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A common anti-inflammatory drug is 75.69% C, 8.80% H, and 15.51% O. What is its empirical formula?

Options
A.C19H16O4
B.C14H13O4
C.C6H9O
D.C14H11O2
E.C13H18O2
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Step-by-Step Analysis
We are given the percent composition of an anti-inflammatory drug: C 75.69%, H 8.80%, O 15.51%. The task is to find the empirical formula. First, convert each percent to moles by assuming a 100 g sample: - Carbon: 75.69 g / 12.01 g·mol−1 ≈ 6.30 mol - Hydrogen: 8.80 g / 1.008 g·mol−1 ≈ 8.73 mol - Oxygen: 15.51 g / 16.00 g·mol−1 ≈ 0.97 mol These are the mole ratios, which we then normalize by dividing by the smallest number of moles among the three elements. The smallest is about 0.97 mol (O): - C: 6.......Login to view full explanation

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