Question

a compound contains 52.2% c, 13.0% h and 34.8% o. which of the following is the correct empirical formula for the compound?
a cho
b c₄h₅o
c c₃h₆o
d ch₇o
e c₂h₆o

Explanation:

Step1: Assume 100g of compound

Masses: \( m_C = 52.2\,\text{g} \), \( m_H = 13.0\,\text{g} \), \( m_O = 34.8\,\text{g} \)

Step2: Calculate moles

Moles of C: \( n_C = \frac{52.2}{12.01} \approx 4.346\,\text{mol} \)
Moles of H: \( n_H = \frac{13.0}{1.008} \approx 12.897\,\text{mol} \)
Moles of O: \( n_O = \frac{34.8}{16.00} \approx 2.175\,\text{mol} \)

Step3: Divide by smallest mole (O)

Ratio C: \( \frac{4.346}{2.175} \approx 2 \)
Ratio H: \( \frac{12.897}{2.175} \approx 5.93 \approx 6 \)
Ratio O: \( \frac{2.175}{2.175} = 1 \)

Step4: Determine empirical formula

From ratios, formula is \( \text{C}_2\text{H}_6\text{O} \) (matches option e).

Answer:

e. \( \text{C}_2\text{H}_6\text{O} \)