# how to work out percentage abundance chemistry

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#1
1 (b) A naturally occurring sample of the element boron has a relative atomic mass of 10.8
In this sample, boron exists as two isotopes, 10B and 11B
1 (b) (i) Calculate the percentage abundance of 10B in this naturally occurring sample of boron.
[2 marks]
1
5 years ago
#2
(Original post by jellyandjam)
1 (b) A naturally occurring sample of the element boron has a relative atomic mass of 10.8
In this sample, boron exists as two isotopes, 10B and 11B
1 (b) (i) Calculate the percentage abundance of 10B in this naturally occurring sample of boron.
[2 marks]
Which exam board is this from?
0
5 years ago
#3
(Original post by ravioliyears)
Which exam board is this from?
AQA
0
5 years ago
#4
it should give you the percentage abundance of other isotope - 11B?
0
5 years ago
#5
(Original post by MA1998)
AQA
Thank you!
(Original post by batoot)
it should give you the percentage abundance of other isotope - 11B?
Yeah, thats what I was thinking... o.O
0
5 years ago
#6
80% boron11, 20% boron10
(0.8*11)+(0.2*10)=10.8
I think...
1
5 years ago
#7
I've seen this very question from an AQA paper and no percentage abundance of either of the isotopes was given.

This can be solved through trial and error or algebraically.
The user above me is correct, there's 20% boron-10 and 80% boron-11
1
5 years ago
#8
It's 80% 11B and 20% 10B, I find it kind of hard to explain but I'll give it a go.

If it was 50/50 the average mass would be 10.5, so to raise the mass you need a higher percentage of 11.

Another way of looking at it is 10x + 11y= 10.8

10x+11y=10.8
x+y=1 (ratio)

10x+11y=10.8
10x+10y=10

Take bottom away from top
y=0.8
x+y=1 therefore x=0.2

then multiply by 100 to get percentages as these are ratios

So there is 80% 11 and 20% 10

Hope this helped
6
5 years ago
#9
(Original post by baz221)
It's 80% 11B and 20% 10B, I find it kind of hard to explain but I'll give it a go.

If it was 50/50 the average mass would be 10.5, so to raise the mass you need a higher percentage of 11.

Another way of looking at it is 10x + 11y= 10.8

10x+11y=10.8
x+y=1 (ratio)

10x+11y=10.8
10x+10y=10

Take bottom away from top
y=0.8
x+y=1 therefore x=0.2

then multiply by 100 to get percentages as these are ratios

So there is 80% 11 and 20% 10

Hope this helped
Yep, these are the ways in which I would have worked the question out. The former is trial and error and the latter is algebraically involving a simultaneous equation. 0
5 years ago
#10
x+y=1 (ratio)
y=1-x

10x+11(1-x)
10x-11x+11=10.8
-x=-0.2
x=0.2

Multiply by 100 to get a percentage and find other percentage:
0.2x100=20%
100-20=80%

20% boron-10 & 80% boron-11
0
#11
(Original post by baz221)
It's 80% 11B and 20% 10B, I find it kind of hard to explain but I'll give it a go.

If it was 50/50 the average mass would be 10.5, so to raise the mass you need a higher percentage of 11.

Another way of looking at it is 10x + 11y= 10.8

10x+11y=10.8
x+y=1 (ratio)

10x+11y=10.8
10x+10y=10

Take bottom away from top
y=0.8
x+y=1 therefore x=0.2

then multiply by 100 to get percentages as these are ratios

So there is 80% 11 and 20% 10

Hope this helped
Thankyouuu!! 0
#12
Thanks everyone!! Merrry christmas! 0
5 years ago
#13
aqa
0
4 years ago
#14
You have to solve this algebraically like this:
10.8= 10 (x) 11 (1-x)
So 10.8= 10x 11 -11x
So now you have to get the x's to one side:11 - 10.8= 10x - 11x ( note because it was 11 so to take it to the other side u have to do the opposite which is to subtract it)
Now solve: 0.2 = -1xNow divide both sides by x so:X= 0.2 0.2 x 100 because it's asking for a percentage so:X = 20 ;/;
so this means tht 10B= 20;/;
This means that 11B is 80;/; ( because 100-20 = 80)
0
6 months ago
#15
I did it in a different way. So what you would do, would be to create a table, with 10B and 11B at the top. Under those, write x, and 100-x. The reason for writing x is because thats the one we need to find out, but for 100-x it's because both of the values for "x" have to equal 100.

Seeing as the RAM is 10.8, you would work backwards.
10.8 = 10 (x) + 11 (100-x) /100. You would multiply the values of the Borons together with the x to get the unknown. Divide it by 100 to get the percentage.

After that, it's basic solving. I got 1080 = 10x + 1100 - 11x. (got the 1080 by multiplying it by 100). 10x-11x is -x.
1080=-x+1110
1080 - 1100 = -20
-20 = -x, therefore, x is equals to 20
0
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