Various formulas will help to find amount of substance which unit of measure is the mol. Also the amount of substance can be found on the equation of the reaction given in a task.
Instruction
1. Having the weight and the name of substance without effort it is possible to find amount of substance: n = m/M where n is amount of substance (mol), m is the mass of substance (g), M - the molar mass of substance (g/mol). For example, the mass of chloride of sodium is equal to 11.7 g, to find amount of substance. To substitute necessary values in a formula, it is necessary to find the molar mass of chloride of sodium: M (NaCl) = 23+35.5 = 58.5 g/mol. We substitute: n(NaCl) = 11.7/58.5 = 0.2 mol.
2. If it is about gases, then such formula takes place: n = V/Vm where n is amount of substance (mol), V - the volume of gas (l), Vm - the molar volume of gas. Under normal conditions (pressure of 101,325 Pas and temperature 273 J) the molar volume of gas is a constant and is equal to 22.4 l/mol. For example, what amount of substance will have nitrogen of 30 l under normal conditions? n(N2) = 30/22.4 = 1.34 mol.
3. One more formula: n = N/NA where n is amount of substance (mol), N is number of molecules, NA is constant Avogadro, it is equal 6.02*10 in 23 degrees (1/mol). For example, what amount of substance consists in 1.204*10 in 23 degrees? We solve: n = 1.204*10 in 23 degrees / 6.02*10 in 23 degrees = 0.2 mol.
4. On any equation of reaction it is possible to find amount of the substances which reacted and formed as a result of it. 2AgNO3 + Na2S = Ag2S + 2NaNO3. From this equation it is visible that 2 mol of nitrate of silver provzaimodeystvovat with 1 mol of sulfide of sodium, as a result of it it was formed by 1 mol of sulfide of silver and 2 mol of nitrate of sodium. By means of these amounts of substances it is possible to find other sizes which are required in tasks. Let's consider on an example. To the solution containing nitrate of silver weighing 25.5 g flowed the solution containing sodium sulfide. What amount of substance of sulfide of silver is formed at the same time? At first we find amount of substance of nitrate of silver, previously having calculated its molar weight. M (AgNO3) = 170 g/mol. n(AgNO3) = 25.5/170 = 0.15 mol. The reaction equation for this task is written above, follows from it that 1 mol of sulfide of silver is formed of 2 mol of nitrate of silver. We define how many mol of sulfide of silver are formed of 0.15 mol of nitrate of silver: n(Ag2S) = 0.15*1/2 = 0.075 mol.