From the time of the discovery, some 110 years ago, that water is a compound body, made up of oxygen and hydrogen, the notion prevailed up to within a quarter of a century that it was composed of even equivalents of the elements named, and all but the youngest students of chemistry well remember how its formula was written HO, the atomic weight of oxygen being expressed by 8, making the molecular weight of water (H=1 + O=8) 9. But the vapor density of water, referred to air, is 0.635, and this number multiplied by the constant 28.87, gives 18 as the molecular weight of water, or exactly twice that accepted by chemists. This discrepancy led to closer observations, and it was eventually found that in decomposing water, by whatever method (excepting only electrolysis), not more than the eighteenth part in hydrogen of the water decomposed was ever obtained, or, in other words, only just one-half the weight deducible from the formula HO = 9. The conclusion was irresistible that in a water molecule two atoms of hydrogen must be assumed, and, as a natural sequence, followed the doubling of the molecular weight of water to 18, represented by the modern formula H_{2}O. Both the theory and the practice of substitution enable us to further prove the presence of two hydrogen atoms in a water molecule. Decomposing water by sodium, only one-half of the hydrogen contained is eliminated, the other half, together with all of the oxygen, uniting with the metal to form sodium hydroxide, H_{2}O + Na = H + NaHO. Doubling the amount of sodium does not alter the result, for decomposition according to the equation H_{2}O + 2Na = H_{2} + Na_{2}O never happens. Introducing the ethyl group into the water molecule and reacting under appropriate conditions with ethyl iodide upon water, the ethyl group displaces one atom of hydrogen, and, uniting with the hydroxyl residue, forms ethyl alcohol, thus: H_{2}O + C_{2}H_{5}I = C_{2}H_{5}OH + HI. Halogens do not act directly on water, hence we may not properly speak of halogen substitution products. By the action, however, of phosphorus haloids on water an analogous splitting of the water molecule is again observed, one-half of the hydrogen uniting with the halogen to form an acid, the hydroxyl residue then forming a phosphorus compound, thus: PCl_{3} + 3H_{2}O = 3HCl + P(OH)_{3}. Now these examples, which might readily be multiplied, prove not only the presence of _two_ hydrogen atoms in the water molecule, but they further demonstrate that these two atoms _differ from each other_ in respect to their form of combination and power of substitution. The two hydrogen atoms are certainly not of equal value, whence it follows that the accepted formula for water: H > O H