Mole Problems Solution Guide.

Slides:

Advertisements

Similar presentations

It’s a beauty mark… It’s a small furry garden pest… No, wait… its how we count ATOMS!

Advertisements

Chapter 10 Chemical Quantities

I. Measuring Matter Chemists need a convenient method for counting accurately the number of atoms, molecules, or formula units in a sample of a substance.

Section Percent Composition and Chemical Formulas

Key: Compound Calculations and Compound Stoichiometry.

1 Chapter 8 Chemical Quantities. 2 How you measure how much? How you measure how much? n You can measure mass, n or volume, n or you can count pieces.

The Mole: A Measurement of Matter OBJECTIVES: Describe how Avogadro’s number is related to a mole of any substance.

Unit 5 CHEMICAL QUANTITIES CHAPTER 10 – THE MOLE.

Chapter 8 Chemical Composition Chemistry B2A. Atomic mass unit (amu) = × g Atomic Weight Atoms are so tiny. We use a new unit of mass:

Percentage Composition

Molar Mass & Conversions. The Mole mole (mol)- SI Unit for the amount of a substance that contains as many particles as there are atoms in exactly 12g.

Unit 6 The Mole: % Composition and Emperical Formula

How many roses are in 3.5 dozen?

IIIIIIIV Topic 6 The Mole I. Molar Conversions A. What is the Mole? n A counting number (like a dozen) n Avogadro’s number (N A ) n 1 mol = 6.02  10.

Mass Conservation in Chemical Reactions Mass and atoms are conserved in every chemical reaction. Molecules, formula units, moles and volumes are not always.

CHAPTER 3b Stoichiometry.

How to Count Atoms What unit is used to count donuts? Would that unit be appropriate for counting the # of people in Jacksonville? Would it be appropriate.

Chapter 4 “Chemical Quantities”

Section 7.1 The Mole: A Measurement of Matter

Chapter 10: Chemical Quantities

Mathematical Chemistry

Empirical Formulas. Empirical formula tells the relative number of atoms of each element in a compound Mole concept provides a way of calculating the.

Chapter 6 Chemical Quantities

1 Chemical Composition Chapter 8. 2 Atomic Masses Balanced equation tells us the relative numbers of molecules of reactants and products C + O 2  CO.

Chapter 11. Mole SI base unit for measuring the amount of substance The number of representative particles in exactly 12 grams of pure carbon-12 1 mole.

Unit 10 – The Mole Essential Questions:

1 Chapter 6 Chemical Quantities Powers of Ten Animation.

Unit 5: The Mole.

Mole Concept. Counting Units  A pair refers to how many shoes?  A dozen refers to how many doughnuts or eggs?  How many pencils are in a gross?  How.

Mole Concept. Counting Units  A pair refers to how many shoes?  A dozen refers to how many doughnuts or eggs?  How many pencils are in a gross?  How.

Unit 9 part 1: The Mole Chemical Measurements Mole Conversions Empirical & Molecular Formulas.

IIIIIIIV The Mole I. Molar Conversions What is the Mole? A counting number (like a dozen or a pair) Avogadro’s number 6.02  mole = 6.02  10.

8 | 1 CHAPTER 8 CHEMICAL COMPOSITION. 8 | 2 Atomic Masses Balanced equations tell us the relative numbers of molecules of reactants and products. C +

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Introductory Chemistry: A Foundation FIFTH EDITION by Steven S. Zumdahl University of.

 Dalton used the percentages of elements in compounds and the chemical formulas to deduce the relative masses of atoms  Unit is the amu(atomic mass.

By Steven S. Zumdahl & Donald J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry,

Using the MOLE. Percentage composition Percentage composition is the mass of individual elements in a compound expressed as a percentage of the mass of.

1 Chapter 10 “Chemical Quantities” Yes, you will need a calculator for this chapter!

Chapter 11 : Matter Notes. Mole (mol) is equal to 6.02x10 23 The mole was named in honor of Amedeo Avogadro. He determined the volume of one mole of gas.

The Mole: A Measurement of Matter Describe how Avogadro’s number is related to a mole of any substance Solve problems involving mass in grams, amount in.

Mole Concept. Counting Units  A pair refers to how many shoes?  A dozen refers to how many doughnuts or eggs?  How many pencils are in a gross?  How.

1 Chemical Quantities or. 2 Representative particles n The smallest pieces of a substance. n For an element it is an atom. (Ex: Na) n For a covalent compound.

Chemical Calculations Mole to Mass, Mass to Moles.

Moles COUNTING BY WEIGHING. Moles (is abbreviated: mol)  It is an amount, defined as the number of carbon atoms in exactly 12 grams of carbon-12.  1.

(4.6/4.7) Empirical and Molecular Formulas SCH 3U.

Chapter 7 Chemical Quantities or How you measure how much? You can measure mass, volume, or you can count pieces of a substance. We measure mass in grams.

Chapter 11 The Mole. I. Measuring Matter A. Counting Particles Chemists needed a convenient method for counting the number of atoms in a sample of a substance.

The Mole Calculating -Molecular Weight -Formula Weight -Molar Mass.

1 Chapter 10 “Chemical Quantities” Yes, you will need a calculator for this chapter!

“Chemical Quantities” Yes, you will need a calculator for this chapter!

1 Chapter 10 The Mole Chemical Quantities. 2 How do you measure how much in Chemistry? How do you measure how much in Chemistry? We count chemical pieces.

Isotope Practice sodium Na 11 protons, 11 electrons, 12 neutrons copper Cu 29 protons, 29 electrons, 35 neutrons silver Ag.

The Mole Honors Chem. -How do we measure chemical quantities? -What units of measure do we use?

