methyl butyrate (component of apple taste and smell): 58.80%c, 9.87%h, and 31.33%o. express your answer as a…

Explanation:

Step1: Assume 100g of the compound

For methyl - butyrate, assume 100g. So we have 58.80g of C, 9.87g of H, and 31.33g of O.

Step2: Calculate the number of moles of each element

For C: $n_{C}=\frac{58.80g}{12.01g/mol}\approx4.90mol$. For H: $n_{H}=\frac{9.87g}{1.01g/mol}\approx9.77mol$. For O: $n_{O}=\frac{31.33g}{16.00g/mol}\approx1.96mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (1.96mol). For C: $\frac{4.90mol}{1.96mol}\approx2.5$. For H: $\frac{9.77mol}{1.96mol}\approx5$. For O: $\frac{1.96mol}{1.96mol}=1$. Multiply by 2 to get whole - numbers. The empirical formula is $C_{5}H_{10}O_{2}$.

For glucose:

Step1: Assume 100g of the compound

Assume 100g of glucose. So we have 40.00g of C, 6.71g of H, and 53.29g of O.

Step2: Calculate the number of moles of each element

For C: $n_{C}=\frac{40.00g}{12.01g/mol}\approx3.33mol$. For H: $n_{H}=\frac{6.71g}{1.01g/mol}\approx6.64mol$. For O: $n_{O}=\frac{53.29g}{16.00g/mol}\approx3.33mol$.

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (3.33mol). For C: $\frac{3.33mol}{3.33mol}=1$. For H: $\frac{6.64mol}{3.33mol}\approx2$. For O: $\frac{3.33mol}{3.33mol}=1$. The empirical formula is $CH_{2}O$.

Answer:

Part A: $C_{5}H_{10}O_{2}$ Part B: $CH_{2}O$