Zinc phosphate is used as a dental cement. A 50.00-mg sample is broken down into its constituent elements and gives 16.58 mg oxygen, 8.02 mg phosphorus, and 25.40 mg zinc. Determine the empirical formula of zinc phosphate.

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Chemistry

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11 March, 12:58

Zinc phosphate is used as a dental cement. A 50.00-mg sample is broken down into its constituent elements and gives 16.58 mg oxygen, 8.02 mg phosphorus, and 25.40 mg zinc. Determine the empirical formula of zinc phosphate.

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Aliya Patton · Answer 1

Zn3P2O8

Explanation:

In this particular question, it is necessary to convert the respective masses to percentages. We convert to percentages by placing each mass over the total mass and multiplying by 100%. Since the total is 50mg, conversion to percentage can be done by multiplying the masses by 2 as 100/50 is 2

For Oxygen = 16.58 * 2 = 33.16%

For phosphorus = 8.02 * 2 = 16.04%

For zinc = 25.40 * 2 = 50.80%

We then proceed to divide these percentages by their respective atomic masses. The atomic mass of oxygen, phosphorus and zinc are 16, 31 and 65 respectively.

O = 33.16/16 = 2.0725

P = 16.04/31 = 0.5174

Zn = 50.80/65 = 0.7815

Now, we divide by the smallest value which is that of the phosphorus

O = 2.0725/0.5174 = 4

P = 0.5174/0.5174 = 1

Zn = 0.7815/0.5174 = 1.5

Now, we need to multiply through by 2. This yields: O = 8, P = 2 and Zn = 3

The empirical formula is thus: Zn3P2O8