isotope question

[joonkimdds]

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Boron found in nature has an atomic weight of
10.811 and is made up of the isotopes 10B (mass
10.013 amu) and 11B (mass 11.0093 amu). What
percentage of naturally occurring boron is made up of
10B and 11B respectively?

*10 from 10B is the superscript.

 

[tranv117]

Boron found in nature has an atomic weight of
10.811 and is made up of the isotopes 10B (mass
10.013 amu) and 11B (mass 11.0093 amu). What
percentage of naturally occurring boron is made up of
10B and 11B respectively?

*10 from 10B is the superscript.

X/100(mass10B) + 100-X/100(mass11B) = 10.811

solve for X.

There may be an easier way to set up the equation, but this is how Id do it.

 

[utdent20]

is that the right answer?

 

[tranv117]

yes. Well X will give you the percentage of one. and the other percentage youd just subtract from 100.

 

[Andre3k]

Here you go:

x= %10B
y= % 11B

x+y=1
y= 1-x

(10.013)(x) + (11.0093)👍 = 10.11
(10.013)(x) + (11.0093)(1-x) = 10.11
10x + 11 - 11x = 10.11
x = .189 = 18.9% 10B

.189 + y = 1
y= .811 = 81.1% 11B

Answer: 18.9% 10B and 81.1% 11B 👍

 

[joonkimdds]

u guys r monsters 😀

 

[klutzy1987]

well percentage is out of 100 where 100 is the highest possible percent. Now The resuting mass is made up on average of the two isotopes where one is in morer abundance than the other. The average would be ((massA*%A)+(massB*resulting %))/100

resulting % = 100%-%A

This is simplified into (massA*%A + massb*100%-%A)/100

The answer when calculated for %A is 18.9%, and the resulting %B is 81.1%

A is Boron mass 10.