How many grams of Cl are in 38.0 g of each sample of chlo-

Problem 67P Chapter 6

Introductory Chemistry | 5th Edition

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Problem 67P

How many grams of Cl are in 38.0 g of each sample of chlo- rofluorocarbons (CFCs)?

(a) CF2Cl2

(b) CFCl3

(d) CF3Cl

Step-by-Step Solution:

Solution 67P

First, you need to find how many moles of chlorine are in 38.0g of each of these compounds:

Cl is 35.453 grams/mol

C is 12.011 grams/mol

F is 18.988 grams/mol

Step 1:

(a) CF2Cl2 = 1 Carbon + 2 Fluorine + 2 Chlorine

[1 $\times{}$ (12.011g/mol)] + [2 $\times{}$ (18.988 g/mol)] + [2 $\times{}$ (35.453 g/mol)] = 120.893 g/mol

So for this particular compound, one mole weighs 120.893 grams

(or in other words 6.0221413 x 1023 of these molecules would weight 120.893 grams)

Now find how many moles of the compound it takes to weigh 38.0 grams:

38.0 grams x (1 mol /120.893 grams) = grams cancel and we get 0.314 moles

( 0.314 moles means we have 0.314 $\times{}$ (6.0221413 $\times{}$ 1023) molecules of compound

In each molecule of compound (CF2Cl2) we have 2 Chlorine molecules

therefore...

In each mole of compound (6.0221413 $\times{}$ 1023 molecules of CF2Cl2)

we have 2 moles of Chlorine (2 x [6.0221413 $\times{}$ 1023] molecules of Cl)

So,

In 0.314 moles of CF2Cl2 there are (2 x 0.314) moles of Chlorine

2 x 0.314 = 0.628 moles of Chlorine in 38.0 grams of CF2Cl2

Chlorine = 35.453 grams/mole

So...

0.628 moles x (35.453 grams / 1 mole) = 22.3 grams of chlorine in 38.0 grams of CF2Cl2

Just to help clarify...

Because there are 0.314 moles of CF2Cl2, there are 2 $\times{}$ 0.314 moles of Fluorine and 1 $\times{}$ 0.314 moles of carbon in the compound as well...

2 $\times{}$ 0.314 moles = 0.628 moles of Fluorine

Fluorine = 18.988 grams/mol

So...

0.628 moles x (18.988 grams / 1 mole) = 11.92 grams of fluorine in 38.0 grams of CF2Cl2

And 1 $\times{}$ 0.314 moles = 0.314 moles of Carbon

Carbon is 12.011 grams/mol

So...

0.314 moles $\times{}$ (12.011 grams / 1 mole) = 3.77 grams of carbon in 38.0 grams of CF2Cl2

In total, in 38 grams of CF2Cl2 there are:

22.3 grams of chlorine

11.92 grams of fluorine

3.77 grams of carbon

= 38.0 grams total

Step 2:

(b) CFCl3 = 1 Carbon + 1 Fluorine + 3 Chlorine

[1 $\times{}$ (12.011 g/mol)] + [1 $\times{}$ (18.988 g/mol)] + [3 $\times{}$ (35.453 g/mol)] = 137.358 g/mol

38.0 grams $\times{}$ (1 mol /137.358 grams) = 0.277 moles CFCl3

0.277 moles CFCl3 $\times{}$ (3 moles Cl / 1 mole CFCl3) = 0.830 moles Cl

Chlorine = 35.453 grams/mole

0.830 moles $\times{}$ (35.453 grams / 1 mole) = 29.4 grams of chlorine in 38.0 grams of CFCl3

Step 3 of 4

Chapter 6, Problem 67P is Solved

Step 4 of 4

Textbook: Introductory Chemistry

Edition: 5

Author: Nivaldo Tro

ISBN: 9780321910295

Introductory Chemistry was written by Patricia and is associated to the ISBN: 9780321910295. Since the solution to 67P from 6 chapter was answered, more than 746 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. This full solution covers the following key subjects: rofluorocarbons, grams, chlo, cfcs, cfcl. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. The answer to “How many grams of Cl are in 38.0 g of each sample of chlo- rofluorocarbons (CFCs)?(a) CF2Cl2(b) CFCl3(c) C2F3Cl3(d) CF3Cl” is broken down into a number of easy to follow steps, and 20 words. The full step-by-step solution to problem: 67P from chapter: 6 was answered by Patricia, our top Chemistry solution expert on 05/06/17, 06:45PM.