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To determine the molecular formula of the compound, we need to follow several steps: calculating the moles of each element, finding their mole ratios, simplifying the ratios to find the empirical formula, and comparing the empirical formula with the given molecular mass to deduce the molecular formula.
### Step 1: Calculating Moles of Each Element
1. Calculate the moles of Carbon (C):
[tex]\[ \text{Moles of C} = \frac{\text{Mass of C}}{\text{Atomic mass of C}} = \frac{283.4 \text{ g}}{12.01 \text{ g/mol}} = 23.597 \text{ moles} \][/tex]
2. Calculate the moles of Hydrogen (H):
[tex]\[ \text{Moles of H} = \frac{\text{Mass of H}}{\text{Atomic mass of H}} = \frac{31.7 \text{ g}}{1.008 \text{ g/mol}} = 31.448 \text{ moles} \][/tex]
3. Calculate the moles of Oxygen (O):
[tex]\[ \text{Moles of O} = \frac{\text{Mass of O}}{\text{Atomic mass of O}} = \frac{377.4 \text{ g}}{16.00 \text{ g/mol}} = 23.588 \text{ moles} \][/tex]
### Step 2: Finding the Mole Ratios
Next, we determine the smallest number of moles among the elements:
[tex]\[ \text{Smallest moles} = 23.588 \text{ (moles of O)} \][/tex]
Then, we divide the moles of each element by the smallest number of moles to get the ratios:
1. Ratio of C:
[tex]\[ \text{Ratio of C} = \frac{23.597}{23.588} \approx 1.000 \][/tex]
2. Ratio of H:
[tex]\[ \text{Ratio of H} = \frac{31.448}{23.588} \approx 1.333 \][/tex]
3. Ratio of O:
[tex]\[ \text{Ratio of O} = \frac{23.588}{23.588} = 1 \][/tex]
So, the empirical formula has the approximate ratios of:
[tex]\[ \text{C:H:O} \approx 1.000:1.333:1 \][/tex]
To convert these ratios to whole numbers, we recognize that 1.333 is very close to [tex]\( \frac{4}{3} \)[/tex], implying the ratios should be proportional to whole numbers:
[tex]\[ \text{C:H:O} \approx 3:4:3 \][/tex]
Thus, the empirical formula is:
[tex]\[ C_3H_4O_3 \][/tex]
### Step 3: Determining the Molecular Formula
We now compare the empirical formula mass to the given molar mass:
1. Empirical formula mass of [tex]\( C_3H_4O_3 \)[/tex]:
[tex]\[ (3 \times 12.01) + (4 \times 1.008) + (3 \times 16.00) = 36.03 + 4.032 + 48.00 = 88.062 \text{ g/mol} \][/tex]
2. Given molar mass of compound: 176.124 g/mol
We find the multiple (n) by dividing the molar mass by the empirical formula mass:
[tex]\[ n = \frac{\text{molar mass}}{\text{empirical formula mass}} = \frac{176.124}{88.062} \approx 2 \][/tex]
Thus, the molecular formula is twice the empirical formula:
[tex]\[ \text{Molecular formula} = (C_3H_4O_3)_2 = C_6H_8O_6 \][/tex]
### Final Answer
The molecular formula of the compound is:
[tex]\[ \boxed{C_6H_8O_6} \][/tex]
Therefore, the correct option is:
D. [tex]\( C_6H_8O_6 \)[/tex]
### Step 1: Calculating Moles of Each Element
1. Calculate the moles of Carbon (C):
[tex]\[ \text{Moles of C} = \frac{\text{Mass of C}}{\text{Atomic mass of C}} = \frac{283.4 \text{ g}}{12.01 \text{ g/mol}} = 23.597 \text{ moles} \][/tex]
2. Calculate the moles of Hydrogen (H):
[tex]\[ \text{Moles of H} = \frac{\text{Mass of H}}{\text{Atomic mass of H}} = \frac{31.7 \text{ g}}{1.008 \text{ g/mol}} = 31.448 \text{ moles} \][/tex]
3. Calculate the moles of Oxygen (O):
[tex]\[ \text{Moles of O} = \frac{\text{Mass of O}}{\text{Atomic mass of O}} = \frac{377.4 \text{ g}}{16.00 \text{ g/mol}} = 23.588 \text{ moles} \][/tex]
### Step 2: Finding the Mole Ratios
Next, we determine the smallest number of moles among the elements:
[tex]\[ \text{Smallest moles} = 23.588 \text{ (moles of O)} \][/tex]
Then, we divide the moles of each element by the smallest number of moles to get the ratios:
1. Ratio of C:
[tex]\[ \text{Ratio of C} = \frac{23.597}{23.588} \approx 1.000 \][/tex]
2. Ratio of H:
[tex]\[ \text{Ratio of H} = \frac{31.448}{23.588} \approx 1.333 \][/tex]
3. Ratio of O:
[tex]\[ \text{Ratio of O} = \frac{23.588}{23.588} = 1 \][/tex]
So, the empirical formula has the approximate ratios of:
[tex]\[ \text{C:H:O} \approx 1.000:1.333:1 \][/tex]
To convert these ratios to whole numbers, we recognize that 1.333 is very close to [tex]\( \frac{4}{3} \)[/tex], implying the ratios should be proportional to whole numbers:
[tex]\[ \text{C:H:O} \approx 3:4:3 \][/tex]
Thus, the empirical formula is:
[tex]\[ C_3H_4O_3 \][/tex]
### Step 3: Determining the Molecular Formula
We now compare the empirical formula mass to the given molar mass:
1. Empirical formula mass of [tex]\( C_3H_4O_3 \)[/tex]:
[tex]\[ (3 \times 12.01) + (4 \times 1.008) + (3 \times 16.00) = 36.03 + 4.032 + 48.00 = 88.062 \text{ g/mol} \][/tex]
2. Given molar mass of compound: 176.124 g/mol
We find the multiple (n) by dividing the molar mass by the empirical formula mass:
[tex]\[ n = \frac{\text{molar mass}}{\text{empirical formula mass}} = \frac{176.124}{88.062} \approx 2 \][/tex]
Thus, the molecular formula is twice the empirical formula:
[tex]\[ \text{Molecular formula} = (C_3H_4O_3)_2 = C_6H_8O_6 \][/tex]
### Final Answer
The molecular formula of the compound is:
[tex]\[ \boxed{C_6H_8O_6} \][/tex]
Therefore, the correct option is:
D. [tex]\( C_6H_8O_6 \)[/tex]
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