Question

The elemental composition of a compound is 54.2%C, 9.2%H, and 36.6%O. If the molar mass of the compound is 132 g/mol, the molecular formula of the compound is:

• A. \( \text{C}_6 \text{H}_{12} \text{O}_6 \)
• B. \( \text{C}_6 \text{H}_{12} \text{O}_3 \)
• C. \( \text{C}_4 \text{H}_9 \text{O}_3 \)
• D. \( \text{C}_4 \text{H}_8 \text{O}_2 \)

Hint:

The empirical formula can help determine the molecular formula by using the molar mass.

Answer 1

The Correct Option is A

The molecular formula of a compound with a molar mass of 132 g/mol and an elemental composition of 54.2% C, 9.2% H, and 36.6% O is determined as follows:

Step 1: Determine the Empirical Formula

Assuming a 100 g sample, the masses of each element are:

• C: 54.2 g
• H: 9.2 g
• O: 36.6 g

Step 2: Convert Mass to Moles

Calculate the moles of each element using their atomic masses:

• For Carbon (C): \[ \frac{54.2 \, \text{g}}{12.01 \, \text{g/mol}} = 4.51 \, \text{mol} \]
• For Hydrogen (H): \[ \frac{9.2 \, \text{g}}{1.008 \, \text{g/mol}} = 9.13 \, \text{mol} \]
• For Oxygen (O): \[ \frac{36.6 \, \text{g}}{16.00 \, \text{g/mol}} = 2.29 \, \text{mol} \]

Step 3: Determine the Simplest Mole Ratio

Divide each mole value by the smallest mole value (2.29 mol for Oxygen):

• C: \[ \frac{4.51}{2.29} = 1.97 \approx 2 \]
• H: \[ \frac{9.13}{2.29} = 3.98 \approx 4 \]
• O: \[ \frac{2.29}{2.29} = 1 \]

Step 4: State the Empirical Formula

The empirical formula, based on the mole ratio, is CH₂O.

CH₂O

Step 5: Calculate the Molecular Formula

The molar mass of the empirical formula (CH₂O) is:

12.01 g/mol (C) + 2(1.008 g/mol) (H) + 16.00 g/mol (O) = 30.026 g/mol

Compare the given molecular molar mass to the empirical formula mass:

\[ \frac{132 \, \text{g/mol}}{30.026 \, \text{g/mol}} \approx 4.4 \approx 4 \]

Multiply the empirical formula by this factor (4) to obtain the molecular formula:

The molecular formula is C₆H₁₂O₆.