104 APPENDIX. 



The symbol of Phosphoric acid isP 2 O u ; hence, there 

 are two equivalents of Phosphorus combined with five 

 equivalents of Oxygen requisite to make Phosphoric acid. 

 An equivalent of Phosphorus is 15.7, multiplied by 2 

 gives 31.4 ; an equivalent of Oxygen is 8, multiplied by 

 5 gives 40 ; and 31.4 added to 40 = 71.4, which is the 

 equivalent of Phosphoric acid, as expressed by its sym- 

 bol P 2 O 5 . 



NAME. BASE. ACID. EQUIVALENT. SYMBOL. 



Sulphate of Lime, 1 eq. -f 1 . . . 68.6 . . CaO, SOs 



" " as gypsum, with 2 eq. of water 18 .86.6. . 



Carbonate of Lame (Marble), . 1 eq. -f 1 eq. . . 50.62 . CaO, COz 

 Sulphate of Soda, . . . . . . . 1 eq. -f 1 eq. . . 71.4 . . NaO, SOz 



" " in crystals, with 10 eq. of water 90. 161.4. . 



Nitrate of Soda, 1 eq. -f 1 eq. . . 85.45 . . AaO, NOs 



Chloride of Sodium (com. salt), 1 eq. -f- 1 eq. of Cl 58.72 . . NaO 

 Nitrate of Potassa (saltpetre), . 1 eq. -j- 1 eq. . 101.3 . . KO> NO?> 

 Carbonate of Magnesia, . . . . 1 eq. + 1 eq. . . 42.82 . . MgO> CO t 

 Sulphate of Ammonia, . . . . 1 eq. -}- 1 eq. . . 66.25 . . NH O, 503 

 Nitrate of (oxide of) Ammonia, 1 eq. -f 1 eq. . . 80.3 . . Hi NO,NOb 

 Muriate of Ammonia, 1 eq. -f 1 eq. . . 53.57 . . H\ JV, HCl 



The teacher should analyze and explain the above sym- 

 bols, showing his class the value of the numerals, and 

 the part they play in the formation of the articles they 

 represent. Thus, Sulphate of Lime is composed of Ca O 

 and SO 3 ; Ca, the symbol of Calcium (20.5), and 0, 

 the symbol of Oxygen (8) , represent that combination 

 called Oxide of Calcium ; one equivalent of this Oxide 

 of Calcium is combined with one equivalent of Sulphuric 

 acid; S, the symbol of Sulphur (16.1), and O 8 , the 

 symbol of Oxygen (8) , multiplied by 3. Now, we may 

 show the formation of the Sulphate of Lime, and the 



