STOICHIOMETRY AND PERCENT PURITY Many samples of chemicals are not pure. We can define percent purity as
mass of pure compound in the impure sample
total mass of impure sample x 100
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
Consider the reaction of magnesium hydroxide with phosphoric acid.
3Mg(OH)2 + 2H3PO4 H 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 mole Mg3(PO4)2
x 58.3 g Mg(OH)2
1 mole Mg(OH)2
3 moles Mg(OH)2 1 mole Mg3(PO4)2
262.9 g Mg3(PO4)2
x
= 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 mole Mg3(PO4)2
x
58.3 g Mg(OH)2
1 mole Mg(OH)2
3 moles Mg(OH) 1 mole Mg3(PO4)2 2
262.9 g Mg3(PO4)2
x
= 84.49 g Mg(OH)2.
mass 88.5% Mg(OH)2 = = 95.5 g x
88.5 g Mg(OH)2
100 g 88.5% Mg(OH)2
84.49 g Mg(OH)2
-5b-
(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 x x
58.3 g Mg(OH)2
1 mole Mg(OH)2
3 moles Mg(OH) 0.6695 mole H3PO4 2
2 moles H3PO4 1000 mL H3PO4
= 2.251 g Mg(OH)2. Percent purity = = 87.7%