Atomic mass is the combined mass of all the protons, neutrons, and electrons that make up an atom. Electrons have nearly 1/2000th of the mass of protons and neutrons, so electrons are typically ignored in calculations involving atomic mass. This means the atomic mass is the sum of the masses of the protons and neutrons in an atom.
The easiest way to find the atomic mass is to look it up on a periodic table. The atomic mass for each element is given in atomic mass units or grams per mole of atoms. This value is the average atomic mass of the element. This is because elements may have more than one naturally occurring isotope.
Example: Find the element copper (Cu or element number 29) on the periodic table. The atomic mass is listed as 63.546. This means the average mass of a mole of copper atoms is 63.546 grams.The average is important since there are two different natural isotopes of copper: Copper-63 and Copper-65. Copper-65 has two additional neutrons than copper-63 and therefore has more mass.
The average is important since there are two different natural isotopes of copper: copper-63 and copper-65. Copper-65 has two additional neutrons than copper-63 and therefore has more mass. The average mass of copper takes account of the natural abundance of each isotope of an element. Copper-63 accounts for just under 70% of all copper found in nature. The other 30% is copper-65. These abundances are used to calculate the atomic mass value found on the periodic table.
How to Calculate Atomic Mass From Natural Abundance Example
This example will show how to find the average atomic mass of an element when given the natural abundance of each of the element’s isotopes.
Magnesium (Mg, element 12) has three natural isotopes: Mg-24, Mg-25, and Mg-26.
Mg-24 has a mass of 23.99 amu and accounts for 78.99% of all natural magnesium.
Mg-25 has a mass of 24.99 amu and accounts for 10.00% of natural magnesium.
Mg-26 has a mass of 25.98 amu and accounts for the final 11.01% of natural magnesium.
What is the atomic mass of magnesium?
Answer: The atomic mass of magnesium is the weighted average of each of these isotopes. Each of the abundances adds up to 100%. Take each isotope and multiply it by its percent abundance in decimal form and add them all together. Since each of the abundances
mass of magnesium = mass Mg-24 ⋅ (0.7899) + mass Mg-25 ⋅ (0.1000) + mass Mg-26 ⋅ (0.1101)
mass of magnesium = (23.99 amu) ⋅ (0.7899) + (24.99 amu) ⋅ (0.1000) + (24.99 amu) ⋅ (0.1101)
mass of magnesium = 18.95 amu + 2.50 amu + 2.86 amu
mass of magnesium = 24.31 amu
This value agrees with the value of 24.305 given on the periodic table.
How to Calculate Natural Abundance from Atomic Mass
A common homework problem involves finding the natural abundance of isotopes from the atomic masses of the isotopes and the element’s atomic mass.
Boron (B, element 5) has an atomic mass of 10.81 amu and has two natural isotopes: B-10 and B-11.
B-10 has an atomic mass of 10.01 amu and B-11 has an atomic mass of 11.01 amu. Find the natural abundance of each isotope.
Answer: Set up the equation the same way as the previous example.
mass of boron = mass of B-10⋅(abundance of B-10) + mass of B-11⋅(abundance of B-10)
10.81 = (10.01)⋅(abundance of B-10) + 11.01⋅(abundance of B-11)
Now our problem is we have too many unknowns. Since we’re working with percent abundances, we know the combined total of the abundance is equal to 100%. In decimal form, this means
1 = (abundance of B-10) + (abundance of B-11)
(abundance of B-10) = 1 – (abundance of B-11)
Let X = abundance of B-11 then
(abundance of B-10) = 1 – X
Plug these values into the above equation
10.81 = (10.01) ⋅ (1 – X) + 11.01 ⋅ (X)
Solve for X
10.81 = 10.01 – 10.01 ⋅ X + 11.01 ⋅ X
10.81 – 10.01 = -10.01 ⋅ X + 11.01 ⋅ X
0.80 = 1 ⋅ X
0.80 = X = abundance of B-11
1 – X = abundance of B-10
1 – 0.80 = abundance of B-10
0.20 = abundance of B-10
Multiply both answers by 100% to get the percent abundance of each isotope.
%abundance of B-10 = 0.20 x 100% = 20%
%abundance of B-11 = 0.80 x 100% = 80%
Solution: Boron is made up of 20% B-10 and 80% B-11.