The major salts used in formulation correspond approximately to the major constituents of ocean water shown on Table 1. To ascertain how large a quantity of salt may be required it is usual practice to convert the natural readings into gram-equivalents, from their original grams per kilo quantities , as shown on the Table. This process is best described by giving an example of the technique.
Column   (a } of Table 1  shows the   weight  of nine final elements required. These g/kg weights are then divided. by the atomic weight of each element concerned to convert them into gram-equivalent weights, eg;

Weight of element in seawater divided by the atomic weight of the same element = Gram-Equivalent weight

Each manufacturer of laboratory chemicals states the molecular weight of a reagent or salt on its packaging and labels, if not also in their catalogues. A simple multiplication will then provide a working result as;- 
Gram-Equivalent weight X Molecular weight =Final weight of salt needed.

(The mathematics involved also of course work in reverse, and formulae stating weights of salt can be converted back into weights of elements per kilo and then compared to Table 1.}
I t is imperative that, if a formula is being 'worked out , one works in reverse order dealing first with the minor constituents, then on up into the major constituents to the   penultimate figure ;the one for sodium chloride.
The salt   containing -bicarbonate is the last to be calculated, enough HCO ions (theoretically) being added to use any spare sodium ions  as yet unpaired  ,remembering that any ion carrying a positive {+) indicator has to be paired with a similar ion carrying a {-) negative indicator and that where an ion carries two symbols of one type, it requires double the quantity of   an   opposite type carrying only one symbol e.g. Na+ plus C1- = NaCl, which is common salt but Na+ plus S04--  =  Na2S04,or sodium .sulphate.
   Thus   only ten gram equivalents of chlorine (Cl} are needed  to neutralise ten gm-equiv. of sodium in the first
example but twenty gram-equivalents of sodium would be used in neutralising ten -gram-equivalents of sulphate in the second.
Let us assume that we wish to produce an extremely simple and not too accurate formula, and follow the complete process: ***

The final solution is to contain the following elements the weights of which are given in grams per kilo.= g/kg
Sodium. Na+)10•9   Chloride(C1-)19•4 Magnesium Mg++)1•25
Calcium Ca++)0•5   Bromide (Br-) 0•1   Potassium K+) 0•4 Sulphate(S04--)2.5 Boric Acid (H3BO3) which is neutral,0•5
and lastly bicarbonate of, as yet, unknown quantity.

2.We intend using the following commonly available salts.
Common salt, Sodium chloride NaCl Molecular weight= 58•44 Anhydrous Magnesium chloride MgC12.   MW = 95•22
Potassium chloride KC1.   MW= 74•56 Magnesium sulphate(crystallino)MgS047H20 MW = 246•48
Calcium chloride. Dry. CaCl 2H20   MW = 110•99
Boric Acid.H3803,which is already a salt and needs no MW. calculations, Potassium bromide. KBr MW = 119•01
lastly Sodium bicarbonate. Na HC03   MW =   84•01

3e The elements and radicals concerned have the   following atomic weights according to the IUPAC 1973 list.
Sodium = 22•989   Chlorine = 35•453   Magnesium = 24•305 Calcium= 40•08   Bromine = 79•904   Potassium = 39.098 Sulphate = (S04 = Sulphur (32906)plus 4xOxygen(15•999))or 96•057
Boric acid, which is a salt requiring no further calculations.
Bicarbonate =(HCO3=1xHydrogen,1xCarbon & 3 x Oxygen ) or 61•017

***The measurements given in this section are as examples and are not intended for use.
Scientifically, elements have atomic weights whereas radicals or combinations (eg:HCO3 ) have molecular weights, which as shown are the totals of the atomic weights of all component elements.This section has intentionally been very much simplified,as has much of this chapter.(See preface)  

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