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Carbon has two naturally occurring isotopes: carbon-12

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

Solution for problem 103P 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 103P

Carbon has two naturally occurring isotopes: carbon-12 (mass = 12.00 amu) and carbon-13 (mass = 13.00 amu). Chlorine also has two naturally occurring isotopes: chlorine-35 (mass = 34.97 amu) and chlorine-37 (mass = 36.97 amu). How many CCl4 molecules of different masses can exist? Determine the mass (in amu) of each of them.

Step-by-Step Solution:
Step 1 of 3

Solution 103P

Based on C-12, C-13 , Cl-35 and Cl-37, 10 different types of CCl4 is 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

C12Cl335Cl37

153.88 amu

C12Cl235Cl237

155.88 amu

C12Cl35Cl337

157.88 amu

C12Cl437

159.88 amu

C12Cl435

151.88 amu

C13Cl437

160.88 amu

C13Cl435

152.88 amu

C13Cl335Cl37

154.88 amu

C13Cl235Cl237

156.88 amu

C13Cl35Cl337

158.88 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) +...

C12Cl335Cl37

Formula mass = 1 x (atomic mass of C-12) + 3 x (atomic mass of Cl-35) + 1 x (atomic mass of Cl-37)

                =  1 x 12 amu + 1 x 36.97 amu + 3 x 34.97 amu

                = 12 amu +104.91 amu + 36.97 amu

                = 153.88 amu

The mass of the compound (1) C12Cl335Cl37 is 157.88 amu.

(2) C12Cl235Cl237

Formula mass = 1 x (atomic mass of C-12) + 2 x (atomic mass of Cl-37) +2 x (atomic mass of Cl-37)

                =...

Step 2 of 3

Chapter 5, Problem 103P is Solved
Step 3 of 3

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

This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The full step-by-step solution to problem: 103P from chapter: 5 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. The answer to “Carbon has two naturally occurring isotopes: carbon-12 (mass = 12.00 amu) and carbon-13 (mass = 13.00 amu). Chlorine also has two naturally occurring isotopes: chlorine-35 (mass = 34.97 amu) and chlorine-37 (mass = 36.97 amu). How many CCl4 molecules of different masses can exist? Determine the mass (in amu) of each of them.” is broken down into a number of easy to follow steps, and 53 words. Since the solution to 103P from 5 chapter was answered, more than 1344 students have viewed the full step-by-step answer. This full solution covers the following key subjects: AMU, mass, carbon, chlorine, naturally. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295.

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