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Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

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Page 1: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4
Page 2: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Thursday, April 7th : “A” DayFriday, April 8th: “A” Day

AgendaFinish Section 7.1: “Avogadro’s Number & Molar

Conversions”In-Class:

Practice pg. 231: #1-4Practice pg. 232: #1-3Concept Review: “Avogadro’s Number &

Molar Conversions”: #6-10

Section 7.1 quiz next time…

Page 3: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Finish 7.1: “Avogadro’s Number and Molar Conversions”

6.022 X 1023 atoms 1 mole

Page 4: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Molar Mass Relates Moles to GramsMolar mass: the mass in grams of one mole of an

element or compound.

Molar mass is numerically equal to the atomic mass of elements (Fe, Al, etc.) and the formula mass of compounds (H2O, CH4, etc. ) and diatomic elements. (O2, F2, etc.)

The units for molar mass are grams mole

Page 5: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Molar Mass Example #1What is the molar mass of water, H2O?To determine the molar mass, look up the

atomic mass of each element present in the compound and add them up. This is the molar mass.

H = 1.01 g/mol H = 1.01 g/molO = 16.0 g/mol

Molar mass of water = 18.02 g H2O

1 mole

Page 6: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Molar Mass Example #2What is the molar mass of sulfuric acid, H2SO4?

H = 1.01 g/molH = 1.01 g/molS = 32.07 g/molO = 16.0 g/molO = 16.0 g/molO = 16.0 g/molO = 16.0 g/mol

Molar mass of sulfuric acid = 98.09 g H2SO4

1 mole

Page 7: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

So, why do we need to know the molar mass…

Molar mass can be used as a conversion factor in problems converting between mass and amount!!!

Page 8: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Converting Number of Particles to MassTo convert from number of particles to mass, you

must use a two-step process: First, convert number of particles to amount

in moles using Avogadro’s number. Second, convert amount in moles to mass in

grams using the molar mass.Two different conversion factors must be used!

Page 9: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Converting Between Mass, Amount, and Number of Particles

Page 10: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Sample Problem C, pg. 231Converting Number of Particles to Mass

Find the mass in grams of 2.44 X 1024 atoms of carbon, whose molar mass is 12.01 g/mol.

Known: 2.44 X 1024 atoms of carbonUnknown: Mass in grams of carbon

Two conversion factors:1 mole C AND 12.01 g C

6.022 X 1023 atoms C 1 mole C

Page 11: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Sample Problem C, continuedConverting Number of Particles to Mass

Step 1: Convert # of particles amount in moles:2.44 X 1024 C atoms X 1 mole C

6.022 X 1023 atoms C= 4.05 moles C

Step 2: Convert amount in moles mass (in grams)4.05 moles C X 12.01 g C

1 mole C

= 48.6 g C

Page 12: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Additional PracticeFind the mass in grams of 6.12 X 1014 formula units of

rhenium dioxide, ReO2, 218.21 g/mol.

Known: 6.12 X 1014 formula units ReO2

Unknown: Mass in grams of ReO2

Two conversion Factors:

1 mole ReO2 AND 218.21 g ReO2

6.022 X 1023 form unit ReO2 1 mole ReO2

Page 13: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Additional Practice

Step 1: Convert # of particles amount in moles:6.12 X 1014 form. unit X _1 mol_ReO2__

6.022 X 1023 form. unit ReO2

= 1.02 X 10 -9 mole ReO2

Step 2: Convert amount in moles mass (in grams)1.02 X 10 -9 mole ReO2 X 218.21 g ReO2

1 mole ReO2

= 2.23 X 10 -7 g ReO2

Page 14: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Converting Mass to Number of ParticlesConverting from mass to number of particles is

simply the opposite of the previous problems.To convert from mass to number of particles, you

must use a two-step process:1. First, convert mass in grams to amount in

moles using molar mass.2. Second, convert amount in moles to number

of particles using Avogadro’s number. Two different conversion factors must be used!

Page 15: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Sample Problem D, pg. 232Converting Mass to Number of Particles

Find the number of molecules present in 47.5 g of glycerol, C3H8O3. The molar mass of glycerol is 92.11 g/mol.

Known: 47.5 g glycerol, C3H8O3

Unknown: # of molecules presentTwo conversion factors:

1 mole C3H8O3 AND 6.022 X 1023 molecules C3H8O3

92.11 g C3H8O3 1 mole C3H8O3

Page 16: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Sample Problem D, cont.Converting Mass to Number of Particles

Step 1: Convert mass (in grams) amount in moles: 47.5 g C3H8O3 X 1 mole C3H8O3 = 0.516 mol C3H8O3

92.11 g C3H8O3

Step 2: Convert amount in moles # of molecules: 0.516 mol C3H8O3 X 6.022 X 1023 molecules C3H8O3

1 mole C3H8O3

= 3.11 X 1023 molecules C3H8O3

Page 17: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Additional Practice

Determine the number of atoms present in 54.3 grams of cobalt, Co.

Known: 54.3 g of cobalt, Co.Unknown: # of atoms

Two conversion Factors:1 mole Co AND 6.022 X 1023 atoms

Co58.93 g Co 1 mole Co

Page 18: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

Additional PracticeStep 1: Convert mass (in grams) amount in moles:54.3 g Co X 1 mol Co = 0.921 mol Co

58.93 g Co

Step 2: Convert amount in moles # of atoms:0.921 moles Co X 6.022 X 1023 atoms Co

1 mole Co

= 5.55 X 1023 atoms Co

Page 19: Thursday, April 7 th : A Day Friday, April 8 th : A Day Agenda Finish Section 7.1: Avogadros Number & Molar Conversions In-Class: Practice pg. 231: #1-4

In-Class Assignments/HomeworkYou Must SHOW WORK!

Practice pg. 231: #1-4Practice pg. 232: #1-3Concept Review: “Avogadro’s Number & Molar

Conversions”: #6-10

Section 7.1 quiz next time…