Secondaire 4 • 4a
I don't understand how to solve this problem:
The atomic mass of isotope # 1 of element Ab is 73 au. The atomic mass of isotope # 2 of the element Ab is 76 au. The atomic mass of isotope # 3 of the element Ab is 74 au. The relative frequencies of each isotope are 10%, 80% and 10%, respectively.
Find the average atomic mass.
Thank you!
Explication d'Alloprof
Cette explication a été donnée par un membre de l'équipe d'Alloprof.
Thank you for your question!
The question refers to the average atomic mass, that is, the average mass of the atoms of the (fictitious) element Ab. The average atomic mass is that shown in the periodic table. It weights the mass of each isotope with its abundance in nature in order to find an average.
To find the average atomic mass of an element, we must multiply its atomic mass by its abundance in nature and then add all the products of this operation.
In the question, the results obtained will then be:
$$73\:ua•0,1=7,3\:ua$$
$$76\:ua•0,8=60,8\:ua$$
$$74\:ua•0,1=7,4\:ua$$
Note: converting percentages into numbers with decimals makes the sum easier .
$$M_{average}=7,3\:ua+60,8\:ua+7,4\:ua=75,5\:ua$$
Here it is!
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