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Many samples of chemicals are not pure. We can define percent purity as
Many samples of chemicals are not pure. We can define percent purity as
mass of pure compound in the impure sample x 100%
total mass of impure sample
If an impure sample of a chemical of known percent purity is used in a chemical reaction, the percent purity has to be used in stoichiometric calculations. Conversely, the percent purity of an impure sample of a chemical of unknown percent purity can be determined by reaction with a pure compound as in an acid-base titration. Percent purity can also be determined, in theory, by measuring the amount of product obtained from a reaction. This latter approach, however, assumes a 100% yield of the product.
Examples
3Mg(OH)2 + 2H3PO4 ----> Mg3(PO4)2 + 6H2O
(a) Calculate the mass of Mg3(PO4)2 that will be formed (assuming a 100% yield) from the reaction of 15.0 g of 92.5% Mg(OH)2 with an excess of H3PO4
mass Mg(OH)2 = 15.0 x 0.925 = 13.875 g
mass Mg3(PO4)2 =
13.875 g Mg(OH)2 x 1 mol Mg(OH)2 x 1 mol Mg3(PO4)2 x 262.9 g Mg3(PO4)2
58.3 g Mg(OH)2 3 mol Mg(OH)2 1 mol Mg3(PO4)2
= 20.9 g Mg3(PO4)2
(b) Calculate the mass of 88.5% Mg(OH)2 needed to make 127 g of Mg3(PO4)2, assuming a 100% yield.mass Mg(OH)2 =
127 g Mg3(PO4)2 x 1 mol Mg3(PO4)2 x 3 mol Mg(OH)2 x 58.3 g Mg(OH)2
262.9 g Mg3(PO4)2 1 mol Mg3(PO4)2 1 mol Mg(OH)2
= 84.49 g Mg(OH)2.
mass 88.5% Mg(OH)2 =
84.49 g Mg(OH)2 x 100 g 88.5% Mg(OH)2 = 95.5 g
88.5 g Mg(OH)2
(c) Calculate the percent purity of a sample of Mg(OH)2 if titration of 2.568 g of the sample required 38.45 mL of 0.6695 M H3PO4.
mass Mg(OH)2 =
38.45 mL H3PO4 x 0.6695 mole H3PO4 x 3 moles Mg(OH)2
1000 mL H3PO4 2 moles H3PO4
x 58.3 g Mg(OH)2
1 mole Mg(OH)2
= 2.251 g Mg(OH)2
Percent Purity = 2.251 x 100% = 87.7%
2.568
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