last question to do Watch

elainedorey
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hi, I'm hoping someone would be able to help. Myself and the rest of my class are all stuck on this last question.

A new element has three isotopes with mass numbers of 288, 290 and 291. The weight averaged atomic mass of this element is 289.2 and the percent abundance of the heaviest isotope is 35%, what are the percent abundances of the other two isotopes?

The element in question is one the tutor has made up. Several of us have all come to different answers, one not even using algebra, we need help to make it as clear as possible and as simple as possible to understand please.
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Medic Mind
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(Original post by elainedorey)
hi, I'm hoping someone would be able to help. Myself and the rest of my class are all stuck on this last question.



The element in question is one the tutor has made up. Several of us have all come to different answers, one not even using algebra, we need help to make it as clear as possible and as simple as possible to understand please.
Hi there!
Firstly, it is important to understand that the Average Atomic Mass = The Sum of (Each Atomic Mass x Percentage Abundance) = (M1 x P1) + (M2 x P2) + (M3 x P3)

289.2 = (288 x P1) + (290 x P2) + (291 x 0.35)
289.2 = 288P1 + 290P2 + 101.85
187.35 = 288P1 + 290P2
P1 + P2 + P3 = 1 (as all percentages add up to 100%)
1-0.35 = 0.65
Hence, P1 + P2 = 0.65
Hence, there are a system of simultaneous equations to solve:

187.35 = 288P1 + 290P2
P1 + P2 = 0.65—> (Rearrange) P1 = 0.65 - P2
187.35 = 288(0.65-P2) + 290P2 —>
187.35 = 187.2 - 288P2 + 290P2 —> 2P2 = 0.15
:. P2 = 0.075
As P1 = 0.65 - P2
P1 = 0.65 - 0.075 = 0.575
Hence, P2 = 7.5% (of 290 isotope)
P1 = 57.5% (of 288 isotope)

Hope this helps! To see more resources on Chemistry and other Sciences (Biology and Mathematics) please check out our website which contains A Level and GCSE Revision Notes written by expert tutors who have attained A*s along with past paper questions with model solutions separated BY TOPIC!
www.medicmind.co.uk/resources

For more questions feel free to contact us! We also offer 1-1 tutoring sessions in GCSEs and A-Levels from only £25 per hour!

All the best!

Dhaval N - Medic Mind Tutor
Best UK Medical Startup 2017 🏆
www.interviewcourse.co.uk
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elainedorey
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Thank you for explaining it. I understand most of it, but maybe I'm having an idiot moment or its my dislexia playing up. there is just this part that I've confused myself on. I'm so sorry to be a bother.

187.35 = 288(0.65-P2) + 290P2 —>
187.35 = 187.2 - 288P2 + 290P2 —> 2P2 = 0.15
:. P2 = 0.075
As P1 = 0.65 - P2
P1 = 0.65 - 0.075 = 0.575

Thank you


(Original post by Medic Mind)
Hi there!
Firstly, it is important to understand that the Average Atomic Mass = The Sum of (Each Atomic Mass x Percentage Abundance) = (M1 x P1) + (M2 x P2) + (M3 x P3)

289.2 = (288 x P1) + (290 x P2) + (291 x 0.35)
289.2 = 288P1 + 290P2 + 101.85
187.35 = 288P1 + 290P2
P1 + P2 + P3 = 1 (as all percentages add up to 100%)
1-0.35 = 0.65
Hence, P1 + P2 = 0.65
Hence, there are a system of simultaneous equations to solve:

187.35 = 288P1 + 290P2
P1 + P2 = 0.65—> (Rearrange) P1 = 0.65 - P2
187.35 = 288(0.65-P2) + 290P2 —>
187.35 = 187.2 - 288P2 + 290P2 —> 2P2 = 0.15
:. P2 = 0.075
As P1 = 0.65 - P2
P1 = 0.65 - 0.075 = 0.575
Hence, P2 = 7.5% (of 290 isotope)
P1 = 57.5% (of 288 isotope)

Hope this helps! To see more resources on Chemistry and other Sciences (Biology and Mathematics) please check out our website which contains A Level and GCSE Revision Notes written by expert tutors who have attained A*s along with past paper questions with model solutions separated BY TOPIC!
www.medicmind.co.uk/resources

For more questions feel free to contact us! We also offer 1-1 tutoring sessions in GCSEs and A-Levels from only £25 per hour!

All the best!

Dhaval N - Medic Mind Tutor
Best UK Medical Startup 2017 🏆
www.interviewcourse.co.uk
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