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Principles of Instrumental Analysis, 6th ed. Chapter 20
 
 6
 where the subscripts s and u designate standard and unknown respectively. Dividing one 
of these equations by the other gives the desired relationship. 
 s u
u
( / ) / s = = 
( / ) / u s
m z VK V
m z K V V
 
 (b) 
u
69.00 = 0.965035
( / )m z
 
 (m/z)u = 71.50 
 (c) The approximately half-integral m/z value suggests that the ion being studied in part 
(b) was doubly charged. This conclusion is in agreement with the fact that the molecular 
mass of the unknown is 143. The second conclusion is that the unknown must contain an 
odd number of nitrogen atoms. 
20-15. The difference in mass between 12C and 13C is 1.00335. Therefore, making the 
assumption that (P + 1)+ is due only to 13C means 
 mass (P + 1)+ = mass P+ + 1.00335 
 In Problem 20-14, we derived the following relationship 
 s u
u
( / ) = 
( / ) s
m z V
m z V
 
 Taking into accound the fact that only singly charged ions were specified, and rewriting 
this equation with V1 representing the standard and V2 the unknown, we find 
 2
1
(P ) (P )
(P 1) (P ) 1.00335
Vm m
m m
+ +
+ += =
+ + V
 
 (b) Substituting the voltage ratio into the last equation allows m(P+) to be calculated 
 2
1
(P ) 0.987753
(P ) 1.00335
Vm
m V
+
+ = =
+