# Average Atomic Mass

The atomic masses of most of the elements are not whole numbers. For example, look at the picture of Carbon's element box below.Carbon's atomic mass is 12.0107. You might wonder how this is possible, since Carbon can't contain a fraction of a proton or a neutron. In this lesson, we'll learn how the atomic mass is calculated, and why it is a decimal.

It is based on the percent natural abundance of the elements. That means that it is a weighted average of

*all*the naturally occuring isotopes of that element. Let's see how to do the math.

Example One:

*What is the average atomic mass of Boron if it exists as 19.90%*

^{10}B (10.013 g/mol) and 80.10%^{11}B (11.009 g/mol)?
Solution:

To solve this, we have to take a weighted average.

First we convert the percentages to decimals (19.90% becomes 0.1990, and 80.10% becomes 0.8010)

Second, we multiply those decimals by the masses, and add the products.

(0.1990) X (10.013 g/mol) = 1.993 g/mol

(0.8010) X (11.009 g/mol) =

10.811 g/mol

Notice that the result, 10.811 g/mol is close to the value reported for Boron on the periodic table.

To solve this, we have to take a weighted average.

First we convert the percentages to decimals (19.90% becomes 0.1990, and 80.10% becomes 0.8010)

Second, we multiply those decimals by the masses, and add the products.

(0.1990) X (10.013 g/mol) = 1.993 g/mol

(0.8010) X (11.009 g/mol) =

__+ 8.818 g/mol__10.811 g/mol

Notice that the result, 10.811 g/mol is close to the value reported for Boron on the periodic table.

Example Two:

*Out of 500 silicon atoms, 460 are*

^{28}Si (27.98 g/mol), 25 are^{29}Si (28.98g/mol) and 15 are^{30}Si (29.97g/mol). What is the average atomic mass?
Solution:

In this problem, they did not give us the percentages. However, since we know the number of atoms, we can easily calculate the percentages. For example:

500

If we do this for all three isotopes, we get 92%, 5%, and 3%. (We'll assume these are absolute numbers for determining our significant figures. (If you don't know about significant figures and how they are used for rounding, don't worry about that right now.)

Now the problem is just like the previous one. First convert the percentages into decimals. Then multiply those decimals by the masses and add. Here's the solution:

(0.92) X (27.98 g/mol) = 25.74 g/mol

(0.05) X (28.98 g/mol) = 1.449 g/mol

(0.03) X (29.97 g/mol) =

28.09 g/mol

In this problem, they did not give us the percentages. However, since we know the number of atoms, we can easily calculate the percentages. For example:

__460__X 100 = 92%500

If we do this for all three isotopes, we get 92%, 5%, and 3%. (We'll assume these are absolute numbers for determining our significant figures. (If you don't know about significant figures and how they are used for rounding, don't worry about that right now.)

Now the problem is just like the previous one. First convert the percentages into decimals. Then multiply those decimals by the masses and add. Here's the solution:

(0.92) X (27.98 g/mol) = 25.74 g/mol

(0.05) X (28.98 g/mol) = 1.449 g/mol

(0.03) X (29.97 g/mol) =

__+ 0.8991 g/mol__28.09 g/mol

Take note that I always move the decimal two places. The 5% become 0.05 as a decimal. Also, note that I paid attention to significant figures in my calculations. Click "Next" to try several problems on your own.

*Site built and maintained by Mr. Fredericks*

jfredericks@dallassd.com

jfredericks@dallassd.com