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

`\large\boxed{\large\boxed{C_3H_3O}}`

Step-by-step explanation:

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

`1.24gCO_2* (12.011gC)/(44.01gCO_2)=0.3384gC`

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

`0.255gH_2O* (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`