Respuesta :

Answer:

Empirical formula of the compound is [tex]HgSO_{4}[/tex].

Explanation:

      Element                                            molar mass (g/mol)

         Hg                                                            200.59

          S                                                              32.065

          O                                                              15.999

100 g of the compound contains 67.6 g Hg, 10.8 g of S and 21.6 g of O.

So ratio of number of moles (n) of all the constituting elements in 100 g of the compound is given as:      

[tex]n_{Hg}:n_{S}:n_{O}[/tex]   = [tex]\frac{67.6}{200.59}:\frac{10.8}{32.065}:\frac{21.6}{15.999}[/tex]

                        = 0.337 : 0.337 : 1.35

                       [tex]\approx[/tex] 1: 1: 4

So empirical formula of the compound is [tex]HgSO_{4}[/tex]

samueladesida43

The empirical formula of a compound that is 67.6% mercury, 10.8% sulfur, and 21.6% oxygen is HgSO4.

HOW TO CALCULATE EMPIRICAL FORMULA:

  • The empirical formula of the compound in this question can be calculated as follows:

  1. Hg = 67.6% = 67.76g
  2. S = 10.8% = 10.8g
  3. O = 21.6% = 21.6g

  • First, we convert mass to moles by dividing each element by its molar mass.

  1. Hg = 67.6g ÷ 200.59g/mol = 0.34mol
  2. S = 10.8g ÷ 32g/mol = 0.34mol
  3. O = 21.6g ÷ 16g/mol = 1.35mol

  • Next, we divide each mole value by the smallest value (0.34)

  • Hg = 0.34mol ÷ 0.34 = 1
  • S = 0.34mol ÷ 0.34 = 1
  • O = 1.35mol ÷ 0.34 = 3.97

  • Approximately, the empirical ratio of Hg, S and O is 1:1:4. Hence, the empirical formula of the compound is HgSO4.

Learn more at: https://brainly.com/question/14044066?referrer=searchResults

Otras preguntas