Guide

1

To find

**mole****material, you need to remember a very simple rule: the mass of one mole of any substance****numerically equal to its molecular weight, only expressed in other terms.What determined the molecular weight?With the periodic table you know the atomic weight of each element, which is part of the molecule****substance**.Next you need to lay down their atomic weights based on the index of each element, and to get an answer. 2

For example, widely used in agricultural fertilizer, ammonium nitrate (or otherwise ammonium nitrate).The formula of this substance

**NH4NO3.How to determine what is it****mole**?First of all, note the empirical (ie common) formula**substance**: N2H4O3. 3

Count its molecular weight based on the index of each element: 12 * 2 + 1 * 4 + 16 * 3 = 76 amu(atomic mass units).Consequently, its molar mass (ie, the mass of one mole) also is 76, but its dimension: g /

**mole**.Answer:**one mole of ammonium nitrate**weighs 76 grams. 4

Suppose you set such a task.It is known that the mass of 179.2 liters of the gas is 352 grams.You must determine how much weight one

**moles of gas.It is known that under normal conditions, one mole****any gas or gas mixture occupies a volume of approximately 22.4 liters.And you 179.2 liters.Carry out the calculation: 179.2 / 22.4 = 8. Therefore, in this volume of gas contained 8 moles.** 5

Dividing known under the terms of a lot of the problem on the number of moles, get: 352/8 = 44. Therefore, one mole

**this gas weighs 44 grams - is carbon dioxide, CO2.** 6

If there is a certain amount of gas mass M enclosed in a volume V at a given temperature T and pressure P. is required to determine its molar mass (that is, to find what it is

**mole**).Solve the problem will help universal equation Mendeleev-Clapeyron: PV = MRT / m, where m - the molar mass is the same, that we need to identify and R - universal gas constant, equal to 8.31.Transforming the equation, we get: m = MRT / PV.Substituting the known values into the formula, you will find what is**mole**gas.# Tip 2: How to find the mole

different formulas to help you find the amount of matter, the unit of which is

**mole**.Also the amount of the substance can be found by the reaction equation given in the task.

Guide

1

Weighing and name of the substance can be easily find a number of substances: n = m / M, where n - number of substances (

**mole**), m - masssubstance (g), M - molar mass of the substance (g /**mole**).For example, the weight of sodium chloride equal to 11.7 g, the amount of substance found.To substitute in the desired values, you need to find the molar mass of sodium chloride: M (NaCl) = 35.5 + 23 = 58.5 g /**mole**.Substitute: n (NaCl) = 11,7 / 58,5 = 0,2**mole**. 2

If it is a gas, then we have this formula: n = V / Vm, where n - number of substances (

**mole**), V - volume of gas (l), Vm - molar volumegas.Under normal conditions (pressure of 101 325 Pa and a temperature of 273 K) molar volume of gas is constant and equal to 22.4 liters /**mole**.For example, a quantity of the substance will have a volume of 30 l nitrogen under normal conditions?n (N2) = 30 / 22.4 = 1.34 mol**.** 3

Another formula: n = N / NA, where n - number of substances (

**mole**), N - number of molecules, NA - the Avogadro constant, equal to 6.02 x 10 to 23 degrees (1 /**mole**) .For example, how much of the substance is 1.204 * 10 to 23 degrees?Solve: n = 1,204 * 10 to 23 degrees / 6.02 * 10 23 degree = 0,2**mole**. 4

For any reaction equation you can find the number of substances, unreacted and formed as a result of it.2AgNO3 + Na2S = Ag2S + 2NaNO3.This equation shows that the 2

**mole of silver nitrate react with 1 mol****sodium sulfide, formed as a result of 1 mol****silver sulfide and 2 mol**

To a solution of silver nitrate weighing 25.5 g, adj solution containing sodium sulfide.What is the amount of substance of silver sulphide is formed at the same time?

First, find the amount of substance of silver nitrate, after calculating its molar mass.M (AgNO3) = 170 g / mol**sodium nitrate.With these quantities of materials may be found in other quantities required zadachah.Rassmotrim example.**To a solution of silver nitrate weighing 25.5 g, adj solution containing sodium sulfide.What is the amount of substance of silver sulphide is formed at the same time?

First, find the amount of substance of silver nitrate, after calculating its molar mass.M (AgNO3) = 170 g / mol

**.n (AgNO3) = 25,5 / 170 = 0,15****mole**.Reaction equation for this problem described above, it implies that 2**mol of silver nitrate produced 1 mol****silver sulfide.Determine how many moles****silver sulfide is formed from 0,15****mole of silver nitrate: n (Ag2S) = 0,15 * 1/2 = 0,075****mole**.

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Note

If the weight measured in kilograms, the amount of matter - in kmol.

Sources:

- "Handbook of Chemistry", GPKhomchenko, 2005.

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Helpful Hint

The calculations are commonly used rounded values of the atomic weights of the elements.If you need more precision, rounding off is unacceptable.