**Isotopy and isotopes**

**Isotopy can be defined as a process/occurence/phenomenon whereby an element has the same atomic number (i.e the number of protons) but different mass number (i.e number of neutrons).**

Proton

Neutron

Electron

Isotopes are the elements that have the same atomic number (number of protons) and different mass number (number of neutrons). Examples of isotopes are;

Chlorine

^{35}

_{17}Cl ,

^{37}

_{17}Cl

Oxygen

^{16}_{8}O ,^{17}_{8}O ,^{18}_{8}O
Hydrogen

^{1}_{1}H ,^{2}_{1 }H ,^{3}_{1}H
Carbon

^{12}_{6}C ,^{13}_{6}C
Sodium

^{23}_{11}Na ,^{24}_{11}Na
Lithium

^{6}_{3}Li ,^{7}_{3}LiRead more: All about elements.

__Worked Examples__:

Question 1

there are two isotopes of chlorine with mass numbers 35 and 37 respectively. if the isotopes exist in the ratio 3:1 , the lighter isotope being more abundant. what is the relative atomic mass of chlorine?

**solution**

1st isotope 35Cl

2nd isotope 37Cl

the isotopic ratio = 3:1

Total ratio = 3 + 1 = 4

% abundance of 35Cl = (3/4 x 100) = 75%

% abundance of 37Cl = (1/4 x 100) = 25%

Relative atomic mass of chlorine (R.A.M) = (75/100 X 35) + (25/100 X 37) = 3550/100

Relative atomic mass of chlorine (R.A.M) = 35.5

Question 2

Boron exist as an isotopic mixture containing 80% of B , 15% of B and 5% of B. Calculate the relative atomic mass of Boron?

**solution**

1st isotope 9B

2nd isotope 10B

3rd isotope 11B

the % abundance ratio = 80:15:5

Relative atomic mass of chlorine (R.A.M) = (80/100 X 9) + (15/100 X 10) + (5/100 X 11) = 925/100

Relative atomic mass of boron (R.A.M) = 92.5

Question 3

If the relative atomic mass of natural copper is 63.54. Calculate the proportions of isotopes 63CU and 65CU in the metal?

**solution**

1st isotope 63Cu

2nd isotope 65 Cu

Let the proportion of cu be x

Let the proportion of cu be y

therefore, x + y = 100 ...................(1)

Relative atomic mass of copper (R.A.M) = (x/100 X 63) + (y/100 X 65) = 63.54

63.54 = 0.63x + 0.65y .............(2)

from equation (1); y = 100 - x

substitute for y in the equation (2) above 63.54 = 0.63x + 0.65(100 - x)

63.54 = 0.63x + 65 - 0.65x

1.46 = 0.02x

x = 0.73 (=73%)

recall; y = 100 - x

y = 100 - 73

y = 27%

Question 4

Two isotopes of z with mass number 18 and 20 are in the ratio 1:2. Determine the relative atomic mass.

**solution**

1st isotope 20Z

2nd isotope 18 Z

the isotopic ratio = 1:2

Total ratio = 2 + 1 = 3

% abundance of 20 Z = (2/3 x 100) = 66.6%

% abundance of 18 Z = (1/3 x 100) = 33.3%

Relative atomic mass of Z (R.A.M) = (66.6/100 X 20) + (33.3/100 X 18) = 1931.4/100

Relative atomic mass of Z (R.A.M) = 19.314.

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