CHEMICAL CALCULATIONS

Accurate numbers of atoms, ions or molecules can be counted by weighing, if a suitable counting unit is

defined. The unit used in Chemistry is the mole. It is one of the seven base units of the SI, the unit of

amount (also known as quantity):

Unit of amount: mole

Abbreviation: mol

Symbol: n

Definition: The mole is the amount of substance that contains as many particles

of the substance as there are atoms in 12 g of carbon-12.

N.B.: the “particles” may be atoms, ions, molecules or formula units, as the case may be.

Examples:0,5 mol Cu

2,4 mol Ag+

1,5 mol H2O

0,05 mol NaCl

5 mol CuSO4.5H2O (containing 5 mol Cu; 5 mol S; 45 mol O and

50 mol H in total)

Molar mass

Since the mole is based upon C-12, it follows that 1 mol C must have a mass of 12 g. This is its molar mass, M. Molar mass has the same value as relative atomic mass, but it has a unit:

M(C) = 12 g.mol1

The relative atomic masses of all the elements are based upon C-12, thus one mole of an element must have a mass equal to its relative atomic mass, taken in grams:

M(H) = 1 g.mol1

M(Na) = 23 g.mol1

M(U) = 238 g.mol1

Furthermore, molar masses of compounds and ions are found in the same way:

M(H2O) = (21) + 16 = 18 g.mol1

M(NaOH) = 23 + 16 + 1 = 40 g.mol1

M(CuSO4.5H2O) = 63,5 + 32 + (416) + 5[(21) + 16] = 249,5 g.mol1

M(SO42) = 32 + (416) = 96 g.mol1

Exercise: Find the molar masses of the following:

1. H2SO46. P2O5

2. KOH7. AgNO3

3. Na2CO38. C6H4(CH3)2

4. KMnO49. (NH4)2Cr2O7

5. Ca(OH)2 10. MgSO4.7H2O

Since molar mass is the mass, in g, of one mole of a substance, we can write

where m = mass, in g

n = amount (or quantity), in mol

M = molar mass, in g.mol1

Rearranging, we get the very useful formula for finding the amount of substance in a given mass:

The mass of an amount of substance can be found using m = n.M

Exercise:

1. Find the amount of NaOH in 5,0 g

2. Calculate the quantity of carbon dioxide gas in 12,5 g CO2

3. What is the mass of 0,55 mol ZnSO4?

4. Find the amount of CuSO4 in 250,0 g

5. Calculate the mass of 18,6 mol KI

Avogadro’s Constant

The mass of a carbon atom is 1,993  1023 g, thus the number of C atoms in 12 g of carbon is given by

This means that one mole of a substance will contain 6,02  1023 particles. This number is referred to as

Avogadro’s constant, NA:

NA = 6,02  1023 mol1

In order to calculate the amount of substance if the number of particles is known, the following formula may be used:

where N = number of particles

NA = 6,02  1023 mol1

n = amount (or quantity), mol

This leads to the formula for the number of particles in a given amount of substance:

N = n.NA

Exercise:

1. Find the amount of H2O in 5  1022 molecules.

2. Calculate the number of O atoms in 2 mol O2.

3. What quantity of Na+ will be contained in 1,5  1024 formula units of NaHSO4?

4. Find the number of molecules in 50 g CO2 gas.

5. If a computer were able to count at the rate of 1  109 “counts” per second, how long

(in years) would it take to count to 6,02  1023?

Molar volume

Amadeo Avogadro stated, in 1811, one of the basic laws in Chemistry:

Equal volumes of gases, at the same temperature and pressure,

contain the same number of molecules.

If we define the standard temperature and pressure (STP) conditions as 273 K (0oC) and 100 kPa,

it can be shown that at STP, the volume of one mole of any gas is 22,4 dm3. This is known as the

molar volume at STP, Vo:

Vo = 22,4 dm3.mol1 (at STP)

Naturally, if the conditions of temperature and pressure were to change, the volume of one mole of the

gas would also change. Thus the molar volume of 22,4 dm3 holds only at STP.

The relationship between amount of a gas, its volume and the molar volume is given by

where V = gas volume at STP, dm3

Vo = molar gas volume, 22,4 dm3.mol1

n = amount (or quantity) of gas, mol

This yields the formula V = n.Vo for the volume of a quantity of gas at STP.

Exercises:

1. What quantity of O2 is there in 10 dm3 of the gas at STP?

2. Find the volume of 24 mol CO2 at STP.

3. What is the volume of 12 g H2 at STP?

4. Calculate the volume of 5  1022 molecules of He at STP.

5. Find the number of N2 molecules in 500 cm3 of N2 at STP.

Using the mole in chemical reaction calculations

Consider the reaction of carbon and oxygen:

C(s) + O2(g)  CO2(g) ………(1)

Thus

reacts with to give

If we have 1 mole of C atoms,

reacts with to give

Equation (1), therefore tells us that the MOLE RATIO is 1:1:1 (i.e. C:O2:CO2)

Thus,

2 mol C + 2 mol O2 2 mol CO2

0,5 mol C + 0,5 mol O2 0,5 mol CO2

120 mol C + 120 mol O2 120 mol CO2 , etc.

Now consider the reaction

2H2+ O2 2H2O , with mole ratio 2:1:2.

Here

2 mol H2+ 1 mol O2 2 mol H2O

4 mol H2+ 2 mol O2 4 mol H2O

1 mol H2 + 0,5 mol O2 1 mol H2O

10 mol H2 + 5 mol O2 10 mol H2O

3 mol H2 + 1,5 mol O2 3 mol H2O

1,7 mol H2 + 0,85 mol O2 1,7 mol H2O, etc.

Once we know the mole ratio for a particular reaction, we can calculate product quantities, masses or volumes that are obtained with given reactant quantities, masses or volumes, or the reactant quantities,

masses or volumes that are required to yield given product quantities, masses or volumes.

The oxidation of ammonia is the first step in the production of nitric acid:

4NH3(g) + 5O2(g)  4NO + 6H2O

Mole ratio 4 : 5 : 4 : 6

To calculate what mass of O2 will react with 5 g NH3, we work as follows:

Amount of NH3 : mol NH3

Since it follows that 0,294 mol NH3 will react with

mol O2

m = nM = 0,368  32 = 11,8 g O2

If we have to find the quantity of H2O that is produced when 0,75 mol O2 reacts:

(Remember that quantity and amount mean moles)