Fundamental E11

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CenCen
Posts: 29
Joined: Fri Sep 25, 2015 3:00 am

Fundamental E11

Postby CenCen » Sat Nov 28, 2015 3:00 pm

The nuclear power industry extracts 6Li but not 7Li from natural samples of lithium. As a result, the molar mass of commercial samples of lithium is increasing. The current abundances of the two isotopes are 7.42% and 92.58%, respectively, and the masses of their atoms are 9.988*10^-24 and 1.165*10^-23 respectively. (A) what is the current molar mass of a natural sample of lithium? (B) what will the molar mass be when the abundance of 6Li is reduced to 5.67?

Can someone just walk me through the process of solving this. Thank you.

Anne Cam 3A
Posts: 71
Joined: Fri Sep 25, 2015 3:00 am

Re: Fundamental E11

Postby Anne Cam 3A » Sun Nov 29, 2015 5:55 pm

To find the average molar mass of an element from the percentage of its isotopes, you would multiply the percentages (in decimals) by their respective molar masses.

In (a), only the masses of 6Li and 7Li atoms are given, so you would need to convert them to g/mol. Multiply each mass value by 6.022*10^23, the number of atoms in a mole, to get the molar mass for each Li isotope. (About 6.0148 g/mol for 6Li, and 7.0156 g/mol for 7Li)

Then multiply each molar mass by its corresponding percentage, and add the results to get the average molar mass. It would look like this:

(0.0742*6.0148) + (0.9258*7.0156) = 6.94 g/mol (rounded with 2 significant figures)

For part (b), you would keep the molar mass values for each isotope, but change the percentage of 6Li to 5.67%, and again multiply each molar mass by its corresponding percentage to get the average molar mass. The process is the same, except one percentage is changed.

Hope this helps!

Nicolette_Canlian_2L
Posts: 77
Joined: Fri Sep 28, 2018 12:25 am
Been upvoted: 1 time

Re: Fundamental E11

Postby Nicolette_Canlian_2L » Sun Sep 30, 2018 11:38 pm

The method I used to solve part a is by multiplying the mass of each with its correlating percentage of abundance and then added these products together.
m=0.0742(9.988x10^-24g)+0.9258(1.165x10^-23g)
m=1.153x10^-23g
Then I multiplied the sum of those two with Avogadro's constant and got 6.94g/mol


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