Question

acetaminophen (tylenol): c, 63.56%, h, 6.00%, n, 9.27%, and o, 21.17%
express your answer as a chemical formula.

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part b
naproxen (aleve): c, 73.03%, h, 6.13%, and o, 20.84%
express your answer as a chemical formula.

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Explanation:

Step1: Assume 100g of the compound.

For acetaminophen:

• Mass of C = 63.56g, Molar mass of C = 12.01g/mol, Moles of C = $\frac{63.56}{12.01} \approx 5.292$
• Mass of H = 6.00g, Molar mass of H = 1.008g/mol, Moles of H = $\frac{6.00}{1.008} \approx 5.952$
• Mass of N = 9.27g, Molar mass of N = 14.01g/mol, Moles of N = $\frac{9.27}{14.01} \approx 0.662$
• Mass of O = 21.17g, Molar mass of O = 16.00g/mol, Moles of O = $\frac{21.17}{16.00} \approx 1.323$

Step2: Divide by the smallest number of moles (0.662 for N).

• Ratio of C: $\frac{5.292}{0.662} \approx 8$
• Ratio of H: $\frac{5.952}{0.662} \approx 9$
• Ratio of N: $\frac{0.662}{0.662} = 1$
• Ratio of O: $\frac{1.323}{0.662} \approx 2$

So the empirical formula (and molecular formula, as acetaminophen's molar mass matches) is $C_8H_9NO_2$.

For naproxen:

Step1: Assume 100g of the compound.

• Mass of C = 73.03g, Molar mass of C = 12.01g/mol, Moles of C = $\frac{73.03}{12.01} \approx 6.081$
• Mass of H = 6.13g, Molar mass of H = 1.008g/mol, Moles of H = $\frac{6.13}{1.008} \approx 6.081$
• Mass of O = 20.84g, Molar mass of O = 16.00g/mol, Moles of O = $\frac{20.84}{16.00} \approx 1.303$

Step2: Divide by the smallest number of moles (1.303 for O).

• Ratio of C: $\frac{6.081}{1.303} \approx 4.667$ (multiply by 3 to get whole number: 14)
• Ratio of H: $\frac{6.081}{1.303} \approx 4.667$ (multiply by 3 to get whole number: 14)
• Ratio of O: $\frac{1.303}{1.303} = 1$ (multiply by 3 to get whole number: 3) Wait, correction: Wait, actually, let's recalculate. Wait, maybe I made a mistake. Wait, let's do it properly. Wait, the moles of C: 73.03/12.01 ≈6.08, H:6.13/1.008≈6.08, O:20.84/16≈1.3025. Now, divide by 1.3025: C:6.08/1.3025≈4.67, H:6.08/1.3025≈4.67, O:1. So multiply by 3: C:14, H:14, O:3? Wait, no, wait naproxen's formula is $C_{14}H_{14}O_3$. Wait, let's check molar masses. Wait, maybe my initial division was wrong. Wait, let's use the correct method. Wait, the empirical formula calculation:

Moles of C: 73.03/12.01 ≈6.08, H:6.13/1.008≈6.08, O:20.84/16≈1.3025. Now, find the ratio. Let's divide by 1.3025:

C: 6.08 / 1.3025 ≈ 4.67 (which is 14/3, since 4.67*3=14), H: 6.08/1.3025≈4.67 (14/3), O:1 (3/3). So multiply all by 3: C:14, H:14, O:3. So empirical formula is $C_{14}H_{14}O_3$.

Answer:

Part A: $C_8H_9NO_2$
Part B: $C_{14}H_{14}O_3$