Home Past Papers of NUST Website
Recommended Post : Online Preparation for NET-3

Tuesday, October 29

Calculating Relative Atomic Mass

Suppose we have an element with two isotopes, let’s name them ‘a’ and ‘b’. Their relative abundances are ‘x’ and ‘y’. So their relative atomic mass is given by : Relative Atomic Mass formula Lets define values for our variable a,b,x and y Let the mass of isotope ‘a’ is 12 and of ‘b’ is 14, their relative abundances ‘x’ and ‘y’ be 50% and 50% respectively. The Relative atomic mass is given by Formula with defined values In exam you don’t have enough time to calculate such equations specially if you’re biology student. But without any calculation I can tell you that the answer is ‘13’, the mid value of ‘12’ and ‘14’ because both of them contribute half or 50%. Now let’s take some odd numbers masses are the same i.e. a = 12 and b = 14 . let their relative abundances be x = 75% and y = 25%. The Relative atomic mass is given by Formula with different values As we see that isotope ‘a’ has greater abundance 75% which means the Relative atomic mass will be closer to 12 but it cannot be 13 because it is mid value and will occur only when both contributes equally. This means the answer must be between 12 and 13. (As I’ve practiced this method a lot so I can tell that answer is 12.5 because if 50% changes the Relative atomic mass value to 1 from the closer value than 25% will cause half of the change i.e. 0.5 so just add 0.5 to 12 and this will be answer.) This was some mathematics for more complex values this probably won’t work, but there is a solution. Let’s draw a line with some calibrations A Simple line to be calibrated A simple line? No! To be more precise I’m going to calibrate it furthermore A calibrated line As you see in the above line that between 12 and 13 there are 10 divisions and each division corresponds to 0.1 amu this means that with each step you move forward or backward the value increase or decrease by 0.1. 12 is marked as 0% and 13 with 50 so the difference is 50 in these 10 division so each division corresponds to 5%. If you move one step forward you actually move 5% forward. Now if you start from 12 and move 1 step you move 5% and 0.1 amu, so at 5% (when Relative abundance of a is 95% and of b is 5%) the relative atomic mass is 12.1 and so on. Now let’s do that example using the line formula. The relative abundance of isotope ‘a’ is 75% . Subtract it from 100. 100 - 75 = 25 , now starting from beginning of line i.e. from 0% creep to 25% at 25% the other calibrated value is 12.5 , which is the answer . You can practice this method by using your own values.

1 comment:

Home Solved Past Papers of NUST Website