[Big-Daddy]

If we take a compound like CBrClF₂, and you ask, what % of this compound naturally occurs as ¹²C⁷⁹Br³⁵Cl[¹⁹F]₂, it's just the fraction of Br which is ⁷⁹Br, multiplied by the fraction of Cl which is ³⁵Cl, multiplied by the fraction of C that is ¹²C, multiplied by the fraction of F that is ¹⁹F squared. This comes round to roughly 12%.

But then if we have CCl₂F₂, and we want the % which is ¹²C[³⁵Cl]₂[¹⁹F]₂, we apparently have to multiply by the fraction of Cl which is ³⁵Cl, squared (as there are 2 Cl), but then double it. Why? This seems to disagree with my procedure for the above example - if we double the Cl as there are 2 Cl atoms, why don't we double it again, as there are 2 F atoms? And why are we doubling it anyway?

[MOTOBALL]

I'm afraid that your first calc. is not correct. Also, you double each atom of Cl2 and F2 has to be taken separately.

In round numbers, 12C & 19F both= 1; 79Br = 1/2; 35Cl =3/4 for ease of arithmetic by longhand

Gives 1x 1 x 1/2 x 3/4 = ca. 38% for CF* BrCl
and    1x 1 x 1/2 x 3/4 = ca. 38% for CF** BrCl
since there are two "distinct" F atoms.

Then total for 12C79Br35ClF2 is 76%

For CCl2F2, there exist the following combos,

12C35Cl35ClF* = 3/4 x 3/4= 9/16
12C35Cl35ClF** = 3/4 x 3/4= 9/16

Then total for 12C35Cl2F2 = 2 x 9/16 = 112%

Similarly,
12C35Cl37ClF* = 1/4 x 3/4 = 3/16
12C35Cl37ClF** = 1/4 x 3/4 = 3/16
12C37Cl35ClF* = 1/4 x 3/4 = 3/16
12C35Cl37ClF** = 1/4 x 3/4 = 3/16

Then total = 4 x 3/16 = 75%

Similarly,
12C37Cl37ClF* = 1/4 x 1/4 = 1/16
12C37Cl37ClF** = 1/4 x 1/4 = 1/16

Then total = 2 x 1/16 = 12.5%

Normalizing to 100% for 12C35Cl2F2 gives 100: 67: 9, compared to the theoretical values of 100: 65: 11

[MOTOBALL]

Sorry, but MY first calc. is not correct.

Since 12C & 19F both = 1 for the purposes of the calculation, they can both be ignored in finding the isotopic abundance distribution.  We then are only concerned with the Br-Cl molecule.

So the combos are,

for m/z 114, 35Cl -79Br = 3/4 x 1/2 = 38%

for m/z 116, 35Cl - 81Br = 3/4 x 1/2= 38%
for m/z 116, 37Cl - 79Br = 1/4 x 1/2= 13%

for m/z 118, 37Cl - 81Br = 1/4 x 1/2= 13%

This gives ratios of 38:51:13, i.e. 76:100:26 compared to theoretical values of 77:100:24.

Similarly for Cl - Cl,

m/z 70,  35Cl - 35Cl = 3/4 x 3/4 = 56%

m/z 72,  35Cl - 37Cl = 3/4 x 1/4 = 19%
m/z 72,  37Cl - 35Cl = 1/4 x 3/4 = 19%

m/z 74,  37Cl - 37Cl = 1/4 x 1/4 = 6%

This gives ratios of 56:38:6, i.e. 100:67:11 compared to theoretical values of 100:65:11.

Again, sorry for the fuzzy thinking and incorrect explanation  !!!!