To determine the molecular formula for a sample with the given masses of carbon (C), hydrogen (H), and oxygen (O), follow these steps:
### Step 1: Calculate the number of moles of each element in the sample
1. Molar masses:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Oxygen (O): 16.00 g/mol
2. Masses of elements in the sample:
- Mass of carbon (C): 1.388 g
- Mass of hydrogen (H): 0.345 g
- Mass of oxygen (O): 1.850 g
3. Calculating moles:
For each element, we use the formula:
[tex]\[
\text{moles} = \frac{\text{mass}}{\text{molar mass}}
\][/tex]
- Moles of carbon (C):
[tex]\[
\text{moles\_C} = \frac{1.388 \, \text{g}}{12.01 \, \text{g/mol}} \approx 0.11557 \, \text{mol}
\][/tex]
- Moles of hydrogen (H):
[tex]\[
\text{moles\_H} = \frac{0.345 \, \text{g}}{1.01 \, \text{g/mol}} \approx 0.34158 \, \text{mol}
\][/tex]
- Moles of oxygen (O):
[tex]\[
\text{moles\_O} = \frac{1.850 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.11563 \, \text{mol}
\][/tex]
### Step 2: Determine the mole ratio of the elements
1. Identify the smallest number of moles among C, H, and O:
[tex]\[
\text{smallest\_mole} = \min(0.11557, 0.34158, 0.11563) = 0.11557 \, \text{mol}
\][/tex]
2. Calculate the mole ratios by dividing each element’s moles by the smallest number of moles:
- Ratio of C:
[tex]\[
\text{ratio\_C} = \frac{0.11557}{0.11557} \approx 1.0
\][/tex]
- Ratio of H:
[tex]\[
\text{ratio\_H} = \frac{0.34158}{0.11557} \approx 2.96
\][/tex]
- Ratio of O:
[tex]\[
\text{ratio\_O} = \frac{0.11563}{0.11557} \approx 1.00
\][/tex]
### Step 3: Approximate the ratios to nearest whole numbers
The ratios should ideally be in whole numbers for the empirical formula:
- C: [tex]\( \approx 1 \)[/tex]
- H: [tex]\( \approx 3 \)[/tex]
- O: [tex]\( \approx 1 \)[/tex]
Thus, the empirical formula is [tex]\( \text{C}_1\text{H}_3\text{O}_1 \)[/tex] or simply [tex]\( \text{CH}_3\text{O} \)[/tex].
### Step 4: Calculate the molecular formula
1. Calculate the empirical formula’s molar mass:
[tex]\[
\text{Empirical mass} = (1 \times 12.01) + (3 \times 1.01) + (1 \times 16.00) = 12.01 + 3.03 + 16.00 = 31.04 \, \text{g/mol}
\][/tex]
2. Use the given molar mass of the sample to find the multiplier:
[tex]\[
\text{Multiplier} = \frac{62.00 \, \text{g/mol}}{31.04 \, \text{g/mol}} \approx 2.0
\][/tex]
3. Apply the multiplier to the empirical formula to get the molecular formula:
- For C: [tex]\(1 \times 2 = 2\)[/tex]
- For H: [tex]\(3 \times 2 = 6\)[/tex]
- For O: [tex]\(1 \times 2 = 2\)[/tex]
Therefore, the molecular formula is [tex]\( \text{C}_2\text{H}_6\text{O}_2 \)[/tex].
### Conclusion
The molecular formula of the sample [tex]\( P \)[/tex] is [tex]\( \text{C}_2\text{H}_6\text{O}_2 \)[/tex].
