[cabaal]

"The mass spec of (CH₃)₄Si has a base peak at m/z=73. Calculate the relative abundance of the isotopic peaks at m/z=74 & m/z=75."

Isotopes are from ²⁹Si and ³⁰Si with the base peak being ²⁸Si. There is also ¹²C and ¹³C.

Relative abundance is # of atoms * natural abundance of isotope 1/natural abundance of isotope 2.

m/z 74 = 1 Si atom * (²⁹Si abundance)/(²⁸Si abundance) = 1 * 0.0467/0.9221 * 100% = 5.06%. I understand this.
m/z 75 = 1 Si atom * (³⁰Si abundance)/(²⁸Si abundance) = 1 * 0.0310/0.9221 * 100% = 3.36%. I understand this.

(CH₃)₄Si = 88 g/mol - 74 m/z ratio (presumably unit charge) = 15 mass unit loss (e.g., 1 CH₃.) This implies that there are 3 remaining methane groups, and consequently 3 isotopes (¹³CH₃.) I understand this.

The solutions manual gives a calculation that I don't understand:

3 C atoms * (¹³C abundance)/(²⁸Si abundance) = 3 * 0.011/0.9221 * 100% = 3.61%

Why am I dividing by the relative abundance of ²⁸Si? Why am I not dividing by the relative abundance of ¹²C?

It also says "the contribution of ¹³C to the m/z=75 peak is negligible." Why?

[MOTOBALL]

Your third line,

RA is # of atoms *natural abundance of isotope 1/nat. abund. of isotope 2

is true only for a single atom of the element.

In general case, the theoretical RA values are derived from probability theory, and are of the form
of the binomial expansion,

(a + b) to the nth power, where a is nat. abund. of isotope 1
                                            b ..  2

and n is the number of atoms of that element.

For example, 35Cl = about 72 % RA, and 37Cl = about 24 % RA (numbers approximated to make the calculation more obvious)

For Cl2, a = 0.72, b = 0.24, n = 2
So, (a + b) squared = a sqd + 2ab + b sqd = ratio is 0.5184:0.3360:0.0576

= 9:6:1 approximately

Here, n = 1 for Si but n = 3 for C, and n = 9 for H.

m/z 73 = (CH3)3Si, so m/z 75 = 12C3 1H9 30Si, RA = 3.36 %
                 and also m/z 75 = 13C2 12C1 1H9 28Si
ratio = (a + b) cubed, where a = 98.90, b = 1.10 % and n = 3 C atoms.[see calc. below]

You can see that the 13C3 contribution (0.00014%) IS negligible compared to the 30Si contribution (3.36%).

for m/z 74 = 29Si gives 5.06 % RA
and also m/z 74 = 13C1 12C2 1H9 28Si, (a + b) to nth power gives
(a cubed )+ (3* a sqd *b )+ (3 * b sqd * a) + (b cubed), since n = 3 C atoms
a = 98.9%, b = 1.10%

so ratio m/z 73:74:75 is 967362:32278 + 359:1.331  =  100:3.37:0.00014.

These probabilities are independent of each other, so are additive.

So if m/z 73 =100 % (base peak),

then m/z 74 = 5.06 for 29Si + 3.37 for the 13C1 12C2 = 8.43%.

The calculation given by the solutions manual is I believe INCORRECT, since the natural relative  abundances of the elements, and therefore the probabiliites of occurrence, are independent of each other.

Your calculation for 30Si = 3.36% should be correct.

It has been quite a few yrs since I have had to grind this out, line by line, so have someone else check it.