Question

determine the correct empirical formula of a compound containing 26.68% carbon, 2.24% hydrogen, and 71.08% oxygen.
a cho
b c₂h₂o₄
c c₀.₅h₀.₅o₁
d cho₂

Explanation:

Step1: Assume 100g of the compound

So, we have \(26.68g\) of \(C\), \(2.24g\) of \(H\), and \(71.08g\) of \(O\).

Step2: Calculate the moles of each element

• Moles of \(C\): \(n_{C}=\frac{26.68g}{12.01g/mol}\approx2.22mol\) (using \(n = \frac{m}{M}\), where \(M_{C}=12.01g/mol\))
• Moles of \(H\): \(n_{H}=\frac{2.24g}{1.008g/mol}\approx2.22mol\) (using \(M_{H} = 1.008g/mol\))
• Moles of \(O\): \(n_{O}=\frac{71.08g}{16.00g/mol}\approx4.44mol\) (using \(M_{O}=16.00g/mol\))

Step3: Find the mole - ratio

Divide each mole value by the smallest mole value (\(2.22mol\))

• For \(C\): \(\frac{2.22mol}{2.22mol}=1\)
• For \(H\): \(\frac{2.22mol}{2.22mol}=1\)
• For \(O\): \(\frac{4.44mol}{2.22mol}=2\)

Answer:

D. \(CHO_{2}\)