Combustion Formula and Hydrogenation Limiting Reagent
An unknown oxygen-containing organic liquid is analyzed by combustion. A 1.92 g sample burns completely in excess oxygen and gives 5.28 g of CO₂ and 2.16 g of H₂O. A mass-spectrum determination gives a molar mass of 128 g/mol. The compound is then hydrogenated over a nickel catalyst: 1.92 g of the compound is treated with 0.0120 mol of H₂, and 1.30 g of the saturated alcohol is isolated after purification. The laboratory report must establish the empirical formula, the molecular formula, the limiting reagent in the hydrogenation, the theoretical mass of alcohol, and the percentage yield. A technician writes that because the combustion used excess oxygen, oxygen is also in excess in the hydrogenation, so the organic compound must be limiting; the report must check this claim by mole ratio rather than by mass. The alcohol is assumed to have one additional H₂ unit per molecule and no other change.
1. Determine the empirical formula of the compound from the combustion data.
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Moles of carbon: 5.28 g CO₂ / 44.01 g/mol ≈ 0.120 mol C, giving 0.120 × 12.01 ≈ 1.44 g C.
Moles of hydrogen: 2.16 g H₂O / 18.02 g/mol ≈ 0.1199 mol H₂O, hence 0.2398 mol H, giving 0.2398 × 1.008 ≈ 0.242 g H.
Oxygen is found by difference: 1.92 − 1.44 − 0.242 ≈ 0.238 g O, which is 0.238/16.00 ≈ 0.0149 mol O.
Divide by the smallest amount, 0.0149: C ≈ 8.05, H ≈ 16.1, O = 1.00. Rounding gives C₈H₁₆O. Checking the mass: 8(12.01) + 16(1.008) + 16.00 ≈ 128.2 g/mol, consistent with the measured molar mass.
2. Find the molecular formula and state whether the empirical and molecular formulas differ.
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The empirical formula C₈H₁₆O has molar mass 8(12.01) + 16(1.008) + 16.00 ≈ 96.08 + 16.13 + 16.00 = 128.21 g/mol.
The measured molar mass is 128 g/mol, so the multiplier is 128/128.21 ≈ 1. The molecular formula is therefore C₈H₁₆O, identical to the empirical formula. The one degree of unsaturation is consistent with a single C=O double bond in an aldehyde or ketone; no C=C unit is required. Hydrogenating that carbonyl to an alcohol adds one H₂ per molecule, giving C₈H₁₈O.
3. Using 1.92 g of compound and 0.0120 mol H₂, identify the limiting reagent, calculate the theoretical mass of alcohol, and find the percentage yield for 1.30 g isolated.
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Moles of compound: 1.92 g / 128.21 g/mol ≈ 0.01498 mol. The hydrogenation consumes one mole of H₂ per mole of compound, so 0.01498 mol would require 0.01498 mol H₂. Only 0.0120 mol H₂ is supplied; therefore H₂ is the limiting reagent and the compound is in excess.
The product is C₈H₁₈O, molar mass 8(12.01) + 18(1.008) + 16.00 ≈ 130.2 g/mol. Theoretical product mass = 0.0120 mol × 130.2 g/mol ≈ 1.56 g.
Percentage yield = (1.30 g / 1.56 g) × 100 ≈ 83.2%.
The technician’s argument is invalid: excess O₂ in combustion says nothing about H₂ in a separate reaction. The 1:1 mole ratio and the smaller amount of H₂ determine the limit, not the mass of the organic sample.