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What is the empirical formula for a compound that contains 12.62% Li, 29.17% S, and 58.21% O?

a) LiSO₂
b) LiS₂O₆
c) Li₂SO₄
d) Li₂SO₃

1 Answer

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Final answer:

The empirical formula for a compound containing Li, S, and O with the given percent composition is calculated using the atomic masses and moles, resulting in the empirical formula Li2SO4.

Step-by-step explanation:

To determine the empirical formula for a compound with the given percent composition of Li, S, and O, we can follow a step-by-step approach based on the percentages given:

  • Convert the percentages to moles by dividing them by the atomic masses of each element: Li (atomic mass 6.94), S (atomic mass 32.07), O (atomic mass 16.00).
  • Divide the number of moles of each element by the smallest number of moles obtained in step one.
  • Round the resulting ratios to the nearest whole number to obtain the subscripts in the empirical formula.

The calculations are as follows:

  • Moles of Li: 12.62% ÷ 6.94 = 1.818 moles
  • Moles of S: 29.17% ÷ 32.07 = 0.910 moles
  • Moles of O: 58.21% ÷ 16.00 = 3.638 moles

The ratio of moles Li:S:O is then 1.818:0.910:3.638. Dividing each by the moles of S (the smallest number), we get approximately 2:1:4.

The empirical formula is Li2SO4, so the answer is (c) Li2SO4.

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