When more than one atom of an element is present this is shown by writing a figure under and after the symbol; thus, FeS_{2} represents a molecule with one atom of iron and two atoms of sulphur, Fe_{2}S_{3} similarly shows one with two atoms of iron and three of sulphur. When a group of atoms is enclosed in brackets, a figure after and under the bracket multiplies all within it; for example, Pb(NO_{3})_{2} is another way of writing PbN_{2}O_{6}. Sometimes it is convenient to represent the atoms of a molecule as divided into two or more groups; this may be done by writing the formulæ of the groups, and separating each simple formula by a full stop. Slaked lime, for instance, has the formula CaH_{2}O_{2}; or, as already explained, we may write it Ca(HO)_{2}; or, if for purposes of explanation we wished to look on it as lime (CaO) and water (H_{2}O), we could write it CaO.H_{2}O. A plus sign (+) has a different meaning; CaO + H_{2}O indicates quantities of two substances, water and lime, which are separate from each other. The sign of equality (=) is generally used to separate a statement of the reagents used from another statement of the products of the reaction; it may be translated into the word "yields" or "becomes." The two statements form an equation. Ignoring the quantitative relation, the meaning of the equation CaO + H_{2}O = CaO.H_{2}O is: "lime and water yield slaked lime." By referring to a table of atomic weights we can elicit the quantitative relations thus:-- CaO + H_{2}O = CaH_{2}O_{2} | | | V V V Ca = 40 H_{2} = 2 = 1×2 Ca = 40 O = 16 O = 16 H_{2} = 2 = 1×2 -- -- O_{2} = 32 = 16×2 56 18 -- 74 Or, putting it in words, 56 parts of lime combine with 18 parts of water to form 74 parts of slaked lime. This equation enables one to answer such a question as this:--How much lime must be used to produce 1 cwt. of slaked lime? for, if 74 lbs. of slaked lime require 56 lbs. of lime, 112 lbs. will require (56×112)/74, or about 84-3/4 lbs. As another example having a closer bearing on assaying take the following question:--"In order to assay 5 grams of 'black tin' (SnO_{2}) by the cyanide process, how much potassic cyanide (KCN) will be required?" The reaction is SnO_{2} + 2KCN = Sn + 2KCNO | | V V Sn = 118 K = 39 O_{2} = 32 C = 12 --- N = 14 150 -- 65×2 = 130 What is sought for here is the relation between the quantities of SnO_{2} and KCN. Note that a figure before a formula multiplies all that follows up to the next stop or plus or equality sign. The question is now resolved to this: if 150 grams of oxide of tin require 130 grams of cyanide, how much will 5 grams require?