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An ionic compound contains 29.08% sodium, 40.56% sulfur and 30.36% oxygen by mass. What...

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lak-86 | Student, Undergraduate | Salutatorian

Posted July 17, 2013 at 11:59 PM via web

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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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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 18, 2013 at 1:06 AM (Answer #1)

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molar mass

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`

 

Mole ratio

`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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