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# Finding percentage composition by mass of isotopes

1. Hi, sorry if this is a really easy question but I don't actually recall doing this topic before. I don't know if I got the right answer or not but I definitely know that my working out wasn't efficient at all and idk what technique I was meant to use. I'll post the question and my working:

Question

The relative atomic mass of neon is 20.18. It consists of three isotopes, 20Ne, 21Ne and 22Ne. It contains 90.5% 20Ne. Find the percentage composition by mass of the other two isotopes.

Working

20Ne: 90.5% = 0.905
21Ne: x%
22Ne: y%

x + y = 9.5% = 0.095

Then I made an equation out of these figures:
(20*0.905) + 21x + 22y = 20.18
21x + 22x = 2.08

And then I used the other equation to make a pair of simultaneous equations (??)
x + y = 0.095
[*22] 22x + 22y = 2.09

If 22x + 22y = 2.09 and 21x + 22y = 2.08, then x = 0.01, and so y = 0.085

So:
20Ne: 90.5%
21Ne: 1%
22Ne: 8.5%

Now, I think that this could be the right answer, but I feel like I took completely the wrong approach, because I've never really used algebra (and definitely not simultaneous equations) in chemistry before How are you actually meant to do these questions?
2. (Original post by surina16)
Hi, sorry if this is a really easy question but I don't actually recall doing this topic before. I don't know if I got the right answer or not but I definitely know that my working out wasn't efficient at all and idk what technique I was meant to use. I'll post the question and my working:

Question

The relative atomic mass of neon is 20.18. It consists of three isotopes, 20Ne, 21Ne and 22Ne. It contains 90.5% 20Ne. Find the percentage composition by mass of the other two isotopes.

Working

20Ne: 90.5% = 0.905
21Ne: x%
22Ne: y%

x + y = 9.5% = 0.095

Then I made an equation out of these figures:
(20*0.905) + 21x + 22y = 20.18
21x + 22x = 2.08

And then I used the other equation to make a pair of simultaneous equations (??)
x + y = 0.095
[*22] 22x + 22y = 2.09

If 22x + 22y = 2.09 and 21x + 22y = 2.08, then x = 0.01, and so y = 0.085

So:
20Ne: 90.5%
21Ne: 1%
22Ne: 8.5%

Now, I think that this could be the right answer, but I feel like I took completely the wrong approach, because I've never really used algebra (and definitely not simultaneous equations) in chemistry before How are you actually meant to do these questions?
No this is all correct. Algebra is probably the only way to tackle these types of problems.
3. (Original post by MathsMonster)
No this is all correct. Algebra is probably the only way to tackle these types of problems.
Oh great aha I just didn't realise because I haven't actually learnt this in class yet it's my homework
Didn't know there'd be this much maths in chem though - which is good because i love it

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