Question

An organic liquid is a mixture of methyl alcohol (CH3OH) and ethyl alcohol (C2H5OH). A 0.220-g sample of the liquid is burned in an excess of O2(g) and yields 0.386 g CO2(g) (carbon dioxide). Set up two algebraic equations, one expressing the mass of carbon dioxide produced in terms of each reagent and the other expressing the mass of sample burned in terms of each reagent.What is the mass of methyl alcohol (CH3OH) in the sample?

Answer 1

Answer: the mass of methyl alcohol (CH3OH) in the sample is 0.0641 g

Explanation:

1. Preliminars;

• Mass  of CH₃ OH: M

• Mass of C₂H₅OH: E

• Molar mass of CH₃ OH: 32.04 g/mol

• Molar mass of C₂H₅OH: 46.07 g/mol

• Molar mass of CO₂: 44.01 g/mol

• Molar mass equation: molar mass = mass in grams / number of moles

2. Algebraic equation # 1: mass of sample burned in terms of each reagent.

• M + E = 0.220 ------ (1)

3. Algebraic equation # 2: mass of carbon dioxide produced in terms of each reagent.

a) Combustion of methyl alcohol

• 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)

• 2 mol CH₃OH → 2 mol CO₂

• 2 × 32.04 g CH₃OH → 2 × 44.01 g CO₂

• Mass of CO₂ produced from M grams of CH₃OH = M / 32.04 × 44.01 = 1.374 M

b) Combustion of ethyl alcohol

• C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)

• 1 mol C₂H₅OH → 2 mol CO₂

• 46.07 g CH₃OH → 2 × 44.01 g CO₂

• Mass of CO₂ produced from E grams of C₂H₅OH = E / 46.07 × 88.02 = 1.911 E

c) Mass of CO₂ (g) obtained, in terms of each reagent

•     1.374 M + 1.911 E = 0.386 ----- (2)

4) System of equations

•        M   +         E = 0.220 ------ (1)

• 1.374 M + 1.911 E = 0.386 ----- (2)

5) Solution

• From equation (1): E = 0.220 - M

• Substituting in equation (2): 1.374 M + 1.911 (0.220 - M) = 0.386

• 1.374 M + 0.4204 - 1.911 M = 0.386

• - 0.537 M = - 0.0344

• M = 0.0344 / 0.537 = 0.0641 g

Answer:  the mass of methyl alcohol (CH3OH) in the sample is 0.0641 g