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

The problem asks for the empirical formula mass of an organic compound (X) based on the mass of the compound combusted and the masses of carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)) produced.

Concept Used:

The solution uses the principle of combustion analysis to determine the empirical formula. In combustion analysis:

1. All the carbon present in the organic compound is converted into \( \text{CO}_2 \).
2. All the hydrogen present in the organic compound is converted into \( \text{H}_2\text{O} \).
3. If the compound contains oxygen, its mass is found by subtracting the masses of carbon and hydrogen from the total mass of the organic compound.
4. The empirical formula is the simplest whole-number ratio of atoms of the elements present in the compound. This ratio is found by converting the mass of each element to moles and then finding their simplest ratio.

Step-by-Step Solution:

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

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

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

Step 2: Calculate 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 \( 2(1) + 16 = 18 \, \text{g/mol} \). In one mole of \( \text{H}_2\text{O} \), there are two moles of H atoms (2 g).

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

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 into 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: Determine the simplest whole-number ratio of the moles.

To find the simplest ratio, divide the number of moles of each element by the smallest value obtained in Step 4 (which is 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. Therefore, the empirical formula of the compound (X) is \( \text{CH}_2\text{O} \).

Step 6: Calculate the empirical formula mass.

The mass of the empirical formula \( \text{CH}_2\text{O} \) is the sum of the atomic masses of the atoms in it.

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

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

Answer 2

Moles of 'C' = \( n_{\text{CO}_2} = \frac{1.46}{44} = 0.033 \) 
Moles of 'C' = \( W_c = 0.033 \times 12 = 0.396 \) 
Moles of 'H' = \( 2 \times n_{\text{H}_2\text{O}} = 2 \times \frac{0.567}{18} = 0.063 \) 
Mass of 'H' = \( 0.0063 \) Mass of Oxygen \( O \) = \( 1 - (W_c + W_h) = 1 - (0.033 \times 12 + 0.063 \times 1) = 0.541 \) 
Moles of 'O' = \( \frac{0.541}{16} = 0.033 \) 
Empirical formula = \( \text{CH}_2\text{O} \) 
Empirical formula mass = 30