MAKING OF FORMULA AND EQUATIONS 101 



the same after the action as before it. Now the skeleton equation 

 shows two atomic weights of oxygen before, and, thus far only one 

 after the action, whereas there ought to be two there also. With 

 O 2 (2 X 16 parts) we have enough oxygen to give 2H 2 0, which 

 contains 2 X 16 parts of oxygen. But this will require us to take 

 2H 2 to " balance " the equation. The final equation is, therefore: 



Balanced Equation: 2H 2 + O 2 -> 2H 2 O. 



Observe, we could not adjust the difficulty by writing H 2 + O 

 > H 2 0, because each substance must be represented by its molec- 

 ular formula, which stands for the weight of the substance in the 

 standard volume of 22.4 liters, or one chemical unit weight, and 

 in the case of oxygen this is 2 ( = 32 g.). Putting this in terms 

 of the hypothesis, each formula must represent 1 molecule, and 

 the molecules of oxygen contain 2 atoms. Hence we could not 

 divide the oxygen molecule. But we could take more than one 

 molecule of hydrogen, so we took 2 molecules of this substance. 



The coefficients in front of formulae multiply the whole formula. 

 2H 2 is equivalent to 2(H 2 0), or two whole molecules of water. 



Balancing Equations. Learning to balance equations 

 correctly comes only by practice. Take, again, the case of iron 

 rusting. The substances are iron (Fe), oxygen (0 2 ) and ferric 

 oxide (Fe 2 3 ). The skeleton equation is 



Fe + 2 - Fe 2 O 3 . 



We are not permitted to alter these formulae themselves, but we 

 may put coefficients in front of any of them to make the number 

 ojt atomic weights alike onjboth sides. A good rule is to pick out 

 the largest formula and reason back from that. Here, this is Fe 2 O 3 . 

 To get oxygen atoms in threes, we must clearly take 3O 2 (= 60). 

 That will give us 2Fe 2 O 3 . This, in turn, will require 4Fe: 



Balanced: 4Fe + 3O 2 > 2Fe 2 O 3 . 



