150 : 130 :: 5 : _x_ _x_ = 4.33 grams. A problem of frequent occurrence is to find the percentage composition of a substance when its formula has been given. For example: "What percentage of iron is contained in a mineral having the formula 2Fe_{2}O_{3}.3H_{2}O?" Bringing this formula together we have Fe_{4}H_{6}O_{9}. Find the molecular weight. Fe_{4} = 224 = 56×4 H_{6} = 6 = 1×6 O_{9} = 144 = 16×9 --- 374 Then we get: 374 parts of the mineral contain 224 of iron. How much will 100 contain? 374 : 224 :: 100 : _x_ _x_ = 59.89. And the answer to the question is 59.89 per cent. Again, suppose the question is of this kind:--"How much crystallised copper sulphate (CuSO_{4}.5H_{2}O) will be required to make 2 litres of a solution, 1 c.c. of which shall contain 0.0010 gram of copper?" A litre is 1000 c.c., so, therefore, 2 litres of the solution must contain 0.001 gram × 2000, or 2 grams. How much crystallised copper sulphate will contain this amount of metal? Cu = 63.3 S = 32.0 O_{4} = 64.0 = 16×4 5H_{2}O = 90.0 = 18×5 ----- 249.3 If 63.3 grams of copper are contained in 249.3 grams of sulphate, in how much is 2 grams contained. 63.3 : 249.3 :: 2 grams : _x_ _x_ = 7.8769 grams. The answer is, 7.8769 grams must be taken. As a sample of another class of problem similar in nature to the last (but a little more complicated) take the following:--"What weight of permanganate of potash must be taken to make 2 litres of a solution, 100 c.c. of which shall be equivalent to 1 gram of iron?" In the first place the 2 litres must be equivalent to 20 grams of iron, for there are 20 × 100 c.c. in two litres. In the titration of iron by permanganate solution there are two reactions. First in dissolving the iron Fe + H_{2}SO_{4} = FeSO_{4} + H_{2} | V 56 and second, in the actual titration, 10FeSO_{4} + 2KMnO_{4} + 9H_{2}SO_{4}= 2MnSO_{4} | + 5Fe_{2}(SO_{4})_{3} + 2KHSO_{4} + 8H_{2}O V K = 39 Mn = 55 O_{4}= 64 --- 158 × 2 = 316 As before, attention is confined to the two substances under consideration--viz., Fe and KMnO_{4}. In the second equation, we find 316 parts of the permanganate are required for 10 molecules of FeSO_{4}; and in the first equation 56 parts of iron are equivalent to one molecule of FeSO_{4}, therefore 560 of iron are equivalent to 316 of permanganate; and the question is, How much of the permanganate will be equivalent to 20 grams of iron? 560 : 316 :: 20 grams : _x_. _x_= 11.286 grams. The answer is 11.286 grams.