(3) A more accurate method is the following. A eudiometer tube is filled with mercury and inverted in a vessel of the same liquid. A convenient amount of air is then introduced into the tube and its volume accurately noted. There is then introduced more than sufficient hydrogen to combine with the oxygen present in the inclosed air, and the volume is again accurately noted. The mixture is then exploded by an electric spark, and the volume is once more taken. By subtracting this volume from the total volume of the air and hydrogen there is obtained the contraction in volume due to the union of the oxygen and hydrogen. The volume occupied by the water formed by the union of the two gases is so small that it may be disregarded in the calculation. Since oxygen and hydrogen combine in the ratio 1: 2 by volume, it is evident that the contraction in volume due to the combination is equal to the volume occupied by the oxygen in the air contained in the tube, plus twice this volume of hydrogen. In other words, one third of the total contraction is equal to the volume occupied by the oxygen in the inclosed air. The following example will make this clear: Volume of air in tube 50.0 cc. Volume after introducing hydrogen 80.0 Volume after combination of oxygen and hydrogen 48.5 Contraction in volume due to combination (80 cc.-48.5 cc.) 31.5 Volume of oxygen in 50 cc. of air (1/3 of 31.5) 10.5 All these methods agree in showing that 100 volumes of dry air contain approximately 21 volumes of oxygen. 2. _Determination of nitrogen._ If the gas left after the removal of oxygen from a portion of air is passed over heated magnesium, the nitrogen is withdrawn, argon and the other rare elements being left. It may thus be shown that of the 79 volumes of gas left after the removal of the oxygen from 100 volumes of air, approximately 78 are nitrogen and 0.93 argon. The other elements are present in such small quantities that they may be neglected. 3. _Determination of carbon dioxide._ The percentage of carbon dioxide in any given volume of air may be determined by passing the air over calcium hydroxide or some other compound which will combine with the carbon dioxide. The increase in the weight of the hydroxide equals the weight of the carbon dioxide absorbed. The amount present in the open normal air is from 3 to 4 parts by volume in 10,000 volumes of air, or about 0.04%. 4. _Determination of water vapor._ The water vapor present in a given volume of air may be determined by passing the air over calcium chloride (or some other compound which has a strong affinity for water), and noting the increase in the weight of the chloride. The amount present varies not only with the locality, but there is a wide variation from day to day in the same locality because of the winds and changes in temperature.