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1 Chapter 6 Chemical Quantities 6.4 Calculations Using Molar Mass Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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Presentation on theme: "1 Chapter 6 Chemical Quantities 6.4 Calculations Using Molar Mass Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings."— Presentation transcript:

1 1 Chapter 6 Chemical Quantities 6.4 Calculations Using Molar Mass Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 2 Molar mass conversion factors Are written from molar mass. Relate grams and moles of an element or compound. Example: Write molar mass factors for methane CH 4 used in gas cook tops and gas heaters. Molar mass: 1 mol CH 4 = 16.04 g Conversion factors: 16.04 g CH 4 and 1 mol CH 4 1 mol CH 4 16.04 g CH 4 Molar Mass Factors

3 3 Acetic acid C 2 H 4 O 2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid. Learning Check Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

4 4 Acetic acid C 2 H 4 O 2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid. Calculate molar mass: 24.02 + 4.032 = 32.00 = 60.05 g/mol 1 mol of acetic acid = 60.05 g acetic acid Molar mass factors 1 mol acetic acid and 60.05 g acetic acid 60.05 g acetic acid 1 mol acetic acid Solution

5 5 Molar mass factors are used to convert between the grams of a substance and the number of moles. Calculations Using Molar Mass Grams Molar mass factor Moles

6 6 Aluminum is used to build lightweight bicycle frames. How many grams of Al are 3.00 mol Al? Molar mass equality: 1 mol Al = 26.98 g Al Setup with molar mass as a factor: 3.00 mol Al x 26.98 g Al = 80.9 g Al 1 mol Al molar mass factor for Al Moles to Grams

7 7 Learning Check Allyl sulfide C 6 H 10 S is a compound that has the odor of garlic. How many moles of C 6 H 10 S are in 225 g C 6 H 10 S? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

8 8 Calculate the molar mass of C 6 H 10 S. (6 x 12.01) + (10 x 1.008) + (1 x 32.07) = 114.21 g/mol Set up the calculation using a mole factor. 225 g C 6 H 10 S x 1 mol C 6 H 10 S 114.21 g C 6 H 10 S molar mass factor(inverted) = 1.97 mol C 6 H 10 S Solution

9 9 Grams, Moles, and Particles A molar mass factor and Avogadro’s number convert Grams to particles molar mass Avogadro’s number (g mol particles) Particles to grams Avogadro’s molar mass number (particles mol g)

10 10 Learning Check How many H 2 O molecules are in 24.0 g H 2 O? 1) 4.52 x 10 23 2) 1.44 x 10 25 3) 8.02 x 10 23

11 11 Solution How many H 2 O molecules are in 24.0 g H 2 O? 3) 8.02 x 10 23 24.0 g H 2 O x 1 mol H 2 O x 6.022 x 10 23 H 2 O molecules 18.02 g H 2 O 1 mol H 2 O = 8.02 x 10 23 H 2 O molecules

12 12 Learning Check If the odor of C 6 H 10 S can be detected from 2 x 10 -13 g in one liter of air, how many molecules of C 6 H 10 S are present?

13 13 Solution If the odor of C 6 H 10 S can be detected from 2 x 10 -13 g in one liter of air, how many molecules of C 6 H 10 S are present? 2 x 10 -13 g x 1 mol x 6.022 x 10 23 molecules 114.21 g 1 mol = 1 x 10 9 molecules C 6 H 10 S


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