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A sample of a hydrocarbon contains 2075 g C and 4.25 g H. Its molar mass is 58.04g/mol What is its empirical formula C_(2)H_(5) is CH_(2) is CH C_(5)H COMPLETE What is its molecular formula? Type in the correct number based on the label C_(A)H_(B) A: square B: square DONE V

Question

A sample of a hydrocarbon contains 2075 g C
and 4.25 g H. Its molar mass is 58.04g/mol What
is its empirical formula
C_(2)H_(5)
is CH_(2)
is CH
C_(5)H
COMPLETE
What is its molecular formula? Type in the correct
number based on the label
C_(A)H_(B)
A: square 
B:
square 
DONE V

A sample of a hydrocarbon contains 2075 g C and 4.25 g H. Its molar mass is 58.04g/mol What is its empirical formula C_(2)H_(5) is CH_(2) is CH C_(5)H COMPLETE What is its molecular formula? Type in the correct number based on the label C_(A)H_(B) A: square B: square DONE V

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OllieProfessional · Tutor for 6 years

Answer

The empirical formula is \(C_{41}H\).<br /><br />The molecular formula does not fit the provided options as it results in fractional values, suggesting a need to re-evaluate the initial data or assumptions.

Explain

## Step 1: <br />Determine the moles of carbon and hydrogen in the sample.<br />### **\(\text{moles of C} = \frac{\text{mass of C}}{\text{atomic mass of C}} = \frac{2075 \text{ g}}{12.01 \text{ g/mol}} \approx 172.68 \text{ mol}\)**<br />### **\(\text{moles of H} = \frac{\text{mass of H}}{\text{atomic mass of H}} = \frac{4.25 \text{ g}}{1.008 \text{ g/mol}} \approx 4.21 \text{ mol}\)**<br /><br />## Step 2: <br />Determine the simplest ratio of moles of C to H.<br />### **\(\text{ratio of C} = \frac{\text{moles of C}}{\min(\text{moles of C}, \text{moles of H})} = \frac{172.68}{4.21} \approx 41\)**<br />### **\(\text{ratio of H} = \frac{\text{moles of H}}{\min(\text{moles of C}, \text{moles of H})} = \frac{4.21}{4.21} \approx 1\)**<br /><br />## Step 3: <br />Form the empirical formula based on the ratio.<br />### **\(\text{Empirical formula: } C_{41}H\)**<br /><br />## Step 4: <br />Calculate the molar mass of the empirical formula.<br />### **\(\text{molar mass of } C_{41}H = 41 \times 12.01 \text{ g/mol} + 1 \times 1.008 \text{ g/mol} = 493.418 \text{ g/mol}\)**<br /><br />## Step 5: <br />Determine the number of empirical formula units in the molecular formula.<br />### **\(\text{number of units } n = \frac{\text{molar mass of compound}}{\text{molar mass of empirical formula}} = \frac{58.04 \text{ g/mol}}{493.418 \text{ g/mol}} \approx 0.1176\)**<br /><br />## Step 6: <br />Form the molecular formula.<br />### **\(\text{molecular formula: } C_{4.82}H_{0.12}\)**<br /><br />#
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