Question

a compound contains 76.6% c, 6.38% h and 17.0% o. which of the following is the correct empirical formula for the compound?
a cho
b c₄h₄o
c c₆h₆o
d ch₂o
e c₂h₂o

Explanation:

Step1: Assume 100g of compound

So, mass of C = 76.6g, H = 6.38g, O = 17.0g.

Step2: Calculate moles

Moles of C: $\frac{76.6}{12.01} \approx 6.38$ mol
Moles of H: $\frac{6.38}{1.008} \approx 6.33$ mol
Moles of O: $\frac{17.0}{16.00} \approx 1.06$ mol

Step3: Divide by smallest mole (O's 1.06)

C: $\frac{6.38}{1.06} \approx 6$
H: $\frac{6.33}{1.06} \approx 6$
O: $\frac{1.06}{1.06} = 1$
So ratio C:H:O ≈ 6:6:1, formula $C_6H_6O$.

Answer:

c. $C_6H_6O$