Question

acetaminophen (tylenol) c, 63.56%. h, 6.00%, n, 9.27%, and o, 21.17% express your answer as a chemical formula. c₁₄h₁₄o₃ a chemical reaction does not occur for this question submit previous answers request answer x incorrect; try again; 2 attempts remaining 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 submit request answer

Explanation:

Step1: Assume 100g of the compound

If we assume we have 100g of naproxen, then the masses of \(C\), \(H\), and \(O\) are \(m_{C}=73.03g\), \(m_{H} = 6.13g\), \(m_{O}=20.84g\)

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 \(C\): \(n_{C}=\frac{m_{C}}{M_{C}}=\frac{73.03g}{12.01g/mol}\approx6.08mol\)
The number of moles of \(H\): \(n_{H}=\frac{m_{H}}{M_{H}}=\frac{6.13g}{1.008g/mol}\approx6.08mol\)
The number of moles of \(O\): \(n_{O}=\frac{m_{O}}{M_{O}}=\frac{20.84g}{16.00g/mol}\approx1.30mol\)

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (\(n_{O}\approx1.30mol\))
For \(C\): \(\frac{n_{C}}{n_{O}}=\frac{6.08}{1.30}\approx4.68\)
For \(H\): \(\frac{n_{H}}{n_{O}}=\frac{6.08}{1.30}\approx4.68\)
For \(O\): \(\frac{n_{O}}{n_{O}} = 1\)
Multiply by 3 to get whole numbers:
\(C\): \(4.68\times3\approx14\)
\(H\): \(4.68\times3\approx14\)
\(O\): \(1\times3 = 3\)

Answer:

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