Question 3.6: Determining an Empirical Formula from Combustion Analysis Da...
Determining an Empirical Formula from Combustion Analysis Data
Vitamin C is essential for the prevention of scurvy. Combustion of a 0.2000 g sample of this carbon hydrogen oxygen compound yields 0.2998 g CO_{2} and 0.0819 g H_{2}O. What are the percent composition and the empirical formula of vitamin C?
Analyze
After combustion, all the carbon atoms from the vitamin C sample are in CO_{2} and all the hydrogen atoms are in H_{2}O. However, the oxygen atoms in CO_{2} and H_{2}O come partly from the sample and partly from the oxygen gas consumed in the combustion. So, in the determination of the percent composition, we focus first on carbon and hydrogen and then on oxygen. To determine the empirical formula, we must calculate the amounts of C, H, and O in moles, and then calculate the mole ratios.
Percent Composition
First, determine the mass of carbon in 0.2988 g CO_2, by converting to mol C,
? mol C = 0.2998 g CO_{2} \times \frac{1 mol CO_{2}}{44.010 g CO_{2}} \times \frac{1 mol C}{1 mol CO_{2}} = 0.006812 mol C
and then to g C.
? g C = 0.006812 mol C \times \frac{12.011 g C}{1 mol C} = 0.08182 g C
Proceed in a similar fashion for 0.0819 g H_{2}O to obtain
? mol H = 0.0819 g H_{2}O \times \frac{1 mol H_{2}O}{18.02 g H_{2}O} \times \frac{2 mol H}{1 mol H_{2}O} = 0.00909 mol H
and
? g H = 0.00909 mol H \times \frac{1.008 g H}{1 mol H} = 0.00916 g H
Obtain the mass of O in the 0.2000 g sample as the difference
? g O = 0.2000 g sample – 0.08182 g C – 0.00916 g H = 0.1090 g O
Finally, multiply the mass fractions of the three elements by 100% to obtain mass percentages.
% C = \frac{0.08182 g C}{0.2000 g sample} \times 100\% = 40.91\% C
% H = \frac{0.00916 g H}{0.2000 g sample} \times 100\% = 4.58\% H
% O = \frac{0.1090 g O}{0.2000 g sample} \times 100\% = 54.50\% O
Empirical Formula
At this point we can choose either of two alternatives. The first is to obtain the empirical formula from the mass percent composition, in the same manner illustrated in Example 3-5. The second is to note that we have already determined the number of moles of C and H in the 0.2000 g sample. The number of moles of O is
? mol O = 0.1090 g O \times \frac{1 mol O}{15.999 g O} = 0.006813 mol O
From the numbers of moles of C, H, and O in the 0.2000 g sample, we obtain the tentative empirical formula
C_{0.006812}H_{0.00909}O_{0.006813}
Next, divide each subscript by the smallest (0.006812) to obtain
CH_{1.33}O
Finally, multiply all the subscripts by 3 to obtain
Empirical formula of vitamin C: C_{3}H_{4}O_{3}
Assess
The determination of the empirical formula does not require determining the mass percent composition as a preliminary calculation. The empirical formula can be based on a sample of any size, as long as the numbers of moles of the different atoms in that sample can be determined.