Question

A compound is composed of only C, H, and O. The combustion of a 0.519-g sample of the compound yields 1.24 g of CO 2 and 0.255 g of H 2O. What is the empirical formula of the compound

Answer 1

Answer:

      $\large\boxed{\large\boxed{C_3H_3O}}$

Explanation:

1. Calculate the mass of C in 1.24 g of CO₂

   $1.24gCO_2\times \dfrac{12.011gC}{44.01gCO_2}=0.3384gC$

2. Calculate the mass of H in 0.255g of H₂O

     $0.255gH_2O\times \dfrac{2.016gH}{18.015gH_2O}=0.02854gH$

3. Calculate the mass of O by difference

     $0.519g-0.3384g-0.02854g=0.15206g$

4. Convert every mass to number of moles:

• C: 0.3384g / (12.011g/mol) = 0.02817 mol

• H: 0.02854g / (1.008g/mol) = 0.02831 mol

• O: 0.15206g / (15.999g/mol) = 0.009504 mol

5. Divide every number of moles by the least number of moles (0.009504):

• C: 0.02817/0.009504 = 2.96 ≈ 3

• H: 0.02831 / 0.009504 ≈ 3

• O: 0.009504 / 0.009504 = 1

6. Those numbers are the respective subscripts in the empirical formula:

     $C_3H_3O$