Instruction

1

First and foremost, you should remember a few simple rules. If we are talking about the metal, its equivalent mass is calculated by the formula: Me=M/V, where M is the atomic mass

**of the metal**, and In his valence. Consider this rule with specific examples.2

Calcium (CA). Its atomic weight 40,08. Rounded take it over 40. Valency equal to 2. Consequently, DOE(Sa) = 40/2 = 20 g/mol. Aluminium (Al). Its atomic weight 26,98. (Rounded 27). Valence equal to 3. Thus, Mae(Al) = 27/3 = 9 g/mol.

3

These methods are applicable if we are talking about pure

**metal**X. But if they are part of a connection, for example, the same oxides? Here we must remember another rule: equivalent mass**of the oxide**is calculated by the formula: Me + Mo, where Mo is the equivalent mass of oxygen. It is, accordingly, calculated according to the formula already considered, that is, 16/2 = 8.4

Suppose you have a primary aluminium oxide, Al2O3. How to calculate equivalent weight? Very simple: 27/3 + 16/2 = 17 g/mol.

5

Is there another way to determine the equivalent mass

**of the metal**and its**oxide**? Yes, and very effective. It is based on the so-called law of equivalents, according to which all substances react with each other in equivalent quantities. For example: metal weight 33.4 grams entered into reaction of oxidation with oxygen. The result is an oxide with a total mass of 43 grams. You want to determine the equivalent*mass*of the**metal**and its**oxide**.6

First, calculate how much oxygen combines with the metal in the course of this reaction: 43 – 33,4 = 9.6 grams. According to the law of equivalents, this mass is as much larger than the equivalent

*mass*of oxygen (which, recall, is equal to 8), many times the equivalent mass**of the metal**is less than its original amount. That is 33,4/DOE(IU) = 9,6/8. Consequently, Me(me) = 33,4*8/9,6 = 27,833 g/mol, or about 27.8 g/mol. This is the equivalent mass**of the metal**.7

Equivalent weight

**of oxide**find the following: 27,8 + 8 = 35,8 g/mol.