Therefore it appears that 8 grains of Oxygen are _equivalent_ to 36 grains of Chlorine and to 126 grains of Iodine, seeing that these quantities all play the same part in combining; and so it is with regard to the other elements,--to every one of them a figure can be assigned which represents the number of parts by weight in which that element unites with others. These figures are the "equivalents" or "combining proportions," and they are denoted by the _symbol_ of the element. A symbol does not stand as a simple representative of an element, but as a representative of _one equivalent_ of an element. Thus "O" indicates 8 parts by weight of Oxygen; "Cl" one equivalent, or 36 parts by weight, of Chlorine; and so with the rest. Observe however that these figures, termed "equivalents," do not refer to the _actual number_ of parts by weight, but only to the _ratio_ which exists between them: if Oxygen is 8, then Chlorine is 36; but if we term Oxygen 100, as some have proposed, then Chlorine would be 442·65. In the scale of equivalents now usually adopted, Hydrogen, as being the lowest of all, is taken as unity, and the others are related to it. _Equivalents of Compounds._--The law of equivalent proportions applies to compounds as well as to simple bodies, the combining proportion of a compound being always the sum of the equivalents of its constituents. Thus Sulphur is 16, and Oxygen 8, therefore Sulphuric Acid, or SO{3}, equals 40. The equivalent of Nitrogen is 14, that of Nitric Acid, or NO{5}, is 54. The same rule applies with regard to salts. Take for instance the Nitrate of Silver: it contains Equivalent. Nitrogen 14 6 Oxygen 48 Silver 108 --- Total of equivalents, or equivalent } 170 of the Nitrate of Silver } _Practical application of the Laws of Combination ._--The utility of being acquainted with the law of combining proportions is obvious when their nature is understood. As bodies both unite with and replace each other in equivalents, a simple calculation shows at once how much of each element or compound will be required in a given reaction. Thus, supposing it be desired to convert 100 grains of Nitrate of Silver into _Chloride_ of Silver, the weight of Chloride of Sodium which will be necessary is deduced thus:--one equivalent, or 170 parts, of Nitrate of Silver, is decomposed by an equivalent, or 60 parts, of Chloride of Sodium. Therefore as 170 : 60 :: 100 : 35·2; that is, 35·2 grains of Salt will precipitate, in the state of Chloride, the whole of the Silver contained in 100 grains of Nitrate.