Combustion Analysis and the Missing Factor of Two
Ms. Alvarez is reviewing a lab report from a student who analyzed an unknown liquid containing only carbon, hydrogen, and oxygen. The student burned a 0.500 g sample in excess oxygen and collected 0.733 g of CO2 and 0.300 g of H2O. The report states: 'moles CO2 = 0.733/44.01 = 0.0167 mol, so moles C = 0.0167; moles H2O = 0.300/18.02 = 0.0167 mol, so moles H = 0.0167. The C:H ratio is 1:1, so the empirical formula is CH. A freezing-point measurement gives a molar mass of 60 g/mol, so the molecular formula is C5H5.' The student also notes that the sample mass is not fully accounted for but assumes the balance was off. Ms. Alvarez has the raw data and must decide what feedback to give before the student revises. The class has already covered combustion analysis and empirical formulas.
1. Identify the specific calculation error in the student's report and determine the correct empirical formula from the data. Show the mass of oxygen by difference.
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Error: The student converted moles of H2O directly to moles of H, but each water molecule contains two hydrogen atoms. The correct calculation is:
- mol C = 0.733 g CO2 / 44.01 g/mol = 0.01665 mol C; mass C = 0.01665 × 12.01 = 0.200 g.
- mol H2O = 0.300 / 18.02 = 0.01665 mol; mol H = 2 × 0.01665 = 0.03330 mol; mass H = 0.03330 × 1.008 = 0.0336 g.
- mass O = 0.500 g − 0.200 g − 0.0336 g = 0.266 g; mol O = 0.266 / 16.00 = 0.0166 mol.
Divide by 0.0166: C 1.00, H 2.00, O 1.00 → CH2O. The student also did not subtract C and H from the sample mass to find oxygen, so the CH formula is not justified.
2. The student's proposed molecular formula C5H5 has a molar mass of about 65 g/mol, not 60. Using the corrected empirical formula and the measured molar mass of 60 g/mol, determine the molecular formula and explain why the student's final answer is inconsistent.
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Empirical formula mass of CH2O = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol. n = 60 / 30.03 ≈ 2.0, so molecular formula = (CH2O)2 = C2H4O2. The student's C5H5 has molar mass 5(12.01) + 5(1.008) = 65.1 g/mol and contains no oxygen, so it is inconsistent with both the measured molar mass and the combustion data. The correct molecular formula C2H4O2 has molar mass 60.05 g/mol. The student's final formula should be rejected even though the numerical molar mass is close to 60; the elemental ratio must match.
3. The student asks whether the 0.500 g sample could have contained residual water. Explain how residual water would affect the calculated empirical formula, and state whether the data would still support CH2O if the sample were wet.
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Residual water adds both H and O to the sample. Because the calculation finds oxygen by difference, water would inflate the apparent O mass and also inflate the H:C ratio, since water contributes two H per O. The calculated empirical formula would be shifted toward higher H and O, not toward CH. If the sample were wet, the true compound could have a lower H content than the raw data suggest. To test, dry the sample (e.g., over a desiccant or gently under vacuum) and repeat the combustion. The present data match CH2O exactly, so residual water is not needed to explain them, but if water were present the empirical formula would not be trustworthy.