Problem calculating the atomic weight of an isotope.

[ihaveabutt]

Boron has two naturally occurring isotopes, lOB and 11B. We know that 80.22% of its atoms are 11B, atomic weight 11.009 amu. From the natural atomic weight given on the inside back cover, calculate the atomic weight of the lOB isotope.

Solution
If 80.22% of all boron atoms are 11B, then 100.00 - 80.22, or 19.78%, are the unknown isotope. We can use W to represent the unknown atomic weight in our calculation:
(0.8022 x 11.009) + (0.1978 x W) = 10.81 amu (natural atomic weight)
W = (10.81-8.831)/0.1978 = 10.01 amu

On the part “(0.8022 x 11.009) + (0.1978 x W) = 10.81 amu (natural atomic weight),” how do I know what to multiply 0.1978 by if W is unknown? How did they arrive at 10.81?

 

[oli4]

Hi youhaveabuttandamazinglgysodoI
They didn't 'arrive' at 10.81, 10.81 is a given, it was measured, presumably.
W is the unknown and is multiplied by its known ratio so as to put the equation that is then solved

Cheers...

 

[Knightycloud]

OK! This is done like this.

80.22% B atoms weight 11.009 amu and the other 19.78% weight let's say "m".

So we can construct a statement using all the data given and that is;

\frac{11.009\;amu\;×\;80.22\;+\;m\;amu\;×\;19.78}{100} = 10.81 amu

11.009 × 80.22 = 883.1

\frac{(883.1\;amu\;+\;m19.78\;amu)\;×\;100}{100} = 1081 amu

m19.78 amu = (1081 - 883.1) amu

∴m = \frac{(1081 - 883.1)\;amu}{19.78}

So m comes as 10.005 ≈ 10.01 amu