It is found by experiment that 88.86 parts by weight of oxygen, combining with 11.14 parts of hydrogen, form 100 parts of water; so that from the weight of water formed it is easy to calculate the amount of oxygen the oxide contained.
Take 1 gram of the dried and powdered oxide and place it in a warm dry combustion tube. Place the tube in a furnace, and connect at one end with a hydrogen apparatus provided with a sulphuric acid bulb for drying the gas, and at the other with a weighed sulphuric acid tube for collecting the water formed. The apparatus required is shown in fig. 62. Pass hydrogen through the apparatus, and, when the air has been cleared out, light the furnace. Continue the heat and current of hydrogen for half an hour (or longer, if necessary). Allow to cool. Draw a current of dry air through the weighed tube. Weigh. The increase in weight gives the amount of water formed, and this, multiplied by 0.8886, gives the weight of the oxygen. The percentage of oxygen thus determined should be compared with that got by the oxidation of the metal. It will be practically the same. The following results can be taken as examples:—
Twenty grams of tin, calcined as described, gave 25.37 grams of oxide.
Two grams of tin, oxidised with nitric acid and ignited, gave 2.551 grams of oxide.
One gram of the oxide of tin, on reduction in a current of hydrogen, gave 0.2360 gram of water (equivalent to 0.2098 gram of oxygen), and left 0.7900 gram of metal.
Ten grams of ferrous sulphate gave, on strong ignition, 2.898 grams of ferric oxide (Fe2O3)[96] instead of 2.877.
The student should similarly determine the percentage of oxygen in oxides of copper and iron. The former oxide may be prepared by dissolving 5 grams of copper in 50 c.c. of dilute nitric acid, evaporating to dryness, and strongly igniting the residue. The oxide of iron may be made by weighing up 10 grams of powdered ferrous sulphate (= to 2.014 grams of iron) and heating, at first gently, to drive off the water, and then at a red heat, until completely decomposed. The weight of oxide, in each case, should be determined; and the percentage of oxygen calculated. Compare the figures arrived at with those calculated from the formula of the oxides, CuO and Fe2O3.
It would be found in a more extended series of experiments that the same metal will, under certain conditions, form two or more oxides differing among themselves in the amount of oxygen they contain. These oxides are distinguished from one another by such names as "higher" and "lower oxides," "peroxides," "protoxides," "dioxides," &c.
The oxides may be conveniently classified under three heads:—