Relative Atomic Mass Calculations Chemistry Tutorial

Key Concepts

• The relative atomic mass of an element is also known as the relative atomic weight of an element, or, the atomic weight of an element.
• Relative atomic mass is often abbreviated as r.a.m.
• The relative atomic mass of an element (its atomic weight) is given in the Periodic Table.
• The relative atomic mass of an element is the weighted average of the masses of the isotopes in the naturally occurring element relative to the mass of an atom of the carbon-12 isotope which is taken to be exactly 12.
• The atomic mass unit (u) is defined as a mass equivalent to ¹/₁₂ of the mass of one atom of carbon-12.

  1 u = 1.66 × 10^-27 kg

• We can estmate the the relative atomic mass (atomic weight) of an element E with the naturally occurring isotopes ^aE, ^bE, ^cE, etc, and with the respective abundances of A%, B%, C% etc,

  relative atomic mass (r.a.m.)

  = (

  A   100

  × a)

  + (

  B   100

  × b)

  + (

  C   100

  × c)

  + etc

• Given the relative atomic mass (r.a.m.) of an element and the estimated mass of each of its isotopes, we can then estimate the relative abundance of each isotope:

  let x = %abundance of isotope-a
  and 100 - x = %abundance of isotope-b
  then, let r.a.m = relative atomic mass of the element:

  r.a.m.

  = (

  x   100

  × mass isotope-a)

  + (

  100 - x 100

  × mass isotope-b)

  and solve for x

• Note that we can measure the mass of each isotope and its abundance using Mass Spectroscopy

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Abundance of Naturally Occurring Isotopes

Most elements occur in nature as a mixture of different isotopes.

The element carbon, for example, exists in nature as a mixture of different isotopes: stable¹ carbon-12 atoms and carbon-13 atoms.

If you were to take a sample of carbon atoms, for example the soot from a chimney or a lump of coal, you would find that most of the carbon atoms are the carbon-12 isotope and only a few would be the carbon-13 isotope.
We call this amount of each isotope found in the naturally occurring element its abundance, or its isotopic abundance to be more precise.
The abundance of the carbon-12 isotope in naturally occurring bulk carbon is 98.90% while the abundance of the carbon-13 isotope in nature is 1.10%
This means that if you take a lump of coal from nature, 98.90% of the carbon in the coal will be atoms of the carbon-12 isotope, while only 1.10% of it will be atoms of the carbon-13 isotope.

The table below gives the isotopic abundances for some elements on Earth:

You can find a more complete list of isotopic abundances at the bottom of this page.

Estimating Isotopic Mass

The relative atomic mass of a carbon-12 atom is defined as 12.00
The relative atomic mass of an atom of carbon-13 is found to be 1.08333 times the mass of a carbon-12 atom, that is 1.083 × 12 = 13.00

We can estimate the mass of any isotope of an element, its isotopic mass, using its mass number (A).

The mass number (A) of an isotope tells us how many protons and neutrons are in the nucleus of an atom of this isotope.
Nucleon is the term used to describe both protons and neutrons.
So, the mass number (A) tells us the number of nucleons in the nucleus of an atom of the isotope of the element.

For example:

• nucleus of an atom of the carbon-12 isotope contains 12 nucleons
• nucleus of an atom of the carbon-13 isotope contains 13 nucleons

The mass of a proton is almost exactly the same as the mass of a neutron.
The mass of a proton is about 1 u (1 atomic mass unit), so the mass of a neutron is also about 1 u.
The mass of an electron is so small compared to the mass of a proton or neutron that it can be ignored when estimating the mass of an isotope of an element.²

We can estimate the mass of an atom of an isotope of an element by adding together the mass of its nucleons:³

isotopic mass = number of nucleons × mass of nucleon

isotopic mass = number of nucleons × 1 u

For example:

• isotopic mass of carbon-12 atom: 12 nucleons × 1 u/nucleon = 12 u
• isotopic mass of carbon-13 atom: 13 nucleons × 1 u/nucleon = 13 u

Isotopic masses can be measured using mass spectroscopy. You will find a discussion of calculating relative atomic mass (atomic weight) using these measured isotopic masses in the Mass Spectroscopy tutorial.

Calculating the Relative Atomic Mass (Atomic Weight) of an Element

The relative atomic mass of an element is the weighted average of the masses of the isotopes in the naturally occurring element relative to the mass of an atom of the carbon-12 isotope which is taken to be exactly 12.

