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relative atomic mass exam question!

RAM= 87.7, sample consists of 3 isotopes 86Sr , 87Sr, 88Sr
the ratio of abundances of isotopes 86Sr, 87Sr is 1:1
calculate % abundance of 88Sr isotope

how do get 80%? my maths is really poor so I would really appreciate an explanation bc I don't get the mark scheme
86x x 87x x 88(100-2x)=87.7/100
why is it 86x? if the abundance is 1 shouldn't it be (86x1)(87x1)88(100-x)?
why is it 100-2x?
(edited 5 years ago)
The ratio of abundances is 1, not the abundances itself

If the abundance was 1 (=100%) then there would only be 1 isotope.

Let % abundance of Sr-86 be x%
then % abundance of Sr-87 is x% (ratio is 1:1, i.e the same)
% sum to 1, so % abundance of Sr-88 = (100-2x)%

Hence 86x+87x+88(100−2x)100=87.7 \dfrac{86x + 87x + 88(100-2x)}{100} = 87.7
hence solve for x
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usernamenew
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Original post by BobbJo
The ratio of abundances is 1, not the abundances itself

If the abundance was 1 (=100%) then there would only be 1 isotope.

Let % abundance of Sr-86 be x%
then % abundance of Sr-87 is x% (ratio is 1:1, i.e the same)
% sum to 1, so % abundance of Sr-88 = (100-2x)%

Hence 86x+87x+88(100−2x)100=87.7 \dfrac{86x + 87x + 88(100-2x)}{100} = 87.7
hence solve for x


thank u so much!!

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