1 Chapter 7 Chemical Quantities 7.3 Calculations Using Molar Mass Basic Chemistry Copyright © 2011 Pearson Education, Inc. Table salt is NaCl.

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 = g Conversion factors: g CH 4 and 1 mol CH 4 1 mol CH g CH 4 Molar Mass Factors Basic Chemistry Copyright © 2011 Pearson Education, Inc.

3 Acetic acid HC 2 H 3 O 2 gives the sour taste to vinegar. Write an equality and two molar mass conversion factors for acetic acid. Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

4 Acetic acid HC 2 H 3 O 2 gives the sour taste to vinegar. Write an equality and two molar mass conversion factors for acetic acid. Equality: 1 mol of acetic acid = g of acetic acid Molar mass conversion factors: 1 mol acetic acid and g acetic acid g acetic acid 1 mol acetic acid Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

5 Guide to Calculation Using Molar Mass Basic Chemistry Copyright © 2011 Pearson Education, Inc.

6 Aluminum is used to build lightweight bicycle frames. How many grams of Al are in 3.00 mol of Al? Moles to Grams Basic Chemistry Copyright © 2011 Pearson Education, Inc.

7 STEP 1 Given 3.00 mol of Al Need grams of Al STEP 2 Plan moles of Al grams of Al STEP 3 Equalities/Conversion Factors 1 mol of Al = g of Al 1 mol Al and g Al g Al 1 mol Al Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

8 STEP 4 Set Up Problem Set up with molar mass as a factor: 3.00 mol Al x g Al = 80.9 g of Al 1 mol Al molar mass factor for Al Solution (continued) Basic Chemistry Copyright © 2011 Pearson Education, Inc.

9 The artificial sweetener aspartame (NutraSweet), C 14 H 18 N 2 O 5, is used to sweeten diet foods, coffee, and soft drinks. How many moles of aspartame are present in 225 g of aspartame? Learning Check Basic Chemistry Copyright © 2011 Pearson Education, Inc.

10 STEP 1 Given 225 g of C 14 H 18 N 2 O 5 Need moles STEP 2 Plan g of aspartame moles of aspartame STEP 3 Equalities/Conversion Factors 1 mol aspartame = g 1 mol aspartame and g aspartame g aspartame 1 mol aspartame STEP 4 Set Up Problem 225 g aspartame x 1 mol aspartame g aspartame mole factor (inverted) = mol of aspartame Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

11 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 of C 6 H 10 S? Basic Chemistry Copyright © 2011 Pearson Education, Inc.

12 STEP 1 Given 225 g of C 6 H 10 S Need moles of C 6 H 10 S STEP 2 Plan g of C 6 H 10 S moles of C 6 H 10 S STEP 3 Equalities/Conversion Factors 1 mol of C 6 H 10 S = g of C 6 H 10 S 1 mol C 6 H 10 S and g C 6 H 10 S g C 6 H 10 S 1 mol C 6 H 10 S STEP 3 Set Up Problem 225 g C 6 H 10 S x 1 mol C 6 H 10 S g C 6 H 10 S = 1.97 mol C 6 H 10 S Solution Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Guide to Converting Grams to Particles 13

14 Learning Check How many molecules of H 2 O are in 24.0 g of H 2 O? 1) 4.52 x ) 1.44 x ) 8.02 x Basic Chemistry Copyright © 2011 Pearson Education, Inc.

15 Solution STEP 1 Given 24.0 g of H 2 O Need molecules of H 2 O STEP 2 Plan g of H 2 O moles of H 2 O molecules of H 2 O STEP 3 Equalities/Conversion Factors 1 mol of H 2 O = g of H 2 O 1 mol H 2 O and g H 2 O g H 2 O 1 mol H 2 O Basic Chemistry Copyright © 2011 Pearson Education, Inc.

16 Solution (continued) STEP 3 Equalities/Conversion Factors (continued) 1 mol of H 2 O = molecules of H 2 O 1 mol H 2 O and x H 2 O molecules H 2 O molecules 1 mol H 2 O STEP 4 Set Up Problem 24.0 g H 2 O x 1 mol H 2 O g H 2 O x x H 2 O molecules 1 mol H 2 O = 8.02 x H 2 O molecules (3) Basic Chemistry Copyright © 2011 Pearson Education, Inc.