# Isotopes And Isotopy

Previous Topic : Atoms And Molecules

# Isotope And Isotopy

## Isotope

Isotope is a phenomenon whereby atoms of an element have the same chemical properties but different physical properties.

## Isotopy

Isotopy is a phenomenon whereby atoms of an element have the same atomic number but different mass number. The difference in mass number is due to the different number of neutrons.

It is the atomic number of an element that determines the position of the element in the periodic table and it is also the chemical reactivity of the atom.top↑

The presence of two or more isotopes of an element explains why the relative atomic masses of elements are not whole numbers

# Calculations Of Relative Atomic Mass

## Example 1

The atomic number of an element is 17. It has different atoms containing 18 neutrons and 20 neutrons, with a relative abundance of 75% and 25% respectively. Calculate the relative atomic mass of the element.

### Solution

75% + 25% = 100
Mass number of the atom with 75% abundance
75% = 17 + 18 = 35

75 × 35
100
= 26.25

Mass number of the atom with 25% abundance
25% = 17 + 20 = 37

25 × 37
100
= 9.25

Relative atomic mass of the element
= 26.25 + 9.25 = 35.5

## Example 2

There are two isotopes of chlorine with mass number 35 and 37 respectively, if the isotopes exist in the ratio of 3 : 1, calculate the relative atomic mass of chlorine? top↑

### Solution

The ratio is 3 : 1 = 3+1 = 4
Contribution by the isotope 35 of abundance 3
35 in abundance of 3 = 35 × 3 = 105.
Contribution by the isotope 37 of abundance 1
37 in abundance of 1 =37 × 1 = 37
Total contribution = 105 + 37 = 142

142 divided by the ratio in which the contribution was made (4)

142
4
= 35.5
.

## Example 3

The atomic number of lithium is 3, it has different atoms containing mass number of 7 and mass number of 6 with a relative abundance of 90% and 10% respectively, calculate the relative atomic mass of lithium.top↑

### Solution

90% + 10% = 100

Relative atomic mass of 7 with abundance of 90%

7 with abundance of 90% = 90 × 7
100
= 6.3

Relative atomic mass of 6 with abundance of 10%

6 with abundance of 10% = 10 × 6
100
= 0.6
Total relative abundance = 6.3 + 0.6 = 6.9

## Example 4

The atomic number of an element is 10, it has two different atoms of 20 and 22 mass number with relative abundance in ratio of 1 : 3 respectively, calculate the relative atomic mass of the element.top↑

### Solution

Relative abundance in ratio = 1 + 3 = 4
The relative atomic mass of 20 with abundance of 1

= 20 × 1 = 20

The relative atomic mass of 22 with abundance of 3

22 × 3 = 66
Total relative atomic mass

= 20 + 66
4
= 21.5
Relative atomic mass ; Relative atomic mass of an element is the number of times that the mass of an element is heavier than one twelfth of carbon 12
Note : Carbon 12 is the standard element for determine relative atomic mass (R.A.M.). top↑

# Mole

A mole is the amount of substance which contains the same number of elementary particles as the number of atoms, contained in 12g of the carbon atom.
Those elementary particles may be atoms, molecules or ions depending on the nature of the substance.
When the particles are atom or ions, the mass in grams gives the relative atomic mass, but if the particles are molecules the mass in grams gives the relative molecular mass.
For example, one mole of carbon iv oxide gas CO2 contains one mole of carbon atom with a mass of 12g and two moles of oxygen atoms with a mass of 32g.
Hence, CO2 = 33 + 12 =44g

Note : One mole of any gas at standard temperature and pressure occupies 22.4dm³ which is known as molar volume.top↑

Avogadro’s constant is the number of moles contained in the atomic mass in grams of any element.
Avogadro’s constant has constant value of 6.02 × 10²³.

