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# Nitrogen has two naturally occurring isotopes: nitrogen-14 ISBN: 9780321910295 34

## Solution for problem 104P Chapter 5

Introductory Chemistry | 5th Edition

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Problem 104P

Nitrogen has two naturally occurring isotopes: nitrogen-14 (mass= 14.00 amu) and nitrogen-15 (mass = 15.00 amu). Bromine also has two naturally occurring isotopes: bromine-79 (mass = 78.92 amu) and bromine-81 (mass = 80.92 amu). How many types of NBr3 molecules of different masses can exist? Determine the mass (in amu) of each of them.

Step-by-Step Solution:
Step 1 of 3

Solution 104P

Based on N-14, N-15 , Br-79 and Br-81, 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 N14Br279Br81 252.76 amu N14Br79Br281 254.76 amu N14Br381 256.76 amu N14Br379 250.76 amu N15Br279Br81 253.76 amu N15Br79Br281 255.76 amu N15Br381 257.76 amu N15Br379 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 N-14) + 2 x (atomic mass of Br-79) + 1 x (atomic mass of Br-81)

= 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 N-14) + 1 x (atomic mass of Br-79) + 2 x (atomic mass of Br-81)

= 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...

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

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