Question

). The molar mass of an organic acid, a compound composed of carbon, hydrogen, and oxygen, is 194.14 g/mol. Combustion of a 1.500-g sample produced 2.0402 g of CO2 and 0.6955 g of H2O. Determine the molecular formula of the acid. The molar masses of C, H, O, CO2, and H2O are 12.01 g/mol, 1.008 g/mol, 16.00 g/mol, 44.01 g/mol, and 18.02, g/mol, respectively.

Answer 1

The empirical formula is C₆H₁₀O₇.

Step-by-step explanation:

1. Calculate the masses of C, H, and O from the masses given.

Mass of C = 2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568 g C

Mass of H = 0.6955 g H₂O × (2.016 H/18.02 g H₂O) = 0.077 81 g H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O

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2. Convert these masses to moles.

Moles C = 0.5568 × 1/12.01 = 0.046 36

Moles H = 0.077 81 × 1/1.008 = 0.077 19

Moles O = 0.8654 × 1/16.00 = 0.054 09

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3. Find the molar ratios.

Moles C = 0.046 36/0.046 36 = 1

Moles H = 0.077 19/0.046 36 = 1.665

Moles O = 0.054 09/0.046 36 = 1.167

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4. Multiply the ratios by a number to make them close to integers

C = 1 × 6 = 6

H = 1.665 × 6 = 9.991

O = 1.167 × 6 = 7.001

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5. Round the ratios to integers

C:H:O =6:10:7

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6. Write the empirical formula

The empirical formula is C₆H₁₀O₇.

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7. Calculate the empirical formula mass

C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00

C₆H₁₀O₇ = 72.01 + 10.08+ 112.0

C₆H₁₀O₇ = 194.09

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8. Divide the molecular mass by the empirical formula mass.

MM/EFM = 194.14/194.09 = 1.000 ≈ 1

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9. Determine the molecular formula

MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