Calculate fraction of relative molecular mass

[jdoyle]

Can anyone help with the following question - see attachment.
Answer C is correct.
I have been told that it can be solved thus:

As our compound contains carbon and hydrogen we need to calculate these masses and deduct them from 128 to give us the mass of the chlorine and bromine combination.

2H = 2 and C = 12.

128 - 14 = 114. This is the remaining mass of the chlorine and bromine.

The only combination of isotopes that give us this value are Cl 35 and Br 79.

We know that the chance of Cl being 35 is 3:4 or ¾ and the chance of Br being 79 is ½.
If we now multiply these fractions we find the required fraction of the compound.

¾ x ½ = 3/8. Answer C is correct.


However what I do not understand is how this gives the correct fraction. The mass of the compound is (12) + (2) + [(3/4 x 35) + (1/4 x 37)] + [(1/2 x 79) + (1/2 + 81)] = 129.5

Therefore the fraction of the mass of 128 to the total mass of 129.5 = 99% approx. Even if I just select the elements with a combined mass of 128 i.e. 12 + 2 + 26.25 + 39.5 = 79.75, this gives me 79.75/129.5 = 61.5% approx. which is not 3/8.

What I also do not understand is why the fractions are multiplied and why the mass of the hydrogen and carbon (which must be added to give a mass of 128) do not affect this fraction.

Can anyone help? All replies greatly appreciated.

Thanks

John

 

[Borek]

However what I do not understand is how this gives the correct fraction. The mass of the compound is (12) + (2) + [(3/4 x 35) + (1/4 x 37)] + [(1/2 x 79) + (1/2 + 81)] = 129.5

Yes, this molar mass, averaged over all possible combinations.

Therefore the fraction of the mass of 128 to the total mass of 129.5 = 99% approx. Even if I just select the elements with a combined mass of 128 i.e. 12 + 2 + 26.25 + 39.5 = 79.75, this gives me 79.75/129.5 = 61.5% approx. which is not 3/8.[q/uote]

No idea what you are doing here. You are apparently confusing something with something generating complete mess 128/129.5 is a ratio of mass of a particular molecule to the average mass of the compound molecules - and it has nothing to do with the isotopic fractions.

What I also do not understand is why the fractions are multiplied and why the mass of the hydrogen and carbon (which must be added to give a mass of 128) do not affect this fraction.

In all molecules CH₂ has an identical mass of 12, so it doesn't change the outcome. But it is not ignored - it is present in all masses involved.

Bromochloromethane is a mixture of CH₂³⁵Cl⁷⁹Br (m=128), CH₂³⁷Cl⁷⁹Br (m=130), CH₂³⁵Cl⁸¹Br (m=130) and CH₂³⁷Cl⁸¹Br (m=132). Fraction of each is given by the product of fractions of the isotopes:

CH₂³⁵Cl⁷⁹Br (m=12+2+35+79=128) - 3/4*1/2=3/8
CH₂³⁷Cl⁷⁹Br (m=12+2+37+79=130) - 1/4*1/2=1/8
CH₂³⁵Cl⁸¹Br (m=12+2+35+81=130) - 3/4*1/2=3/8
CH₂³⁷Cl⁸¹Br (m=12+2+37+79=132) - 1/4*1/2=1/8

Sum of fractions is 1, as expected.

These fractions can be determined in a different way, for example preparing pieces of paper pretending to be isotopes of Br and Cl (just remember to have them in correct ratios), drawing them form the bag, calculating mass of the 'obtained molecule' and noting its occurrence. After many trials you will see that fractions of different molecules are identical to these calculated.

Molar mass is the weighted average:

$$M = 128\times\frac 3 8 + 130\times\frac 1 8 + 130\times\frac 3 8 + 132\times\frac 1 8 = 129.5$$

 

[jdoyle]

Hi PFAdmin,

Thanks very much. I had become so confused I couldn't see the wood for the trees. Your explanation is very clear. Thanks again.

John