) Moles of \(\mathrm{H}_{2}\mathrm{O} = \frac{4.37 \: \mathrm{mg}}{18.02 \: \mathrm{g/mol}} = 0.2426 \: \mathrm{mol}\)

Determine the moles of C and H in the compound

Use the amount of moles of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) to find the moles of carbon (C) and hydrogen (H) in the compound. Moles of C: 1 mole of \(\mathrm{CO}_{2}\) has 1 mole of C, so moles of C = 0.3635 \(\mathrm{mol}\) Moles of H: 1 mole of \(\mathrm{H}_{2}\mathrm{O}\) has 2 moles of H, so moles of H = 0.2426 \(\mathrm{mol} \times 2 = 0.4852 \: \mathrm{mol}\)

Calculate the mass of C and H in the compound

Use the moles of C and H found previously to calculate their respective masses. Mass of C = 0.3635 \(\mathrm{mol} \times 12.01 \: \mathrm{g}/\mathrm{mol} = 4.364 \: \mathrm{mg}\) Mass of H = 0.4852 \(\mathrm{mol} \times 1.01 \: \mathrm{g}/\mathrm{mol} = 0.490 \: \mathrm{mg}\)

Calculate the mass of O in the compound

Use the initial mass of the compound and subtract the masses of C and H to find the mass of oxygen (O). Mass of O = Total mass - Mass of C - Mass of H = 10.68 \: \mathrm{mg} - 4.364 \: \mathrm{mg} - 0.490 \: \mathrm{mg} = 5.826 \: \mathrm{mg}$

Calculate the moles of O in the compound

Use the mass of O to determine the moles of O in the compound. Moles of O = \(\frac{5.826 \: \mathrm{mg}}{16.00 \: \mathrm{g}/\mathrm{mol}} = 0.3641 \: \mathrm{mol}\)

Find the empirical formula

Determine the ratio of moles of C, H, and O in the compound by dividing each by the smallest mole value and rounding to the nearest whole number. Mole ratio = \(\frac{0.3635 \: \mathrm{mol} \: C}{0.3635 \: \mathrm{mol}} : \frac{0.4852 \: \mathrm{mol} \: H}{0.3635 \: \mathrm{mol}} : \frac{0.3641 \: \mathrm{mol} \: O}{0.3635 \: \mathrm{mol}} = 1 : 1.33 : 1\) We round this to the nearest whole number ratio, which gives: Empirical Formula = \(\mathrm{CH}_{2}\mathrm{O}\)

Calculate the molecular formula

Use the molar mass of the empirical formula and the given molar mass of the compound to determine the molecular formula. Molar mass of empirical formula, \(\mathrm{CH}_{2}\mathrm{O} = 12.01 + (2 \times 1.01) + 16.00 = 30.03 \: \mathrm{g}/\mathrm{mol}\) N = \(\frac{Molar \: mass \: of \: molecular \: formula}{Molar \: mass \: of \: empirical \: formula} = \frac{176.1 \: \mathrm{g}/\mathrm{mol}}{30.03 \: \mathrm{g}/\mathrm{mol}} = 5.86\) We round this to the nearest whole number, which is 6. Multiply the empirical formula by 6 to get the molecular formula: \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\) The empirical formula of the compound is \(\mathrm{CH}_{2}\mathrm{O}\), and the molecular formula is \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\).

Key concepts

Combustion Analysis

Combustion analysis is a fundamental technique used in chemistry to determine the elemental composition of an organic compound. This method involves burning a known mass of the compound in excess oxygen, producing carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)) as combustion products. By measuring the amounts of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) produced, chemists can back-calculate to determine the amounts of carbon (C) and hydrogen (H) present in the original compound.

• Step-by-Step: Start with the measured mass of the sample combusted.
• Determine the mass of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) produced.
• Convert these masses into moles knowing their respective molar masses.
• Each mole of \(\mathrm{CO}_{2}\) corresponds to one mole of carbon and each mole of \(\mathrm{H}_{2}\mathrm{O}\) corresponds to two moles of hydrogen.

The process of combustion analysis allows for the precise calculation of carbon and hydrogen. For compounds only containing carbon, hydrogen, and oxygen, the amount of oxygen can be inferred based on the total sample mass minus the mass attributed to carbon and hydrogen.

Molar Mass Calculation

Calculating molar mass is a key component in determining the quantitative relationships being examined. It involves summing the atomic masses of the elements present in a molecule based on its formula. Knowing the molar mass of a compound is crucial when converting between grams and moles, an essential step in solving many chemical problems.

• Molar Mass of \(\mathrm{CO}_{2}\): Carbon (C) has an atomic mass of 12.01 g/mol and oxygen (O) has an atomic mass of 16.00 g/mol. Thus, \(\mathrm{CO}_{2}\) = 12.01 + (2 x 16.00) = 44.01 g/mol.
• Molar Mass of \(\mathrm{H}_{2}\mathrm{O}\): Hydrogen (H) has an atomic mass of 1.01 g/mol, and oxygen (O) is 16.00 g/mol. Therefore, \(\mathrm{H}_{2}\mathrm{O}\) = (2 x 1.01) + 16.00 = 18.02 g/mol.

With the molar masses known, you can calculate the amounts in moles of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) to find the number of moles of C and H in the original compound. This understanding is a cornerstone in the calculation of empirical and molecular formulas, providing a base measurement unit for the reactions.

Molecular Formula Determination

Once the empirical formula is obtained, which represents the simplest whole-number ratio of atoms in a compound, the molecular formula can be determined. The molecular formula gives the actual number of each type of atom in a molecule of the compound. It is found by comparing the empirical formula mass to the known molar mass of the compound.

• Empirical Formula: Start by deriving the simplest formula based on the mole ratios of the elements involved. For the compound in question, with a mole ratio of 1:1.33:1, the empirical formula is derived as \(\mathrm{CH}_{2}\mathrm{O}\).
• Comparison with Molar Mass: Calculate the mass of this empirical formula and compare it to the given molar mass of the compound. For example, \(\mathrm{CH}_{2}\mathrm{O}\) has a molar mass of 30.03 g/mol.
• Finding the Molecular Formula: Determine how many times the empirical formula mass fits into the molar mass of the compound. This number will tell you how many "units" of the empirical formula are in the molecular formula.

In our exercise, this calculation revealed that the molar mass is 5.86 times that of the empirical formula. Since only whole numbers make sense chemically, we round to 6. This allows for multiplication of the subscripts in \(\mathrm{CH}_{2}\mathrm{O}\) by 6 to yield the molecular formula, \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\). Understanding these steps is crucial for interpreting the full complexity and structure of a compound in chemical analysis.