What is a "weighted average" ?

First lets look at what the "average weight" of carbon would be:

mass of carbon-12 isotope is 12 u

mass of carbon-13 isotope is 13 u

So we can calculate the average mass (average weight) of carbon as:

If we look up the atomic weight of carbon in the Periodic Table we find that it is 12.01 NOT 12.5

This is because most of the atoms found in naturally occurring carbon are atoms of the carbon-12 isotope while very few of the atoms will be of the carbon-13 isotope.
We need to take this into account when we calculate our "average", and when we do this we refer to the result as a "weighted average".

We need to know the abundance of each isotope, that is we need to know how much of the mass (weight) of the naturally occuring bulk carbon is due to each of the isotopes (carbon-12 and carbon-13).
From the table in the section above we find that the isotopic abundance of carbon-12 is 98.90% and the isotopic abundance of carbon-13 is 1.10%

This means that if I had 100 atoms of bulk carbon (like in coal or soot), then:

• 98.90% of the 100 carbon atoms are carbon-12 atoms with a mass of 12 u
  That is, 98.90 of the carbon atoms have a mass of 12 u
• 1.10% of the 100 carbon atoms are carbon-13 atoms with a mass of 13 u
  That is, 1.10 of the carbon atoms have a mass of 13 u

So, the total mass of 100 naturally occurring carbon atoms is:

So the mass of 100 naturally occurring carbon atoms is 1201.1 u
Therefore the "weighted average mass" of 1 carbon atom is 1201.1 ÷ 100 = 12.011 u
This value for the atomic weight of carbon agrees with the value in the Periodic Table.

Note that this does NOT mean that the mass of 1 atom of carbon is 12.011 u
It DOES mean that if we take a sample of naturally occurrring carbon we would find that the average mass of a carbon atom would be 12.011 u

Let's review how we calculated the "weighted average" mass of bulk carbon atoms:

• abundance of the first isotope multiplied by the mass of this isotope given by the mass number
• abundance of the second isotope multiplied by the mass of this isotope given by the mass number
• add these two values together then divide by 100

In general, to calculate the "weighted average" mass of an element that occurs naturally as two different isotopes, isotope 1 and isotope 2, then:

If we estimate the mass of each isotope by using its mass number (A), then we can re-write the expression as:

Or, put another way:

Calculating Isotopic Abundance from Atomic Weight

The Periodic Table gives us the weighted average for the mass of an element, referred to as the element's atomic weight (or relative atomic mass).

If we know the mass of each isotope making up this naturally occurring element (estimated by its mass number), then we can calculate the abundance of each isotope in nature.

We wrote a general mathematical expression (mathematical equation) above for calculating the "weighted average" mass, also known as the relative atomic mass or atomic weight, of an element, which was:

Let's say I wanted to find the abundance (%) of each isotope of nitrogen.

Nitrogen has two naturally occurring stable isotopes: nitrogen-14 and nitrogen-15.

So, substituting these into the mathematical equation above I get:

I can look up the atomic weight (relative atomic mass, or, "weighted average" mass) of nitrogen in the Periodic Table:

atomic weight of nitrogen is 14.01

I can estimate the mass of an atom of each isotope by using its mass number (A):

The mass of an atom of nitrogen-14 = its mass number = 14 u

The mass of an atom of nitrogen-15 = its mass number = 15 u

I can substitute these values into the mathematical equation above:

I can multiply both sides of the mathematical equation by 100:

But how can I solve this equation when there are 2 unknowns, the abundance of nitrogen-14 is unknown and the abundance of nitrogen-15 is unknown.

The trick is to remember that we are talking about percentage abundance! Which means that:

%abundance of nitrogen-14 + %abundance of nitrogen-15 = 100

Or, put a different way:

%abundance of nitrogen-14 = 100 - %abundance of nitrogen-15

So, if I let the %abundance of nitrogen-14 be equal to x then:

%abundance of nitrogen-14 = x

%abundance of nitrogen-15 = 100 - x

If I substitute these into the mathematical equation above I will have only 1 unknown value, x:

To solve for x I will clear the brackets first:

Next I collect like terms, starting with x:

Then subtract 1500 from both sides of the equation:

Note that the abundance of an isotope must be a positive number (not a negative number), so I divide both sides of the equation by -1 to find the value of x :

Then substitute this value for x back into the expressions we wrote for the abundance of each isotope:

%abundance of nitrogen-14 = x = 99 %

%abundance of nitrogen-15 = 100 - x = 100 - 99 = 1%

In general, if an element occurs in nature in 2 isotopic forms, isotope 1 and isotope 2, then we can estimate the percentage abundance of each isotope using the mass number (A) of each isotope because the %abundance of isotope 2 equals 100 - %abundance of isotope 1:

let abundance isotope 1 = x

So,

• %abundance isotope 1 = x
• %abundance isotope 2 = 100 - x

Worked Example 1: Calculating the Atomic Weight of an Element

Question: Naturally occurring silver is 51.84% silver-107 and 48.16% silver-109.

