Question 3.6: Determining an Empirical Formula from Combustion Analysis Da...

$Percent  Composition$
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.