Question

1. a certain alcohol contains only three elements, carbon, hydrogen, and oxygen. combustion of a 20.00 gram sample of the alcohol produced 38.20 grams of co₂ and 23.48 grams of h₂o. what is the empirical formula of the alcohol? a) c₂h₆o b) c₃h₆o c) c₄h₁₀o d) ch₄o e) c₅h₁₀o16) methane and oxygen react to form carbon dioxide and water. what mass of water is formed if 9.6 g of methane reacts with 38.4 g of oxygen to produce 26.4 g of carbon dioxide? a) 21.6 g b) 26.4 g c) 44.4 g d) 48.0 g17) combustion analysis of an unknown compound containing only carbon and hydrogen produced 4.554 g of co₂ and 2.322 g of h₂o. what is the empirical formula of the compound? a) ch₂ b) c₂h₅ c) c₄h₁₀ d) c₅h₂18) combustion analysis of 2.400 g of an unknown compound containing carbon, hydrogen, and oxygen produced 4.171 g of co₂ and 2.268 g of h₂o. what is the empirical formula of the compound? a) c₂h₅o b) c₂h₅o₂ c) c₂h₁₀o₃ d) c₃h₈o₂

Explanation:

Question 15

Step1: Calculate moles of C from CO₂

Molar mass of \( \text{CO}_2 \) is \( 44.01 \, \text{g/mol} \). Moles of \( \text{CO}_2 = \frac{38.20 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.868 \, \text{mol} \). Moles of C = moles of \( \text{CO}_2 \approx 0.868 \, \text{mol} \). Mass of C = \( 0.868 \, \text{mol} \times 12.01 \, \text{g/mol} \approx 10.43 \, \text{g} \).

Step2: Calculate moles of H from H₂O

Molar mass of \( \text{H}_2\text{O} \) is \( 18.02 \, \text{g/mol} \). Moles of \( \text{H}_2\text{O} = \frac{23.48 \, \text{g}}{18.02 \, \text{g/mol}} \approx 1.303 \, \text{mol} \). Moles of H = \( 2 \times 1.303 \, \text{mol} = 2.606 \, \text{mol} \). Mass of H = \( 2.606 \, \text{mol} \times 1.008 \, \text{g/mol} \approx 2.627 \, \text{g} \).

Step3: Calculate mass and moles of O

Mass of O = \( 20.00 \, \text{g} - 10.43 \, \text{g} - 2.627 \, \text{g} \approx 6.943 \, \text{g} \). Moles of O = \( \frac{6.943 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.434 \, \text{mol} \).

Step4: Find mole ratios

Divide moles of C, H, O by smallest moles (O: 0.434).
C: \( \frac{0.868}{0.434} = 2 \), H: \( \frac{2.606}{0.434} \approx 6 \), O: \( \frac{0.434}{0.434} = 1 \).
Ratio: C₂H₆O.

Step1: Apply Law of Conservation of Mass

Total mass of reactants = \( 9.6 \, \text{g} + 38.4 \, \text{g} = 48.0 \, \text{g} \).
Mass of \( \text{CO}_2 = 26.4 \, \text{g} \).
Mass of \( \text{H}_2\text{O} = 48.0 \, \text{g} - 26.4 \, \text{g} = 21.6 \, \text{g} \).

Step1: Moles of C from \( \text{CO}_2 \)

Moles of \( \text{CO}_2 = \frac{4.554 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.1035 \, \text{mol} \). Moles of C = \( 0.1035 \, \text{mol} \). Mass of C = \( 0.1035 \times 12.01 \approx 1.243 \, \text{g} \).

Step2: Moles of H from \( \text{H}_2\text{O} \)

Moles of \( \text{H}_2\text{O} = \frac{2.322 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.1289 \, \text{mol} \). Moles of H = \( 2 \times 0.1289 = 0.2578 \, \text{mol} \). Mass of H = \( 0.2578 \times 1.008 \approx 0.2599 \, \text{g} \).

Step3: Mole ratios

Divide by smallest moles (C: 0.1035).
C: \( \frac{0.1035}{0.1035} = 1 \), H: \( \frac{0.2578}{0.1035} \approx 2.5 \). Multiply by 2: C₂H₅.

Answer:

A) \( \text{C}_2\text{H}_6\text{O} \)

Question 16