CaO+H2O=CaH2O2
↓↓↓
Ca = 40H2 = 2 = 1×2Ca = 40
O = 16O = 16H2 = 2 = 1×2
————O2 = 32 = 16×2
5618——
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' (SnO2) by the cyanide process, how much potassic cyanide (KCN) will be required?" The reaction is

SnO2+2KCN= Sn + 2KCNO
↓↓
Sn = 118K = 39
O2 = 32C = 12
——N = 14
150——
65×2 = 130

What is sought for here is the relation between the quantities of SnO2 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?

150 : 130 :: 5 : x
x = 4.33 grams.

A problem of frequent occurrence is to find the percentage composition of a substance when its formula has been given. For example: "What percentage of iron is contained in a mineral having the formula 2Fe2O3.3H2O?" Bringing this formula together we have Fe4H6O9. Find the molecular weight.

Fe4= 224= 56×4
H6= 6= 1×6
O9= 144= 16×9
——
374

Then we get: 374 parts of the mineral contain 224 of iron. How much will 100 contain?