An ionic compound contains 29.08% sodium, 40.56% sulfur and 30.36% oxygen by mass. What is the formula of the sulfur-containing anion in the compound?
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Na = 23
Sulphur = 32
Oxygen = 16
If we assume we have 100g of the compound then we should have 29.08g of Na, 40.56g of S and 30.36g of O in it.
Moles of Na in the compound `= 29.08/23 = 1.264`
Moles of S in the compound `= 40.56/32 = 1.268`
Moles of O in the compound `= 30.36/16 = 1.898`
`Na:S:O = 1.264:1.268:1.898 = 1.264/1.264:1.268/1.264:1.898/1.264`
`Na:S:O = 1:1:1.5 = 2:2:3`
So in the anion the S:O ratio is 2:3. So the possible anion is `S_2O_3^(2-)`
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