Question

If the abundance of Li (6.015121amu) is 7.500% and the abundance of Li (7.016003 amu )is 92.500% what is the average atomic mass
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Answer 1

Answer:

solving in another way this this is correct

Answer 2

The relative atomic mass of element having two isotopes with mass number 6.015121 and 7.016003 is  6.940937 amu.

Given,

Percentage(percentage abundance) of element with mass number 6.015121=7.5%

Percentage(percentage abundance) of element with mass number 7.016003=92.5%.

To find,

the relative(average) atomic mass of the element.

Solution:

• The relative atomic mass or average atomic mass of an element is given by the sum of the masses of isotope, each multiplied by its relative abundance percentage.
• $AAM=\frac{M1(abundance)1+M2(abundance)2}{100}$.
• Its unit can be amu(atomic mass unit) or grams.

The relative atomic mass of the element will be,

$AAM=\frac{M1(abundance)1+M2(abundance)2}{100} \\\\AAM=\frac{6.015121X7.51+7.016003X92.5}{100}\\ AAM=\frac{45.1134075+648.9802775}{100} \\AAM=\frac{694.093685}{100} \\AAM=6.94093685\\AAM=6.940937 amu.$

Hence, the relative atomic mass is 6.940937 amu.

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