Answer :
To determine the molecular formula of the compound given its empirical formula and molecular weight, we need to follow a series of steps:
1. Calculate the empirical formula weight: This is the sum of the atomic weights of all atoms in the empirical formula [tex]\(C_3H_4O\)[/tex].
- Carbon (C) has an atomic weight of 12.011.
- Hydrogen (H) has an atomic weight of 1.008.
- Oxygen (O) has an atomic weight of 16.00.
Using these atomic weights, the empirical formula weight can be calculated as follows:
[tex]\[ 3 \times 12.011 + 4 \times 1.008 + 16.00 \][/tex]
[tex]\[ = 36.033 + 4.032 + 16.00 \][/tex]
[tex]\[ = 56.065 \text{ amu} \][/tex]
2. Determine the ratio of the molecular weight to the empirical formula weight:
The given molecular weight is 112.124 amu. We divide the molecular weight by the empirical formula weight:
[tex]\[ \text{ratio} = \frac{112.124}{56.065} \][/tex]
[tex]\[ \approx 2 \][/tex]
3. Calculate the molecular formula:
The molecular formula is [tex]\(n\)[/tex] times the empirical formula, where [tex]\(n\)[/tex] is the ratio we just calculated (rounded to the nearest integer, which is 2). Therefore, we multiply each subscript in the empirical formula [tex]\(C_3H_4O\)[/tex] by 2:
[tex]\[ \begin{aligned} C &: 3 \times 2 = 6 \\ H &: 4 \times 2 = 8 \\ O &: 1 \times 2 = 2 \end{aligned} \][/tex]
Thus, the molecular formula is [tex]\(C_6H_8O_2\)[/tex].
4. Verify among given options:
We need to find which option matches this molecular formula [tex]\(C_6H_8O_2\)[/tex]:
- Option A: [tex]\(C_6H_8O\)[/tex]
- Option B: [tex]\(C_9H_{12}O_3\)[/tex]
- Option C: [tex]\(C_8H_4O_2\)[/tex]
- Option D: [tex]\(C_4H_8O_2\)[/tex]
- Option E: [tex]\(C_6H_8O_2\)[/tex]
The correct match is Option E: [tex]\(C_6H_8O_2\)[/tex].
Therefore, the molecular formula of the compound is [tex]\(C_6H_8O_2\)[/tex], and the correct answer is:
Option E: [tex]\(C_6H_8O_2\)[/tex]
1. Calculate the empirical formula weight: This is the sum of the atomic weights of all atoms in the empirical formula [tex]\(C_3H_4O\)[/tex].
- Carbon (C) has an atomic weight of 12.011.
- Hydrogen (H) has an atomic weight of 1.008.
- Oxygen (O) has an atomic weight of 16.00.
Using these atomic weights, the empirical formula weight can be calculated as follows:
[tex]\[ 3 \times 12.011 + 4 \times 1.008 + 16.00 \][/tex]
[tex]\[ = 36.033 + 4.032 + 16.00 \][/tex]
[tex]\[ = 56.065 \text{ amu} \][/tex]
2. Determine the ratio of the molecular weight to the empirical formula weight:
The given molecular weight is 112.124 amu. We divide the molecular weight by the empirical formula weight:
[tex]\[ \text{ratio} = \frac{112.124}{56.065} \][/tex]
[tex]\[ \approx 2 \][/tex]
3. Calculate the molecular formula:
The molecular formula is [tex]\(n\)[/tex] times the empirical formula, where [tex]\(n\)[/tex] is the ratio we just calculated (rounded to the nearest integer, which is 2). Therefore, we multiply each subscript in the empirical formula [tex]\(C_3H_4O\)[/tex] by 2:
[tex]\[ \begin{aligned} C &: 3 \times 2 = 6 \\ H &: 4 \times 2 = 8 \\ O &: 1 \times 2 = 2 \end{aligned} \][/tex]
Thus, the molecular formula is [tex]\(C_6H_8O_2\)[/tex].
4. Verify among given options:
We need to find which option matches this molecular formula [tex]\(C_6H_8O_2\)[/tex]:
- Option A: [tex]\(C_6H_8O\)[/tex]
- Option B: [tex]\(C_9H_{12}O_3\)[/tex]
- Option C: [tex]\(C_8H_4O_2\)[/tex]
- Option D: [tex]\(C_4H_8O_2\)[/tex]
- Option E: [tex]\(C_6H_8O_2\)[/tex]
The correct match is Option E: [tex]\(C_6H_8O_2\)[/tex].
Therefore, the molecular formula of the compound is [tex]\(C_6H_8O_2\)[/tex], and the correct answer is:
Option E: [tex]\(C_6H_8O_2\)[/tex]