Question 3.11: Determining a Molecular Formula from Combustion Analysis Pro......
Determining a Molecular Formula from Combustion Analysis
Problem Vitamin C (ℳ = 176.12 g/mol) is a compound of C, H, and O found in many natural sources, especially citrus fruits. When a 1.000-g sample of vitamin C is burned in a combustion apparatus, the following data are obtained:
\text{Mass of} CO_2 \text{absorber before combustion = 83.85 g}
\text{Mass of} H_2O \text{absorber after combustion = 37.96 g}
\text{Mass of} H_2O \text{absorber before combustion = 37.55 g}
What is the molecular formula of vitamin C?
Plan We find the masses of CO_2 and H_2O by subtracting the mass of each absorber before the combustion from its mass after combustion. From the mass of CO_2, we use Equation 3.7 to find the mass of C. Similarly, we find the mass of H from the mass of H_2O. The mass of vitamin C (1.000 g) minus the sum of the masses of C and H gives the mass of O, the third element present. Then, we proceed as in Sample Problem 3.10: calculate the amount (mol) of each element using its molar mass, construct the empirical formula, determine the whole-number multiple from the given molar mass, and construct the molecular formula.
\text{Mass of element = mass of compound} \times \frac{\text{mass of element in 1 mol of compound}}{\text{mass of 1 mol of compound}} (3.7)
Solution Finding the masses of combustion products:
\text{Mass (g) of }CO_2 = \text{mass of }CO_2 \text{ absorber after − mass before}
= 85.35 g − 83.85 g = 1.50 g CO_2
\text{Mass (g) of }H_2O = \text{mass of }H_2O \text{ absorber after − mass before}
= 37.96 g − 37.55 g = 0.41 g H_2O
Calculating masses (g) of C and H using Equation 3.7:
\text{Mass of element = mass of compound} \times \frac{\text{mass of element in 1 mol of compound}}{\text{mass of 1 mol of compound}}
\text{Mass (g) of C = mass of} CO_2 \times \frac{\text{1 mol C × }ℳ \text{of C}}{\text{mass of 1 mol }CO_2} = 1.50 \cancel{g CO_2} \times \frac{12.01 g C}{44.01 \cancel{g CO_2}}
= 0.409 g C
\text{Mass (g) of H = mass of} H_2O \times \frac{\text{2 mol H × }ℳ \text{of H}}{\text{mass of 1 mol }H_2O} = 0.41 \cancel{g H_2O} \times \frac{2.016 g H}{18.02 \cancel{g H_2O}}
= 0.046 g H
Calculating mass (g) of O:
= 1.000 g − (0.409 g + 0.046 g) = 0.545 g O
Finding the amounts (mol) of elements: Dividing the mass (g) of each element by its molar mass gives 0.0341 mol of C, 0.046 mol of H, and 0.0341 mol of O.
Constructing the preliminary formula: C_{0.0341}H_{0.046}O_{0.0341}
Determining the empirical formula: Dividing through by the smallest subscript gives
C_{\frac{0.0341}{0.0341}}H_{\frac{0.046}{0.0341}}O_{\frac{0.0341}{0.0341}} = C_{1.00}H_{1.3}O_{1.00}We find that 3 is the smallest integer that makes all subscripts into integers:
C_{(1.00×3)}H_{(1.3×3)}O_{(1.00×3)} = C_{3.00}H_{3.9}O_{3.00} ≈ C_3H_4O_3Determining the molecular formula:
\text{Whole-number multiple} = \frac{ℳ\text{ of vitamin C}}{ℳ\text{ of empirical formula}} = \frac{176.12 \cancel{\text{g/mol}}}{88.06 \cancel{\text{g/mol}}} = 2.000 = 2
C_{(3×2)}H_{(4×2)}O_{(3×2)} = C_6H_8O_6
Check The element masses seem correct: carbon makes up slightly more than 0.25 of the mass of CO_2 (12 g/44 g > 0.25), as do the masses in the problem (0.409 g/1.50 g > 0.25). Hydrogen makes up slightly more than 0.10 of the mass of H_2O (2 g/18 g > 0.10), as do the masses in the problem (0.046 g/0.41 g > 0.10). The molecular formula has the same ratio of subscripts (6/8/6) as the empirical formula (3/4/3) and the preliminary formula (0.0341/0.046/0.0341), and it gives the known molar mass:
(6 × ℳ \text{of} C) + (8 × ℳ \text{of} H) + (6 × ℳ \text{of} O) = ℳ \text{of vitamin C}
(6 × 12.01 \text{g/mol}) + (8 × 1.008 \text{g/mol}) + (6 × 16.00 \text{g/mol}) = 176.12 \text{g/mol}
Comment The subscript we calculated for H was 3.9, which we rounded to 4. But, if we had strung the calculation steps together, we would have obtained 4.0:
\text{Subscript of H }= 0.41 g H_2O \times \frac{2.016 g H}{18.02 g H_2O} \times \frac{1 \text{mol} H}{1.008 g H} \times \frac{1}{0.0341 \text{mol}} \times 3 = 4.0