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Solved: Use the law of constant composition to complete

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

Solution for problem 30P 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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1
Problem 30P

Use the law of constant composition to complete the table summarizing the amounts of iron and chlorine produced upon the decomposition of several samples of iron(III) chloride.

 

Mass FeCI3

Mass Fe

Mass Cl

Sample A

3.785 g

1.302 g

2.483 g

Sample B

2.175 g

________

________

Sample C

_________

2.012 g

________

Sample D

_________

________

2.329 g

Step-by-Step Solution:

Solution 30P

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:

Use the law of constant composition to complete the table summarizing the amounts of iron and chlorine produced upon the decomposition of several samples of iron(III) chloride.

Let’s consider, sample B :

Given : Mass of FeCl3 : 2.175 g

First, let’s calculate the molar mass of FeCl3 :

        Molar mass of Iron (Fe) = 55.845

        Molar mass of Chloride (Cl) = 35.453

Hence molar mass of FeCl3 = 55.845 + 3(35.453)

                               = 55.845 + 106.359

                               = 162.204 g/mol.

If 162.204 g of FeCl3 contains 55.845 g of Fe, then mass of Fe in 2.175g of FeCl3 will be :

=  2.175

=  0.3442  2.175

= 0.748 g

Hence, mass of Fe in sample B is 0.748 g

Now we have the mass of Fe and FeCl3, let’s calculate mass of Cl :

Using law of constant composition, we can calculate as :

                 Mass of Cl = Mass of FeCl3  Mass ratio of Cl

If 162.204 g of FeCl3 contains 106.359 g of Cl, then mass of Cl in any sample will be :

         = 0.6557

Therefore , Mass of Cl = 2.175 g  0.6557

= 1.426 g

Hence, mass of Cl in sample B is 1.426 g

Step 3:

Let’s consider, sample C :

Given : Mass of Fe = 2.012 g

We know the molar mass of FeCl3 from the above calculation = 162.204 g/mol

Now, let’s calculate mass of FeCl3 in the sample :

If 162.204 g of FeCl3 contains 55.845 g of Fe, then mass of Fe in sample will be :

= = 0.3442

Let’s consider mass of FeCl3 as ‘x’

                unknown mass of FeCl3 in sample C(x)  mass ratio of Fe = mass of Fe

                 x  0.3442 = 2.012 g

                x =

                x = 5.845 g

Hence, mass of FeCl3 in sample C is  5.845 g

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

If 162.204 g of FeCl3 contains 106.359 g of Cl, then mass of Cl in any sample will be :

         = 0.6557

Therefore mass of Cl will be :

        Mass of  FeCl3   mass ratio of Cl = mass of Cl

        5.845g  0.6557 = mass of O

 3.832 g 

Hence, mass of Cl in sample C is 3.832 g

Step 4 of 5

Chapter 5, Problem 30P is Solved
Step 5 of 5

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

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Solved: Use the law of constant composition to complete

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