Question

On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO₂ and 0.567 g of H₂O. The empirical formula mass of compound (X) is:
(Given molar mass in g mol\(^{-1}\): C: 12, H: 1, O: 16)

• A. 30
• B. 45
• C. 60
• D. 15

Hint:

When calculating the empirical formula, first determine the moles of each element from the given combustion data. Then, divide each element’s mass by its atomic mass to obtain the ratio of atoms.

Answer 1

The Correct Option is A

This problem requires the determination of the empirical formula mass for an organic compound, designated as X. The calculation is based on the mass of the combusted compound and the resultant masses of carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)).

Concept Used:

The methodology employed is combustion analysis, which operates on the following principles:

1. All carbon atoms within the organic compound are oxidized to \( \text{CO}_2 \).
2. All hydrogen atoms within the organic compound are oxidized to \( \text{H}_2\text{O} \).
3. Any oxygen present in the compound is quantified by subtracting the combined masses of carbon and hydrogen from the total mass of the organic compound.
4. The empirical formula represents the most simplified whole-number ratio of elements within the compound. This ratio is derived by converting the mass of each element into moles and subsequently determining their minimal whole-number proportion.

Step-by-Step Solution:

Step 1: Quantify the mass of carbon (C) from the mass of \( \text{CO}_2 \) generated.

The molar mass of \( \text{CO}_2 \) is \( 12 + 2(16) = 44 \, \text{g/mol} \). One mole of \( \text{CO}_2 \) contains one mole of C atoms, contributing 12 g of carbon.

\[ \text{Mass of C} = \left( \frac{\text{Molar mass of C}}{\text{Molar mass of CO}_2} \right) \times \text{Mass of CO}_2 \] \[ \text{Mass of C} = \left( \frac{12}{44} \right) \times 1.46 \, \text{g} = 0.39818 \, \text{g} \approx 0.398 \, \text{g} \]

Step 2: Determine the mass of hydrogen (H) from the mass of \( \text{H}_2\text{O} \) produced.

The molar mass of \( \text{H}_2\text{O} \) is \( 2(1) + 16 = 18 \, \text{g/mol} \). One mole of \( \text{H}_2\text{O} \) contains two moles of H atoms, equivalent to 2 g of hydrogen.

\[ \text{Mass of H} = \left( \frac{\text{Molar mass of 2H}}{\text{Molar mass of H}_2\text{O}} \right) \times \text{Mass of H}_2\text{O} \] \[ \text{Mass of H} = \left( \frac{2}{18} \right) \times 0.567 \, \text{g} = 0.063 \, \text{g} \]

Step 3: Calculate the mass of oxygen (O) by difference.

The total mass of the compound is the sum of the masses of C, H, and O.

\[ \text{Mass of O} = \text{Mass of compound} - (\text{Mass of C} + \text{Mass of H}) \] \[ \text{Mass of O} = 1.0 \, \text{g} - (0.398 \, \text{g} + 0.063 \, \text{g}) = 1.0 \, \text{g} - 0.461 \, \text{g} = 0.539 \, \text{g} \]

Step 4: Convert the mass of each element to moles.

\[ \text{Moles of C} = \frac{\text{Mass of C}}{\text{Molar mass of C}} = \frac{0.398}{12} \approx 0.0332 \, \text{mol} \] \[ \text{Moles of H} = \frac{\text{Mass of H}}{\text{Molar mass of H}} = \frac{0.063}{1} = 0.063 \, \text{mol} \] \[ \text{Moles of O} = \frac{\text{Mass of O}}{\text{Molar mass of O}} = \frac{0.539}{16} \approx 0.0337 \, \text{mol} \]

Step 5: Ascertain the simplest whole-number ratio of the moles.

To determine the simplest ratio, divide the mole quantity of each element by the smallest mole value calculated in Step 4 (approximately 0.0332).

• For C: \( \frac{0.0332}{0.0332} = 1 \)
• For H: \( \frac{0.063}{0.0332} \approx 1.89 \approx 2 \)
• For O: \( \frac{0.0337}{0.0332} \approx 1.01 \approx 1 \)

The simplest whole-number ratio of C : H : O is 1 : 2 : 1. Consequently, the empirical formula for compound (X) is \( \text{CH}_2\text{O} \).

Step 6: Compute the empirical formula mass.

The mass of the empirical formula \( \text{CH}_2\text{O} \) is calculated by summing the atomic masses of its constituent atoms.

\[ \text{Empirical Formula Mass} = (1 \times \text{Molar mass of C}) + (2 \times \text{Molar mass of H}) + (1 \times \text{Molar mass of O}) \] \[ = (1 \times 12) + (2 \times 1) + (1 \times 16) = 12 + 2 + 16 = 30 \, \text{g/mol} \]

The empirical formula mass of compound (X) is determined to be 30 g mol\(^{-1}\).