QuestionJanuary 7, 2026

What is the empirical formula for a molecule containing 86% C and 14% H? CH_(2) CH_(8) C_(3)H_(7) C_(7)H_(16)

What is the empirical formula for a molecule containing 86% C and 14% H? CH_(2) CH_(8) C_(3)H_(7) C_(7)H_(16)
What is the empirical formula for a molecule containing 86% C and 14% H? CH_(2) CH_(8) C_(3)H_(7) C_(7)H_(16)

Solution
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Answer

A. CH_{2} Explanation 1. Assume 100 g Sample and Convert Percentages to Grams Assume a 100\,g sample so that the mass of each element equals its percentage: 86\,g C and 14\,g H. 2. Convert Masses to Moles Calculate moles using atomic masses: - For C: \frac{86\,g}{12.01\,g/mol} \approx 7.16\,mol - For H: \frac{14\,g}{1.008\,g/mol} \approx 13.89\,mol 3. Find Simplest Whole Number Ratio Divide both mole values by the smallest number (7.16): - C: \frac{7.16}{7.16} = 1 - H: \frac{13.89}{7.16} \approx 1.94 \approx 2 So, the ratio is 1:2 for C:H. 4. Write Empirical Formula The empirical formula is CH_2.

Explanation

1. Assume 100 g Sample and Convert Percentages to Grams<br /> Assume a $100\,g$ sample so that the mass of each element equals its percentage: $86\,g$ C and $14\,g$ H.<br />2. Convert Masses to Moles<br /> Calculate moles using atomic masses: <br />- For C: $\frac{86\,g}{12.01\,g/mol} \approx 7.16\,mol$<br />- For H: $\frac{14\,g}{1.008\,g/mol} \approx 13.89\,mol$<br />3. Find Simplest Whole Number Ratio<br /> Divide both mole values by the smallest number ($7.16$):<br />- C: $\frac{7.16}{7.16} = 1$<br />- H: $\frac{13.89}{7.16} \approx 1.94 \approx 2$<br />So, the ratio is $1:2$ for C:H.<br />4. Write Empirical Formula<br /> The empirical formula is $CH_2$.
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