Suppose that a laboratory analysis of white powder showed 42.59% Na, 12.02% C, and 44.99% oxygen. Would you report that the compound is sodium oxalate or sodium carbonate? (Use 43.38% Na, 11.33% C, and 45.29% O for sodium carbonate, and 34.31% Na, 17.93% C, and 47.76% O for sodium oxalate.)

Question

Chemistry

Adonis Marks

9 March, 23:20

Suppose that a laboratory analysis of white powder showed 42.59% Na, 12.02% C, and 44.99% oxygen. Would you report that the compound is sodium oxalate or sodium carbonate? (Use 43.38% Na, 11.33% C, and 45.29% O for sodium carbonate, and 34.31% Na, 17.93% C, and 47.76% O for sodium oxalate.)

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Answers (1)

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Deanna Gomez · Answer 1

The white powder is sodium carbonate

Explanation:

If you supposed an ammount of 100 grams of the white powder, it means there are 42.59 grams of Na, 12.02 grams of C and 44.99 grams of O. If you divided every compound by its molar mass, you will know the moles of every compound. (According to moles=mass (grams) / molar mass (grams/mole)

C=12.01 g/mole, Na=23 g/mole, O=16 g/mole

So:

C=12.02/12.01 = 1.0 moles, Na=42.59 / 23 = 1.85 moles, O = 44.99/16 = 2.81 moles

Knowing this, we can stimated that the white powder has 1 mol of C and approximately 2 and 3 moles of N and O, respectively

Chemical formula for sodium carbonate is Na2CO3 and the formula for sodium oxalate is Na2C2O4. So the white powder is Na2CO3