Problem 104P
Nitrogen has two naturally occurring isotopes: nitrogen14 (mass= 14.00 amu) and nitrogen15 (mass = 15.00 amu). Bromine also has two naturally occurring isotopes: bromine79 (mass = 78.92 amu) and bromine81 (mass = 80.92 amu). How many types of NBr3 molecules of different masses can exist? Determine the mass (in amu) of each of them.
Solution 104P
Based on N14, N15 , Br79 and Br81, eight different types of NBr3 are possible having different masses.
Here we have to calculate the formula mass of each compound.
Different types of compound are given in the table.
Compound 
Formula mass in amu 

252.76 amu 

254.76 amu


256.76 amu 

250.76 amu


253.76 amu 

255.76 amu 

257.76 amu


251.76 amu

Formula mass can be calculated as follows.
Formula mass = (no of atoms of the element ) x (atomic mass of 1st element) + (no of atoms of the element ) x (atomic mass of 2nd element) +...
 N14Br279Br81
Formula mass = 1 x (atomic mass of N14) + 2 x (atomic mass of Br79) + 1 x (atomic mass of Br81)
= 1 x 14 amu + 2 x 78.92 amu + 80.92 amu
= 252.76 amu
The mass of the compound (1) N14Br279Br81 is 252.76 amu
(2) N14Br79Br281
Formula mass = 1 x (atomic mass of N14) + 1 x (atomic mass of Br79) + 2 x (atomic mass of Br81)
= 1 x 14 amu + 1 x 78.92 amu + 2x 80.92 amu
= 14 amu + 78.92 + 161.84 amu
= 254.76 amu
The mass of the compound (2) N14Br79Br281 is 254.76 amu.
(3) N14Br381
Formula mass = 1 x (atomic mass of N14) + 3 x (atomic mass of Br81)
= 1 x 14 amu + 3x 80.92 amu
= 14 amu + 142.76 amu
= 256.76 amu
The mass of the compound (3) N14Br381is 256.76 amu.
(4) N14Br379
Formula mass = 1 x (atomic mass of N14) + 3 x (atomic mass of Br79)
= 1 x 14 amu + 3x 78.92 amu =14 amu +236.76 amu
= 250.76 amu
The mass of the compound (4)N14Br379 is 250.76 amu.
(5) N15Br279Br81
Formula mass = 1 x (atomic mass of N15) + 2 x (atomic mass of Br79) + 1 x (atomic mass of Br81)
= 1 x 15 amu + 2 x 78.92 amu + 80.92 amu
= 253.76 amu
The mass of the compound ((5) N15Br279Br81 is 253.76 amu.
(6) N15Br79Br281
Formula mass = 1 x (atomic mass of N15) + 1 x (atomic mass of Br79) + 2 x (atomic mass of Br81)
= 1 x 15 amu + 1 x 78.92 amu + 2x 80.92 amu
= 15 amu + 78.92 + 161.84 amu
= 255.76 amu
The mass of the compound (6) N15Br79Br281 is 255.76 amu
(7) N15Br381
Formula mass = 1 x (atomic mass of N15) + 3 x (atomic mass of Br81)
= 1 x 15 amu + 3x 80.92 amu
= 15 amu + 142.76 amu
= 257.76 amu
The mass of the compound (7) N15Br381 is 257.76 amu
(8) N15Br379
Formula mass = 1 x (atomic mass of N15) + 3 x (atomic mass of Br79)
= 1 x 15 amu + 3x 78.92 amu =15 amu +236.76 amu
= 251.76 amu
The mass of the compound (8)N15Br379 is 251.76 amu