An organic compound contains 60% carbon, 13.3% hydrogen and 26.7% oxygen. Calculate the empirical formula. (C=12, H=1, O=16)

Correct answer: C₃H₈O

Explanation: Assume 100 g of the compound.

• C: 60 g → \( \frac{60}{12} = 5 \) mol
• H: 13.3 g → \( \frac{13.3}{1} = 13.3 \) mol
• O: 26.7 g → \( \frac{26.7}{16} \approx 1.66875 \) mol

Divide by the smallest (1.66875):

C: \( \frac{5}{1.66875} \approx 3 \),  H: \( \frac{13.3}{1.66875} \approx 8 \),  O: \( \frac{1.66875}{1.66875} = 1 \).

Empirical ratio = C₃H₈O.

emperical formular=    C            H            O

                                    60%        13.3%      26.7%

Divide by the mass/no=  60/12    13.3/1  26.7/16

                                    =5         13.3      1.66875

Divivde by the smallest=  5/1.66875  13.3/1.66875  1.66875/1.66875

                                      =3        8          1

Emperical formular  =C3H8O

                                    =3             8              1