## Mole Formulae

1) Mole = molarity × volume

2)

Mole = mass
molar mass
3)

Mole = volume
Gas molar volume (G.M.V)

G.M.V = 22.4 dm³ or 22400 cm³
Liquid at S.T.P has volume of 1 dm³ or 1000 cm³

4) Atoms = molecules = particles
No. of atoms = amount of atoms × avogadro’s K

Avogadro’s K = 6.02 × 10²³.

# Relative Molecular Mass

The molar mass of a compound is the mass of the correctly written formula or one molecule of that compound, obtained by adding together the appropriate relative atomic masses of all the atoms of the elements present in the molecule.top↑

## Example 1

Calculate the relative molecular mass of Calcium trioxocarbonate iii (CaCO3)

### Solution

R.M.M of Ca = 40, R.M.M of C = 12 and R.M.M of O = 16
One molecule of CaCO3 has one molecule of Calcium (Ca), one molecule of carbon (C) and three molecules of oxygen (O)

Hence, relative molecular mass (R.M.M) =
= 1 × R.M.M of Ca + 1 × R.M.M of C + 3 × R.M.M of O
= (1 × 40) + (1 × 12) + (3 × 16)
= 40 + 12 + 48
= 100g

## Example 2

Calculate the relative molecular mass of sodium trioxocarbonate (iv) (NaCO3).top↑

### Solution

R.M.M of Na = 23, R.M.M of C = 12 and R.M.M of O = 16
One atom of Na2CO3 has 2 atom of sodium (Na), one atom of carbon (C) and three atom of oxygen (O)

Hence, its relative molecular mass =
= 2 × R.M.M of Na + 1 × R.M.M of C + 3 × R.M.M of O
= (2 × 23) + (1 × 12) + (3 × 16)
= 46 + 12 + 48
= 106g.

## Example 3

Calculate the relative molecular mass of hydrogen tetraoxosulphate (vi) H2SO4.top↑

### Solution

R.M.M of H = 1, R.M.M of S = 32, and R.M.M of O = 16
One atom of H2SO4 contains two atoms of Hydrogen (H), one atom of sulphur (S) and 4 atoms of Oxygen (O)
R.M.M of H2SO4 =
2 × R.M.M of H + 1 × R.M.M of S + 4 × R.M.M of O
= (2 × 1) + (1 × 32) + (4 × 16)
= 2 + 32 + 64
= 98g.

## Example 4

Calculate the number of molecules in 10g of oxygen gas, Avogadro’s constant = 6.02 × 10²³

### Solution

Mass of oxygen = 10g
Molar mass of oxygen = 16g top↑
Number of molecules = mole × avogadro’ constant

Mole = mass
molar mass

mole = 10
16
= 0.625
Number of molecules = mole × avogadro’s constant
Number of molecules = 0.625 × 6.02 × 10²³
Number of molecules = 3.76 × 10²³

## Example 5

Calculate the number of molecules in 14g or Nitrogen gas. [N = 14, avogadro’s constant = 6.02 × 10²³] top↑

### Solution

Molar mass of Nitrogen = 14g/mol
Mass of Nitrogen = 14g
Number of molecules = mole × avogadro’ constant

Mole = mass
molar mass

mole = 14
14
= 1
Number of molecules = mole × avogadro’s constant
Number of molecules = 1 × 6.02 × 10²³
Number of molecules of Nitrogen = 6.02 × 10²³ top↑

# Empirical Formula And Molecular Formula

## Empirical Formula

Empirical formula : The empirical formula of a compound is the simplest formula of it which shows the simplest ratio of the number of atoms present in it.

## Molecular Formula

Molecular formula : the molecular formula of a compound shows the actual number of atoms of the different elements in one molecule of it.

