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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

Respuesta :

Edufirst

Answer:

      [tex]\large\boxed{\large\boxed{C_3H_3O}}[/tex]

Explanation:

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

   [tex]1.24gCO_2\times \dfrac{12.011gC}{44.01gCO_2}=0.3384gC[/tex]

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

     [tex]0.255gH_2O\times \dfrac{2.016gH}{18.015gH_2O}=0.02854gH[/tex]

3. Calculate the mass of O by difference

     [tex]0.519g-0.3384g-0.02854g=0.15206g[/tex]

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:

     [tex]C_3H_3O[/tex]

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