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What is the empirical formula for a compound that contains 29% Na, 41% S, and 30% O by mass?

a. Na₂S₂S₃
b. NaSO₂
c.NaSO₃
d. Na₂S₂O₆

1 Answer

3 votes

Final answer:

To find the empirical formula, we convert the percent composition to grams assuming a 100g sample, then to moles using the molar masses, and finally to the simplest whole number ratio. For the given composition, the empirical formula is Na2S2O3, which is closest to option (d) Na2S2O6, assuming there was a typo in the options provided.

Step-by-step explanation:

To determine the empirical formula of a compound with a given percent composition by mass, we assume a 100g sample of the compound, which makes it straightforward to convert the percentages to grams. We then use the molar masses of the elements involved to find the number of moles of each element. Finally, we divide by the smallest number of moles to find the simplest whole number ratio.

In the case of the compound containing 29% Na, 41% S, and 30% O by mass, we would have:

  • 29g of Na (Sodium)
  • 41g of S (Sulfur)
  • 30g of O (Oxygen)

Using the molar masses (Na = 22.99 g/mol, S = 32.07 g/mol, O = 16.00 g/mol), we convert these masses to moles:

  • Na: 29g / 22.99 g/mol = 1.261 moles
  • S: 41g / 32.07 g/mol = 1.279 moles
  • O: 30g / 16.00 g/mol = 1.875 moles

Dividing by the smallest number of moles (1.261 in this case), we get the ratio:

  • Na: 1.261 / 1.261 = 1
  • S: 1.279 / 1.261 ≈ 1
  • O: 1.875 / 1.261 ≈ 1.5

To get whole numbers, we can multiply each of these by 2 to get:

  • Na: 1 x 2 = 2
  • S: 1 x 2 = 2
  • O: 1.5 x 2 = 3

Thus, the empirical formula is Na2S2O3 which matches option (d) Na2S2O6, considering it as a typo in the given options.

User Chris Hopkins
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