# Problem calculating the atomic weight of an isotope.

1. Jun 9, 2012

### ihaveabutt

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?

2. Jun 9, 2012

### 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...

3. Jun 9, 2012

### Knightycloud

OK!!! This is done like this.

80.22% B atoms weight 11.009 amu and the other 19.78% weight lets 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