Question

part b
naproxen (aleve) c, 73.03% h, 6.13% and o, 20.84%
express your answer as a chemical formula.
□ a chemical reaction does not occur for this question.
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Explanation:

Step1: Assume 100g of the compound

Assume we have 100g of naproxen. Then the masses of \(C = 73.03g\), \(H=6.13g\), \(O = 20.84g\)

Step2: Calculate the number of moles of each element

The molar mass of \(C\) is \(M_{C}=12.01g/mol\), the molar mass of \(H\) is \(M_{H} = 1.008g/mol\), and the molar mass of \(O\) is \(M_{O}=16.00g/mol\)
The number of moles of \(C\), \(n_{C}=\frac{73.03g}{12.01g/mol}\approx6.08mol\)
The number of moles of \(H\), \(n_{H}=\frac{6.13g}{1.008g/mol}\approx6.08mol\)
The number of moles of \(O\), \(n_{O}=\frac{20.84g}{16.00g/mol}\approx1.30mol\)

Step3: Divide by the smallest number of moles

Divide each number of moles by \(n_{O}\approx1.30mol\)
For \(C\): \(\frac{6.08mol}{1.30mol}\approx4.68\)
For \(H\): \(\frac{6.08mol}{1.30mol}\approx4.68\)
For \(O\): \(\frac{1.30mol}{1.30mol} = 1\)
Multiply by 2 to get whole - number ratios. So \(C:4.68\times2\approx9.36\approx9\), \(H:4.68\times 2\approx9.36\approx9\), \(O:1\times2 = 2\)

Answer:

\(C_{14}H_{14}O_{3}\)