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Nitrogen has two naturally occurring isotopes: nitrogen-14 (mass= 14.00 amu) and

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 104P Chapter 5

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 104P

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

  1. N14Br279Br81

252.76 amu

  1. N14Br79Br281

254.76 amu

  1. N14Br381

256.76 amu

  1. N14Br379

250.76 amu

  1. N15Br279Br81

253.76 amu

  1. N15Br79Br281

255.76 amu

  1. N15Br381

257.76 amu

  1. 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) +...

  1. 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 x (atomic mass of N-14) + 3 x (atomic mass of Br-81)

                = 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 N-14) + 3 x (atomic mass of Br-79)

                = 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 N-15) + 2 x (atomic mass of Br-79) + 1 x (atomic mass of Br-81)

                = 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 N-15) + 1 x (atomic mass of Br-79) + 2 x (atomic mass of Br-81)

                = 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 N-15) + 3 x (atomic mass of Br-81)

                = 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 N-15) + 3 x (atomic mass of Br-79)

                = 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

Step 2 of 3

Chapter 5, Problem 104P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Nitrogen has two naturally occurring isotopes: nitrogen-14 (mass= 14.00 amu) and