Question

a compound that is composed of carbon, hydrogen, and oxygen contains 70.6% c, 5.9% h, and 23.5% o by mass. the molecular weight of the compound is 136 amu. what is the molecular formula?

c8h4o

c9h12o

c4h4o

c8h8o2

c5h6o2

Explanation:

Step1: Calculate moles of each element

Assume \(100\) g of the compound.

• Moles of \(C\): \(n_{C}=\frac{70.6\space g}{12.01\space g/mol}\approx5.88\space mol\)
• Moles of \(H\): \(n_{H}=\frac{5.9\space g}{1.008\space g/mol}\approx5.85\space mol\)
• Moles of \(O\): \(n_{O}=\frac{23.5\space g}{16.00\space g/mol}\approx1.47\space mol\)

Step2: Find the mole ratio

Divide each mole value by the smallest mole value (\(n_{O} = 1.47\space mol\))

• Ratio of \(C\): \(\frac{5.88}{1.47}\approx4\)
• Ratio of \(H\): \(\frac{5.85}{1.47}\approx4\)
• Ratio of \(O\): \(\frac{1.47}{1.47} = 1\)

Empirical formula is \(C_{4}H_{4}O\), empirical formula weight \(=4\times12.01+4\times1.008 + 16.00=48.04+4.032+16.00 = 68.072\space g/mol\)

Step3: Find the multiple \(n\)

\(n=\frac{\text{molecular weight}}{\text{empirical formula weight}}=\frac{136}{68.072}\approx2\)

Step4: Determine molecular formula

Multiply sub - scripts in empirical formula by \(n = 2\)
Molecular formula \(=(C_{4}H_{4}O)\times2=C_{8}H_{8}O_{2}\)

Answer:

\(C_{8}H_{8}O_{2}\)