Average Atomic Mass (AAM): Calculation — lesson. Science State Board, Class 10.

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Atomic masses of some elements:

Atomic number	Name	Symbol	Atomic mass
$1$	Hydrogen	$H$	$1.008$
$2$	Helium	$He$	$4.003$
$3$	Lithium	$Li$	$6.941$
$4$	Beryllium	$Be$	$9.012$
$5$	Boron	$B$	$10.811$

How to calculate the average atomic mass?

Let us understand this with the help of the following examples.

Example: 1

Oxygen is the most abundant element in both the Earth’s crust and the human body. In nature, it exists as a combination of three stable isotopes, as shown in the table:

Isotopes of oxygen:

Isotope	Mass	% abundance
$_{8}\textrm{O}^{16}$	$15.9949$	$99.757$
$_{8}\textrm{O}^{17}$	$16.9991$	$0.038$
$_{8}\textrm{O}^{18}$	$17.9992$	$0.205$

$\begin{array}{l} Atomic mass of oxygen \\ = (15.9949 \times 0.99757) + (16.9991 \times 0.00038) + (17.9992 \times 0.00205) \\ = 15.999 amu \end{array}$

Hence, the atomic mass of oxygen is

$15.999$

$amu$ .

Example: 2

Boron naturally found as a mixture of boron-

$10$ (

$5$ protons +

$5$ neutrons) and boron-

$11$ (

$5$ protons +

$6$ neutrons) isotopes. The percentage abundance of boron-

$10$ is

$20$ , and that of boron-

$11$ is

$80$ . The atomic mass of boron is then determined as follows:

$\begin{array}{l} Atomic mass of boron \\ = (10 \times \frac{20}{100}) + (11 \times \frac{80}{100}) \\ = (10 \times 0.2) + (11 \times 0.8) \\ = 2 + 8.8 \\ = 10.8 amu \end{array}$

Hence, the atomic mass of boron is

$10.8$

$amu$ .

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4. Average Atomic Mass (AAM): Calculation

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