select all the correct answers.\na sample of an unknown compound has a percent composition of 52.14% carbon…

Explanation:

Step1: Calculate molar masses

Let's assume we have 100 g of the compound. So we have 52.14 g of C, 13.13 g of H and 34.73 g of O. The molar mass of C is $M_{C}=12.01\ g/mol$, of H is $M_{H} = 1.01\ g/mol$ and of O is $M_{O}=16.00\ g/mol$. The number of moles of C: $n_{C}=\frac{52.14\ g}{12.01\ g/mol}\approx4.34\ mol$. The number of moles of H: $n_{H}=\frac{13.13\ g}{1.01\ g/mol}\approx13.0\ mol$. The number of moles of O: $n_{O}=\frac{34.73\ g}{16.00\ g/mol}\approx2.17\ mol$. The mole - ratio of C : H : O is $\frac{4.34}{2.17}:\frac{13.0}{2.17}:\frac{2.17}{2.17}=2:6:1$. So the empirical formula of the compound is $C_{2}H_{6}O$.

Step2: Calculate molar masses of given compounds

• For $CH_{3}CH_{3}CH_{2}O_{2}$: The molar mass $M = 3\times12.01+8\times1.01 + 2\times16.00=36.03+8.08 + 32.00 = 76.11\ g/mol$. The empirical formula is $C_{3}H_{8}O_{2}$.
• For $C_{2}H_{5}OH$: The molar mass $M=2\times12.01 + 6\times1.01+16.00=24.02+6.06 + 16.00 = 46.08\ g/mol$. The empirical formula is $C_{2}H_{6}O$.
• For $C_{4}H_{10}O_{2}$: The molar mass $M = 4\times12.01+10\times1.01+2\times16.00=48.04+10.10+32.00 = 90.14\ g/mol$. The empirical formula is $C_{2}H_{5}O$.
• For $C_{4}H_{12}O_{2}$: The molar mass $M=4\times12.01 + 12\times1.01+2\times16.00=48.04+12.12+32.00 = 92.16\ g/mol$. The empirical formula is $C_{2}H_{6}O$.
• For $CH_{3}CH_{3}CH_{2}OOH$: The molar mass $M=3\times12.01+8\times1.01+3\times16.00=36.03+8.08+48.00 = 92.11\ g/mol$. The empirical formula is $C_{3}H_{8}O_{3}$.

Answer:

B. $C_{2}H_{5}OH$, D. $C_{4}H_{12}O_{2}$