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Use the law of constant composition to complete the table

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

Solution for problem 29P 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 29P

Use the law of constant composition to complete the table summarizing the amounts of nitrogen and oxygen produced upon the decomposition of several samples of dinitrogen monoxide.

 

Mass N2O

Mass N

Mass O

Sample A

2.85 g

1.82 g

1.03 g

Sample B

4.55 g

_____

_____

Sample C

_____

_____

1.35 g

Sample D

_____

1.11 g

_____

Step-by-Step Solution:

Solution 29P

Step 1:

First, let’s see what is the law of constant composition :

Law of constant composition states that ‘samples of a pure compound always contain the same elements in the same mass proportion.’

Now, let’s solve the blanks using the given data .

Step 2:

Let’s consider, sample B :

Given : Mass of N2O : 4.55 g

First, let’s calculate the molar mass of N2O :

        Molar mass of Nitrogen (N) = 14.0067

        Molar mass of Oxygen (O) = 15.9994

Hence molar mass of N2O = 2(14.0067) + 15.9994

                               = 28.0134 + 15.9994

                               = 44.0128 g

If 44.0128g of N2O contains 28.0134g of N, then mass of N in 4.55g of N2O will be :

=  4.55

=  0.6364  4.55

= 2.89 g 2.9 g

        

Hence, mass of N in sample B is 2.9 g

Now we have the mass of N and N2O, let’s calculate mass of O :

Using law of constant composition, we can calculate as :

                 Mass of O = Mass of N2O  Mass ratio of O

If 44.0128g of N2O contains 15.9994g of O, then mass ratio of O in any sample will be :

         = 0.3635

Therefore , Mass of O = 4.55 g  0.3635

= 1.65 g

Hence, mass of O in sample B is 1.65 g

Step 3:

Let’s consider, sample C :

Given : Mass of O = 1.35 g

We know the molar mass of N2O from the above calculation =  44.0128 g

Now, let’s calculate mass of N2O in the sample :

If 44.0128g of N2O contains 15.9994g of O, then mass ratio of O in any sample will be :

         = 0.3635

Let’s consider mass of N2O as ‘x’

                unknown mass of N2O in sample C(x)  mass ratio of O = mass of O

                 x  0.3635 = 1.35 g

                x =

                x = 3.71 g

Hence, mass of N2O in sample C is  3.71 g

Now, let’s calculate mass of N in the sample :

If 44.0128g of N2O contains 14.0067g of N, then mass ratio of N in any sample will be :

         = 0.6364

Therefore mass of N will be :

        Mass of  N2O   mass ratio of N = mass of N

        3.71g  0.6364 = mass of N

 2.36 g

Hence, mass of N in sample C is 2.36 g        

Step 4 of 5

Chapter 5, Problem 29P is Solved
Step 5 of 5

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

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Use the law of constant composition to complete the table

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