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.

Question

Natasha Klein

Chemistry

14 May, 09:08

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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Answers (1)

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Sydnee Branch · 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