Question 3.5.2: Use combustion analysis to determine empirical and molecular......
Use combustion analysis to determine empirical and molecular formulas (C, H, and O).
A 1.155-g sample of butyric acid, an organic compound containing carbon, hydrogen, and oxygen, is analyzed by combustion and 2.308 g of CO_{2} and 0.9446 g of H_{2}O are produced. In a separate experiment, the molar mass is found to be 88.11 g/mol. Determine the empirical and mo-lecular formulas of butyric acid.
You are asked to determine the empirical and molecular formulas of a compound.
You are given the mass of compound analyzed, data from a combustion analysis experiment (mass of H_{2}O and CO_{2}), and the molar mass of the compound.
When combustion analysis involves a compound containing carbon, hydrogen, and oxygen, the amount of oxygen in the compound cannot be determined directly from the amount of CO_{2} or H_{2}O produced. Instead, the amounts of CO_{2} and H_{2}O are used to calculate the mass of carbon and hydrogen in the original sample. The remaining mass of the original sample is oxygen.
Step 1. Use the mass of CO_{2} and H_{2}O produced in the combustion analysis to calculate moles of C and moles of H present in the original sample. Note that there are 2 mol of H present in 1 mol of H_{2}O.
2.308 g \mathrm{CO_{2}}\times\,\frac{1\;\mathrm{mol\;CO_{2}}}{44.010\;g}\times\frac{1\;\mathrm{mol\;C}}{1\;\mathrm{mol\;CO_{2}}} = 0.05244 mol C
0.9446 g \mathrm{H}_{2}\mathrm{O}\times{\frac{1\;\mathrm{mol\;H}_{2}\mathrm{O}}{18.015\,\mathrm{g}}}\times{\frac{2\;\mathrm{mol\;H}}{1\;\mathrm{mol\;H}_{2}\mathrm{O}}} = 0.1049 mol H
Step 2. Use the amount (in mol) of C and H to calculate the mass of C and H present in the original sample.
0.05244 mol C \times\;{\frac{12.011\ {\mathrm{g}}}{1\;\mathrm{mol}\ {\mathrm{G}}}} = 0.6299 g C
0.1049 mol H \times\;\frac{1.0079\mathrm{~g}}{1\;\mathrm{mol\;II}} = 0.1057 g H
Step 3. Subtract the amounts of C and H from the mass of the original sample to determine the mass of O present in the original sample. Use this value to calculate the moles of O present in the original sample
1.155 g sample – 0.6299 g C – 0.1057 g H = 0.419 g O
0.419 g O \times{\frac{1{\bmod{\mathrm{G}}}}{16.00{\mathrm{g}}}} = 0.0262 mol O
Step 4. Divide each amount by the smallest value to obtain a whole-number ratio of elements.
\frac{0.05244\;{\mathrm{mol}}\,{\mathrm{C}}}{0.0262} = 2.00 mol C = 2 mol C
\frac{0.1049{\bmod{11}}}{0.0262} = 4.00 mol H = 4 mol H
\frac{0.0262 {\mathrm{mol}}\;\mathrm{O}}{0.0262} = 1.00 mol O = 1 mol O
The empirical formula of butyric acid is C_{2}H_{4}O.
Step 5. Compare the molar mass of the compound to the molar mass of the empirical formula to determine the molecular formula.
The molecular formula is (C_{2}H_{4}O)_{2}, or C_{4}H_{8}O_{2}.
Is your answer reasonable? The molar mass of the compound must be a whole-number multiple of the molar mass of the empirical formula, which is true in this case.