Question 6.11: Calculating an Empirical Formula from a Percent Composition ......

General Chemistry: Atoms First [1015060]

Calculating an Empirical Formula from a Percent Composition

Vitamin C (ascorbic acid) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?

STRATEGY

Assume that you have 100.00 g of ascorbic acid, and then carry out the procedure outlined in Figure 6.3.

6.3

worked example 6.11

Step-by-Step

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First, find the number of moles of each element in the sample:

40.92 \ \cancel{g \ C} \times \frac{1 \ mol \ C}{12.0 \ \cancel{g \ C}} = 3.41 \ mol \ C \\ 4.58 \ \cancel{g \ H} \times \frac{1 \ mol \ H}{1.01 \ \cancel{g \ H}} = 4.53 \ mol \ H \\ 54.50 \ \cancel{g \ O} \times \frac{1 \ mol \ O}{16.0 \ \cancel{g \ O}} = 3.41 \ mol \ O

Dividing each of the three numbers by the smallest (3.41 mol) gives a C:H:O mole ratio of 1:1.33:1 and a temporary formula of $C_1H_{1.33}O_1$ . Multiplying the subscripts by small integers in a trial-and-error procedure until whole numbers are found then gives the empirical formula: $C_{(3 \times 1)}H_{(3 \times 1.33)}O_{(3 \times 1)} = C_3H_4O_3$ .

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