# Chemistry help

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Hi,

I don't really understand how to do this question, any help?

"A sample of rubidium contains the isotopes ^85Rb and ^87Rb only. The isotope ^85Rb has an abundance 2.5 times greater than that of ^87Rb. Calculate the relative atomic mass of rubidium in the sample."

I don't really understand how to do this question, any help?

"A sample of rubidium contains the isotopes ^85Rb and ^87Rb only. The isotope ^85Rb has an abundance 2.5 times greater than that of ^87Rb. Calculate the relative atomic mass of rubidium in the sample."

0

reply

Report

#2

(Original post by

Hi,

I don't really understand how to do this question, any help?

"A sample of rubidium contains the isotopes ^85Rb and ^87Rb only. The isotope ^85Rb has an abundance 2.5 times greater than that of ^87Rb. Calculate the relative atomic mass of rubidium in the sample."

**Vanilla Twilight**)Hi,

I don't really understand how to do this question, any help?

"A sample of rubidium contains the isotopes ^85Rb and ^87Rb only. The isotope ^85Rb has an abundance 2.5 times greater than that of ^87Rb. Calculate the relative atomic mass of rubidium in the sample."

how to work it out : the relative atomic mass is worked out using the following formula, illustrated for two isotopes, where the abundances are given in percentage values.

Attachment 703790

0

reply

(Original post by

definition of what it is : the relative atomic mass of an element is a weighted average of the masses of the atoms of the isotopes - because if there is much more of one isotope then that will influence the average mass much more than the less abundant isotope will

how to work it out : the relative atomic mass is worked out using the following formula, illustrated for two isotopes, where the abundances are given in percentage values.

Attachment 703790

**clouddbubbles**)definition of what it is : the relative atomic mass of an element is a weighted average of the masses of the atoms of the isotopes - because if there is much more of one isotope then that will influence the average mass much more than the less abundant isotope will

how to work it out : the relative atomic mass is worked out using the following formula, illustrated for two isotopes, where the abundances are given in percentage values.

Attachment 703790

0

reply

Report

#4

(Original post by

Thanks, I know how to work out relative atomic mass on simple calculations, but this one requires you to work out the relative abundances yourself first, and I don't know how to do that?

**Vanilla Twilight**)Thanks, I know how to work out relative atomic mass on simple calculations, but this one requires you to work out the relative abundances yourself first, and I don't know how to do that?

0

reply

Report

#5

Essentially what you're doing is calculating an average between the 85 isotope and 87 isotope of Rubidium.

To do this, you start off with figuring out the ratio of 85:87 isotopes, which as the question told you is:

2.5:1 (because the 85 isotope is 2.5x greater in abundance than the 87 isotope). This then equals:

5:2 (I just prefer looking at it this way as there are no decimals, but you can work it out from the previous ratio just as easily).

Now you have an exact ratio, you just solve to find the average/mean. So,

[(5 x 85) + (2 x 87)]/7

I believe this should give you the correct answer. Similarly, you should be able to use this method for every relative abundance calculation you come across.

To do this, you start off with figuring out the ratio of 85:87 isotopes, which as the question told you is:

2.5:1 (because the 85 isotope is 2.5x greater in abundance than the 87 isotope). This then equals:

5:2 (I just prefer looking at it this way as there are no decimals, but you can work it out from the previous ratio just as easily).

Now you have an exact ratio, you just solve to find the average/mean. So,

[(5 x 85) + (2 x 87)]/7

I believe this should give you the correct answer. Similarly, you should be able to use this method for every relative abundance calculation you come across.

3

reply

Report

#6

Well the whole sample adds up to 100% right?

And 85 Rb is 2.5 times more. So you could say the abundance of Rubidium 85 is 2.5x (%) , and the abundance of Rubidium 87 is x (%)

Therefore you could say that 3.5x = 100 (%)

Solve to find x

Sub in : (85 times (x times 2.5)) + (87 times x) / 100

And 85 Rb is 2.5 times more. So you could say the abundance of Rubidium 85 is 2.5x (%) , and the abundance of Rubidium 87 is x (%)

Therefore you could say that 3.5x = 100 (%)

Solve to find x

Sub in : (85 times (x times 2.5)) + (87 times x) / 100

1

reply

(Original post by

Essentially what you're doing is calculating an average between the 85 isotope and 87 isotope of Rubidium.

To do this, you start off with figuring out the ratio of 85:87 isotopes, which as the question told you is:

2.5:1 (because the 85 isotope is 2.5x greater in abundance than the 87 isotope). This then equals:

5:2 (I just prefer looking at it this way as there are no decimals, but you can work it out from the previous ratio just as easily).

Now you have an exact ratio, you just solve to find the average/mean. So,

[(5 x 85) + (2 x 87)]/7

I believe this should give you the correct answer. Similarly, you should be able to use this method for every relative abundance calculation you come across.

**_RobbieL_**)Essentially what you're doing is calculating an average between the 85 isotope and 87 isotope of Rubidium.

To do this, you start off with figuring out the ratio of 85:87 isotopes, which as the question told you is:

2.5:1 (because the 85 isotope is 2.5x greater in abundance than the 87 isotope). This then equals:

5:2 (I just prefer looking at it this way as there are no decimals, but you can work it out from the previous ratio just as easily).

Now you have an exact ratio, you just solve to find the average/mean. So,

[(5 x 85) + (2 x 87)]/7

I believe this should give you the correct answer. Similarly, you should be able to use this method for every relative abundance calculation you come across.

0

reply

Thanks for replying! Sorry if this seems obvious, but where did you get the 3.5 from?

0

reply

(Original post by

oh yes - ive forgotten how to work out abundance but i can go and check my books from last year if you want ahah

**clouddbubbles**)oh yes - ive forgotten how to work out abundance but i can go and check my books from last year if you want ahah

0

reply

Report

#10

(Original post by

Thanks for replying! Sorry if this seems obvious, but where did you get the 3.5 from?

**Vanilla Twilight**)Thanks for replying! Sorry if this seems obvious, but where did you get the 3.5 from?

And together, they make 100% of the sample, so you could say

x + 2.5x = 100% of the sample

so 3.5x = 100%

x = 28.571%

We've solved for the unknown now. Then plug in x into the standard abundance equation

(85 times ( 2.5 times 28.571)) + (87 times 28.571) /100

0

reply

Oh ok, thanks for the help! Do you know whether these sorts of questions where they require you to work out the abundances are common in exams? I've never come across them before.

0

reply

Report

#12

(Original post by

Oh ok, thanks for the help! Do you know whether these sorts of questions where they require you to work out the abundances are common in exams? I've never come across them before.

**Vanilla Twilight**)Oh ok, thanks for the help! Do you know whether these sorts of questions where they require you to work out the abundances are common in exams? I've never come across them before.

0

reply

I guess it's just simple maths when you know how. Thanks for the help.

0

reply

Report

#14

**Vanilla Twilight**)

Oh ok, thanks for the help! Do you know whether these sorts of questions where they require you to work out the abundances are common in exams? I've never come across them before.

1

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top