(Empirical formula)n = Molecular formula

## Examples Of Empirical Formula And Molecular Formula

### Example 1

Analysis of a sample of an organic compound showed that, the compound contains 39.9% of carbon (C), 6.9% of hydrogen (H) and 53.2% of oxygen (O). Calculate the empirical formula and the molecular formula of the compound if the relative molecular mass is 60? [C = 12, H = 1, O = 16] top↑

#### Solution

Carbon Hydrogen Oxygen
% composition

39.9        6.9             53.2
Dividethrough by their atomic mass
Carbon 1st

39.9
12
= 3.33
Hydrogen =

6.9
1
= 6.9
Oxygen =

53.2
16
= 3.33
Divide through by lowest number
The lowest number is 3.33

Carbon 1st =

3.33
3.33
= 1.00
Hydrogen =

6.9
3.33
= 2.04
Oxygen =

3.33
3.33
= 1.00
Approximate to the nearest whole number =

1        2      1

The empirical formula = CH2O

ii) to calculate the molecular formula,
the relative molecular mass of the compound = 60
(Empirical formula)n = Molecular formula top↑

(CH2O)n = 60
(12 + 2 + 16)n = 60
30n = 60
n = 60
30
= 2

# Calculations Of Percentage Composition

The mass of one mole of a compound is the sum of the masses of the moles of its components element.

Mass of 1 mole of = mass of 1 mole + 1 mole of
(CO3) Carbon Oxygen
60g = 12g + 48g

## Example 1

Calculate the percentage by mass of oxygen in copper (ii) tetraoxosulphate (iv) salt.
[Cu = 64, S = 32, O = 16]

### Solution

Molecular formula of CuSO4
In one mole of CuSO4, there are one mole of Cu, one mole of S and 4 moles of O atoms.
Molar mass of CuSO4 =
= 1 × R.M.M of Cu + 1 × R.M.M of S + 4 × R.M.M of O
= (1 × 64) + (1 × 32) + (4 × 16)
= 64 + 32 + 64
= 160g/mol

160g of CuSO4 contains 64g of oxygen
% by masss of oxygen

=  64
160
= 0.4

% composition of oxygen = 0.4 × 100
= 40%.

## Example 2

Calculate the percentage by mass of Hydrogen in hydrogen tetraoxosulphate (iv).
[H = 1, S = 32, O = 16]

### Solution

In one mole of H2SO4, there are two moles of H, one mole of S and 4 moles of O atoms.
Molar mass of CuSO4 =
= 2 × R.M.M of H + 1 × R.M.M of S + 4 × R.M.M of O
= (2 × 1) + (1 × 32) + (4 × 16)
= 2 + 32 + 64
= 98g/mol

98g of H2SO4 contains 2g of hydrogen
% by masss of hydrogen

= 2
98
= 0.02

% by mass of hydrogen = 0.02 × 100
= 2%

## Example 3

How many atoms are there in 10g of CaCO3?
[Ca = 40, C = 12, O = 16]

### Solution

Number of atoms = mole × avogadro’s constant

Mole = mass
molar mass

Molar mass of CaCO3
= 1 × R.M.M of Ca + 1 × R.M.M of C + 3 × R.M.M of O
= (1 × 40) + (1 × 12) + (3 × 16)
= 40 + 12 +48
= 100g/mol

Mole = mass
molar mass
Mole = 10
100
= 0.1
0.1 mole of CaCO3 contains 6.02 × 10²³
= 0.1 × 6.02 × 10²³
= 6.02 × 10²²top↑

## Example 4

How many moles of NaOH are there in 4.0g of the substance. [Na = 23, H = 1 O = 16]

### Solution

The molar mass of NaOH =
= 1 × R.M.M of Na + 1 × R.M.M of O + 1 × R.M.M of H
= (1 × 23) + (1 × 16) + (1 × 1)
= 23 + 16 + 1
= 40g
40g of NaOH = 1 Mole

4.0g of NaOH = 1 mole × 4.0g
40g
= 0.1 mole

See you next time, please if you don’t understand, feel free to ask me any question via the comment box or ask question page and I will be happy to answer your question and take the session again if necessary, your comment notifications is like a bank alert to me so don’t forget to write something.top↑ you can also use Our Contact Page or will you like to know more About Our Tutors

Next Topic : Chemical Laws