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:
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- \( \text{C}_6 \text{H}_{12} \text{O}_6 \)
- \( \text{C}_6 \text{H}_{12} \text{O}_3 \)
- \( \text{C}_4 \text{H}_9 \text{O}_3 \)
- \( \text{C}_4 \text{H}_8 \text{O}_2 \)
The Correct Option is A
Approach Solution - 1
Step 1: Calculate the Empirical Formula
Given the percentage composition, we can assume we have 100 g of the compound. This makes the mass of each element in the compound directly equal to the percentage values:
- C: 54.2 g
- H: 9.2 g
- O: 36.6 g
Step 2: Calculate the Moles of Each Element
To find the moles of each element, divide the mass of each element by its atomic mass:
- 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: Find the Mole Ratio
Next, divide the moles of each element by the smallest number of moles (which is 2.29 in this case, corresponding to 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: Write the Empirical Formula
The mole ratio of the elements is approximately C2H4O1, so the empirical formula is:
CH2O
Step 5: Find the Molecular Formula
The empirical formula mass is:
CH2O: 12 + 2 + 16 = 30 g/mol
The molecular formula mass is given as 132 g/mol. To find the ratio of the molecular mass to the empirical formula mass, divide the molar mass by the empirical formula mass:
\[ \frac{132 \, \text{g/mol}}{30 \, \text{g/mol}} = 4.4 \approx 4 \]
So, multiply the empirical formula by 4 to get the molecular formula:
The molecular formula is C6H12O6.
Approach Solution -2
The given elemental composition of the compound is:
- 54.2% C
- 9.2% H
- 36.6% O
To determine the empirical formula, we first assume 100 g of the compound. This gives:
- 54.2 g of C
- 9.2 g of H
- 36.6 g of O
Now, we calculate the moles of each element:
- Moles of C = \( \frac{54.2 \, \text{g}}{12 \, \text{g/mol}} = 4.5167 \, \text{mol} \)
- Moles of H = \( \frac{9.2 \, \text{g}}{1 \, \text{g/mol}} = 9.2 \, \text{mol} \)
- Moles of O = \( \frac{36.6 \, \text{g}}{16 \, \text{g/mol}} = 2.2875 \, \text{mol} \)
Next, divide the moles of each element by the smallest number of moles (2.2875):
- Moles of C = \( \frac{4.5167}{2.2875} \approx 1.97 \)
- Moles of H = \( \frac{9.2}{2.2875} \approx 4.02 \)
- Moles of O = \( \frac{2.2875}{2.2875} = 1 \)
Thus, the empirical formula is approximately \( \text{C}_2 \text{H}_4 \text{O} \).
Step 2: Calculate the molecular formula.
Now, we use the molar mass of the compound (132 g/mol) to find the molecular formula. The molar mass of the empirical formula \( \text{C}_2 \text{H}_4 \text{O} \) is:
\[ \text{Molar mass of empirical formula} = (2 \times 12) + (4 \times 1) + (1 \times 16) = 24 + 4 + 16 = 44 \, \text{g/mol}. \]
The ratio of the molar mass of the compound to the empirical formula mass is:
\[ \frac{132 \, \text{g/mol}}{44 \, \text{g/mol}} = 3. \]
Thus, the molecular formula is 3 times the empirical formula:
\[ \text{Molecular formula} = 3 \times \text{C}_2 \text{H}_4 \text{O} = \text{C}_6 \text{H}_{12} \text{O}_6. \]
Final Answer:
\[ \boxed{\text{C}_6 \text{H}_{12} \text{O}_6}. \]
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