Calculate the atomic weight of silver.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Calculate the "weighted average" mass (relative atomic mass, or, atomic weight) of silver

2. What data (information) have you been given in the question?

   Extract the data from the question:

   isotope name	mass number (A)	isotopic abundance (%)
   silver-107	107	51.84
   silver-109	109	48.16

3. What is the relationship between what you know and what you need to find out?

   (a) mass of an isotope can be estimated using its mass number (A) expressed in atomic mass units, u

   (b)

   "weighted average" mass of silver

   = (

   %abundance silver-107 100

   × A(silver-107))

   + (

   %abundance silver-109 100

   × A(silver-109))

4. Substitute in the values and solve:

   "weighted average" mass of silver	= (	51.84 100	× 107)	+ (	48.16 100	× 109)
   	=	55.469		+	52.494
   	=	107.96

5. Is your answer plausible?

   Do a rough calculation:
   Since about half (≈50%) of the atoms of silver are silver-107 and the other half (≈50%) are silver-109, the weighted average mass for silver will lie about half-way between the mass of silver-107 (107) and silver-109 (109):
   atomic weight ≈ ½ × (107 + 109) = 108
   Since our calculated value of 107.96 is close to 108 we are reasonably confident that our answer is correct.
   You can also look up the atomic weight of silver in the periodic table and find that it is 107.9.
   Since our calculated value of 107.96 is about the same as the value in the periodic table we are reasonably confident that our answer is plausible.

6. State your solution to the problem "atomic weight of silver":

   atomic weight of silver is 107.96 u

Worked Example 2: Calculating an Element's Isotopic Abundance

Question: Copper consists of two isotopes, copper-63 and copper-65.
Its atomic weight (relative atomic mass) is given as 63.62 u

Find the % abundance of each isotope.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   (a) Calculate the % abundance of copper-63

   (b) Calculate the % abundance of copper-65

2. What data (information) have you been given in the question?

   Extract the data from the question:

   isotope name	mass number (A)	approximate mass of isotope (u)
   copper-63	63	63
   copper-65	65	65

   atomic weight of copper = 63.62 u

3. What is the relationship between what you know and what you need to find out?

   (a)

   "weighted average" mass of copper

   = (

   %abundance copper-63 100

   × A(copper-63))

   + (

   %abundance copper-65 100

   × A(copper-65))

   (b) let %abundance copper-63 = x
       and %abundance copper-65 = 100 - x

4. Substitute the values into the equation and solve:

   (a)

   63.62	= (	x 100	× 63)	+ (	100 - x 100	× 65)
   63.62	=	63x 100		+	(65 × 100) - 65x 100
   63.62	=	63x 100		+	6500 - 65x 100
   63.62	=	63x + 6500 - 65x 100
   100 × 63.62	=	100 × 63x + 6500 - 65x 100
   6362	=	63x + 6500 - 65x
   6362	=	-2x + 6500
   6362 - 6500	=	-2x
   -138	=	-2x
   -138 -2	=	-2x -2
   69	=	x

   (b) %abundance copper-63 = x = 69 %
       %abundance copper-65 = 100 - x = 100 - 69 = 31 %

5. Is your answer plausible?

   The atomic weight of copper is given as 63.62.
   Because this weight is closer to that of isotopic mass of copper-63 than it is to that of the isotopic mass of copper-65 we know that copper-63 must be more abundant (greater than 50%).
   Our calculated value for the abundance of copper-63 is 69% which is greater than 50% so we are reasonably confident that our answer is plausible.
   Note that the %abundance of both isotopes must add up to 100%.
   69% + 31% = 100%
   So we are reasonably confident that our calculated values for the abundance of each isotope is plausible.

6. State your solution to the problem "abundance of each isotope of copper":

   abundance of copper-63 = 69%

   abundance of copper-65 = 31%