Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: kevinkevin on December 08, 2012, 09:07:23 PM

I have been forced to take a year break from chemistry (the class was completely full before my registration date). I am keeping my mind sharp by working out of a chemistry book and i have ran across a problem that i have no idea why the answer is not working out. The question is as fallows:
The mineral spodumene has the empirical formula LiAlSi2O6. Given the percentage of lithium6 atoms in naturally occurring lithium is 7.40%, how many Li6 atoms are present in a 518g sample of spodumene?
Since I know the flaw lies in my approach and not my calculations I will tell you step by step my thanking and hopefully someone can tell me where the flaw or flaws are.
Since it is known that I have 518g of the mineral, I multiplied that number by 7.40% (0.0740) to get the mass of lithium6 in the sample. Then I converted this mass the moles of lithium6, then I converted to atoms of lithium6. I hypothesized that this would be the correct answer but the book says it's wrong. What am I doing wrong? Thanks!

Your sample is not lithium alone. Use formula and molar masses to calculate amount of lithium in the sample first.

Given the percentage of lithium6 atoms in naturally occurring lithium is 7.40%, how many Li6 atoms are present in a 518g sample of spodumene?
There's your problem. As Borek said, The 7.4% data is for naturally occuring Li, not spodumene. Spodumene doesn't contain 7.4% Li6.

Thank you. That was a really carless mistake on my part. So what I have to do then is convert to grams of lithium and then multiply that by 7.5% to get the amount of lithium6 in the sample?

You're sort of hopping around without doing anything systematically. Try following this procedure:
1) Calculate how many moles of spodumene you have in your 518 gram sample.
2) Determine how many moles of lithium are present in each mole of spodument; use that and the result from 1) to determine how many moles of lithium you have in your sample.
3) Use the answer from 2) to determine how many grams of lithium are present in your sample.
4) Determine how much of the grams of lithium you get from 3) are Li6

Ok so I calculated the mass of lithium6 to be 1.43g. But the question is asking for atoms of lithium6. In order to do that conversion I need to know the mass of lithium6 and i'm assuming it's not the same as the mass on the periodic table. The only method I can think of to calculate the mass of lithium6 is to add the mass of three protons and three neutrons. I just have a feeling that's not the method the book has in mind.

Ok so I calculated the mass of lithium6 to be 1.43g.
Correct answer, but you don't need this number. You are right to convert it to number of atoms would require knowing mass of the lithium6 atom, which can be found in the isotope tables (and it is not just sum of masses of neutrons and protons, it is lower than that because of the mass deficit).
However, if you know mass of Li6, I suppose somewhere earlier you have calculated total mass of lithium in the sample. If you will convert this mass to number of moles of lithium, you will be able to calculate number of atoms of all isotopes. Then find 7.3% of this number. It will work, because molar mass of elements  as given in the periodic table  is already taking into account fact we are dealing with the mixture of isotopes (it is so called weighted average of their masses).

Ooops  yes, I lost track of what the question was asking. You use the number of moles of total lithium from number 2 to find the number of moles of lithium6 present, then use that number to find the total number of atoms of lithium6.
sorry about that

I've been studying a bit of chemistry over the past month and this question just so happened to be one that it looked like I might be able to understand, so I figured I would try solving it using the methods I've learned so far. Here goes:
1.) 518g of Spodumene (LiAlSi_{2}O_{6})
Breaks down to:
 310.8g of O (518g / 10 = 51.8g   considering 6 parts oxygen, we have 51.8*6 = 310.8g)
 103.6g of Si (number obtained in a manner analogous to the above example)
 51.8g of Li (as above)
 51.8g of Al (as above)
2.)Using an atomic weight of 6.94 for Li, we then have 7.46 moles of Li (51.8g / 6.94)
3.)To obtain the number of atoms present in the 7.46 moles of Li we take 7.46 * 6.02x10^{23} which comes out to 4.49x10^{24} which is the number of atoms of Li present in our 518g sample of Spodumene.
4.)Considering the question specifically asked for the number of Li6 atoms, I believe (per Borek's informative post) that we would take 7.3% of this above number. So, 4.49x10^{24} * .073 = 3.2777x10^{23}, which should be the number of Li6 atoms in the sample?
I'm reasonably certain I did okay up until step 4. I don't really understand how to isolate the number of a certain isotope of an element as is requested in the problem, but using Borek's post I took a whack at it.

1.) 518g of Spodumene (LiAlSi_{2}O_{6})
Breaks down to:
 310.8g of O (518g / 10 = 51.8g   considering 6 parts oxygen, we have 51.8*6 = 310.8g)
 103.6g of Si (number obtained in a manner analogous to the above example)
 51.8g of Li (as above)
 51.8g of Al (as above)
It appears to me that what you have done here is find the total number of atoms in a molecule of LiAlSi_{2}O_{6} (1+1+2+6=10), then assigned a weight to each atom based on its proportion by number in the molecule. Unfortunately, that assumes that each atom has the same weight, which is not the case.
Instead of adding the number of atoms, you need to add the atomic weight of the atoms. This gives you the molecular weight of the spodumene. Then you can assign the proportions based on weight. Your calculation becomes (1*(AW_{Li})+1*(AW_{Al})+2*(AW_{Si})+6*(AW_{O})=MW).

19.3+75.1+156.4+267.2 (so that I don't have to calculate it again ;))

Thanks everyone. I got the correct answer now. I can't believe how easy of a problem that was but it really helped me with other problems because I take the time to really understand what they are asking of me in the question. That was my main problem I think, not fully understanding the problem.