It is evident from the circumstance that in all the higher types the _acid hydroxides_ (for example, HClO_{4}, H_{2}SO_{4}, H_{3}PO_{4}) and salts with a single atom of one element contain, like the higher saline type RO_{4}, _not more than four atoms of oxygen_; that the formation of the saline oxides is governed by a certain common principle which is best looked for in the fundamental properties of oxygen, and in general of the most simple compounds. The hydrate of the oxide RO_{2} is of the higher type RO_{2}2H_{2}O = RH_{4}O_{4} = R(HO)_{4}. Such, for example, is the hydrate of silica and the salts (orthosilicates) corresponding with it, Si(MO)_{4}. The oxide R_{2}O_{5}, corresponds with the hydrate R_{2}O_{5}3H_{2}O = 2RH_{3}O_{4} = 2RO(OH)_{3}. Such is orthophosphoric acid, PH_{3}O_{3}. The hydrate of the oxide RO_{3} is RO_{3}H_{2}O = RH_{2}O_{4} = RO_{2}(OH)_{2}--for instance, sulphuric acid. The hydrate corresponding to R_{2}O_{7} is evidently RHO = RO_{3}(OH)--for example, perchloric acid. Here, besides containing O_{4}, it must further be remarked that _the amount of hydrogen in the hydrate is equal to the amount of hydrogen in the hydrogen compound_. Thus silicon gives SiH_{4} and SiH_{4}O_{4}, phosphorus PH_{3} and PH_{3}O_{4}, sulphur SH_{2} and SH_{2}O_{4}, chlorine ClH and ClHO_{4}. This, if it does not explain, at least connects in a harmonious and general system the fact that _the elements are capable of combining with a greater amount of oxygen, the less the amount of hydrogen which they are able to retain_. In this the key to the comprehension of all further deductions must be looked for, and we will therefore formulate this rule in general terms. An element R gives a hydrogen compound RH_{_n_}, the hydrate of its higher oxide will be RH_{_n_}O_{4}, and therefore the higher oxide will contain 2RH_{_n_}O_{4} - _n_H_{2}O = R_{2}O_{8 - _n_}. For example, chlorine gives ClH, hydrate ClHO_{4}, and the higher oxide Cl_{2}O_{7}. Carbon gives CH_{4} and CO_{2}. So also, SiO_{2} and SiH_{4} are the higher compounds of silicon with hydrogen and oxygen, like CO_{2} and CH_{4}. Here the amounts of oxygen and hydrogen are equivalent. Nitrogen combines with a large amount of oxygen, forming N_{2}O_{5}, but, on the other hand, with a small quantity of hydrogen in NH_{3}. _The sum of the equivalents of hydrogen and oxygen_, occurring in combination with an atom of nitrogen, is, as always in the higher types, equal to _eight_. It is the same with the other elements which combine with hydrogen and oxygen. Thus sulphur gives SO_{3}; consequently, six equivalents of oxygen fall to an atom of sulphur, and in SH_{2} two equivalents of hydrogen. The sum is again equal to eight. The relation between Cl_{2}O_{7} and ClH is the same. This shows that the property of elements of combining with such different elements as oxygen and hydrogen is subject to one common law, which is also formulated in the system of the