# An Elementary Quantitative Chemistry Course

Lesson 1

Chemical reactions take place on an atomic level. For this reason the amount of substance (atoms, molecules) is of great importance. Amount units are known from everyday life. An example is the dozen. One dozen of atoms is equal to twelve atoms, one dozen of molecules is equal to twelve molecules or simply we say that 1 dozen is equal to 12.

The amount unit used in chemistry is 1 mol. 1 mol is equal to 602214000000000000000000. That is a very large number. Suppose that you shall count 1 mol atoms at the rate of 1 atom per second. How long will it take?

With the numbers given in scientific notation we get

1 dozen = 1.2 101
1 mol = 6.02214 1023

The number 6.02 1023 (here given with 3 significant digits) is known as Avogadro's number and the constant 6.02 1023 mol-1 is known as Avogadro's constant (mol-1 means per mol).

As an example 3 mol atoms is equal to 18.06 1023 atoms. This was calculated as

3 mol atoms 6.02 1023 mol-1

If N represents the amount and n represents the amount given in the unit mol then we can express this calculation in the general formula:

N = n 6.02 1023 mol-1

With Avogadro's constant 6.02 1023 mol-1 written as NA we finally get:

N = n NA

Lesson 2

At normal temperature and pressure the forces between particles (atoms, molecules) in gases are negligible weak. The volume of the particles is also very small in comparison with the volume of the gas. For these two reasons 1 mol of a gas occupy about the same volume independent of what kind of particles the gas consists of. The volume of 1 mol of a gas is called the molar volume and is expressed by Vm. To calculate the volume, V, of say 5 mol of a gas we simply multiply the molar volume by 5 mol.

We can express this calculation in the general formula

V = n Vm

Lesson 3

The concentration of a solution may be expressed as the amount of substance in one litre of the solution. Then it is called the molarity, has the unit 1 mol dm-3, and is expressed by the letter c.

A concentrated solution of hydrocloric acid HCl in water has the molarity 13 mol dm-3. It can be bought in barrels containing 5 dm3. The amount of HCl molecules in such a barrel is 65 mol which was calculated as 13 mol dm-3 5 dm3. This calculation can be expressed in a general way by the formula:

n = c V

Lesson 4

If this hydrocloric acid solution is diluted by the addition of more water, the amount of HCl molecules is the same in the dilute solution as it was in the concentrated solution because no HCl molecules were added or removed, only water was added. The amount of HCl molecules in the concentrated solution (number 1) was

n1 = c1 V1

and the amount of HCl molecules in the dilute solution (number 2) is

n2 = c2 V2

Because n1 = n2 it follows that:

c1 V1 = c2 V2

This formula is used whenever a solution, number 1, is diluted to give a new solution, number 2.

Lesson 5

The particles of a substance are too small and too many to be counted. For that reason the substance is weighed. We need to know the ratio between the mass of the substance and the amount of the substance. This ratio is known as the molar mass of the substance. It is conventionally expressed by the character M (maybe mm had been better because the subscript m stands for molar as in the molar volume Vm and the molar heat capacity Cm to mention a pair of examples).

If the molar mass of a substance is say 40 g mol-1 it means that 1 mol of the substance has the mass 40 g. Then 2 mol has the mass 80 g and so on. We simply calculate the mass by multiplying the amount of the substance by the molar mass of the substance as this formula tells us:

m = n M

Lesson 6

Solutions with known volumes and concentrations are often made in the laboratory. The mass of the substance to be solved must then be calculated. We can derive a formula for this by substituting n in

m = n M

by c V because n = c V as we have seen here above.

This gives us:

m = c V M

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