IIIIII II. % Composition and Formula Calculations Ch. 3 – The Mole.

1 Chemical Quantities Coach Williams Chemistry. 2 Section 7.1 The Mole: A Measurement of Matter n OBJECTIVES: –Describe how Avogadro’s number is related.

The Mole Concept.

Chemical Quantities Part I

Chemical Compostition

Chapter 8 The Mole.

Molar Conversions (p.80-85, )

The Mole Chapter 10.1.

EMPIRICAL FORMULA The empirical formula represents the smallest ratio of atoms present in a compound. The molecular formula gives the total number of atoms.

II. Percent composition

Chapter 6 Chemical Composition.

II. Percent composition

Chapter 9 Key Terms Mole Molar Mass Avogadro's Number Percent Composition Stoichiometry Limiting Reactant Excess Reactant Actual Yield Theoretical Yield.

III. Formula Calculations (p )

Presentation transcript:

Mole Problems Solution Guide

How many water molecules are found in 18.0g water? 602000000000000000000000 6.02x1023 water molecules If you had 6.02x1023 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x1023 = $238,616,184,677,188.51 (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

How many water molecules are found in 18.0g water? 602000000000000000000000 6.02x1023 water molecules If you had 6.02x1023 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x1023 = $238,616,184,677,188.51 (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

How many water molecules are found in 18.0g water? 602000000000000000000000 6.02x1023 water molecules If you had 6.02x1023 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x1023 = $238,616,184,677,188.51 (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

How many water molecules are found in 18.0g water? 602000000000000000000000 6.02x1023 water molecules If you had 6.02x1023 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x1023 = $238,616,184,677,188.51 (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

How many water molecules are found in 18.0g water? 602000000000000000000000 6.02x1023 water molecules If you had 6.02x1023 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x1023 = $238,616,184,677,188.51 (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

How many water molecules are found in 18.0g water? 602000000000000000000000 6.02x1023 water molecules If you had 6.02x1023 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x1023 = $238,616,184,677,188.51 (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min) In any measurable sample of matter there are too many particles to count directly! Atoms are exceedingly small!!!

1 12 One H has a relative mass of ______ amu. One C has a relative mass of _______ amu 1 6 10 H have a relative mass of ______ amu. 10 C have a relative mass of ______ amu 1 6 1 12 100 H have a relative mass of _____ amu. 100 C have a relative mass of ______ amu 1 12 106 H have a relative mass of ______ amu. 106 C have a relative mass of ______ amu 1 6

1 12 One H has a relative mass of __1___ amu. One C has a relative mass of _______ amu 1 6 10 H have a relative mass of ______ amu. 10 C have a relative mass of ______ amu 1 6 1 12 100 H have a relative mass of _____ amu. 100 C have a relative mass of ______ amu 1 12 106 H have a relative mass of ______ amu. 106 C have a relative mass of ______ amu 1 6

1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu 1 6 10 H have a relative mass of ______ amu. 10 C have a relative mass of ______ amu 1 6 1 12 100 H have a relative mass of _____ amu. 100 C have a relative mass of ______ amu 1 12 106 H have a relative mass of ______ amu. 106 C have a relative mass of ______ amu 1 6

1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu 1 6 10 H have a relative mass of _10___ amu. 10 C have a relative mass of ______ amu 1 6 1 12 100 H have a relative mass of _____ amu. 100 C have a relative mass of ______ amu 1 12 106 H have a relative mass of ______ amu. 106 C have a relative mass of ______ amu 1 6

1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu 1 6 10 H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu 1 6 1 12 100 H have a relative mass of ____ amu. 100 C have a relative mass of ______ amu 1 12 106 H have a relative mass of ______ amu. 106 C have a relative mass of ______ amu 1 6

1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu 1 6 10 H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu 1 6 1 12 100 H have a relative mass of 100 amu. 100_ C have a relative mass of ______ amu 1 12 106 H have a relative mass of ______ amu. 106 C have a relative mass of ______ amu 1 6

1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu 1 6 10 H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu 1 6 1 12 100 H have a relative mass of 100 amu. 100_ C have a relative mass of 1200_ amu 1 12 106 H have a relative mass of ______ amu. 106 C have a relative mass of ______ amu 1 6

1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu 1 6 10 H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu 1 6 1 12 100 H have a relative mass of 100 amu. 100_ C have a relative mass of 1200_ amu 1 12 106 H have a relative mass of 1million amu. 106 C have a relative mass of ______ amu 1 6

1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu 1 6 10 H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu 1 6 1 12 100 H have a relative mass of 100 amu. 100_ C have a relative mass of 1200_ amu 1 12 106 H have a relative mass of 1million amu. 106 C have a relative mass of 12 million amu 1 6

1a) The mass of the sample of carbon is always ______ more than an equivalent number of hydrogens

Mole Unit Questions 1a) The mass of the sample of carbon is always _12x__ more than an equivalent number of hydrogens

Mole Unit Question 1 6.02x1023 atoms of hydrogen have a mass of 1.0 g that equals the atomic mass of hydrogen found on the periodic table 6.02x1023 atoms of carbon have a mass of 12.0 g that equals the atomic mass of carbon found on the periodic table

Mole Unit Question 1 6.02x1023 atoms of hydrogen have a mass of 1.0 g that equals the atomic mass of hydrogen found on the periodic table 6.02x1023 atoms of carbon have a mass of 12.0 g that equals the atomic mass of carbon found on the periodic table

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 2. What is the mass of 3.01 x 1023 carbon–12 atoms? 3.01 x 1023 is one half of 6.02 x 1023 therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x 1023 carbon–12 atoms? 1.51 x 1023 is one fourth of 6.02 x 1023 therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of 12.04 x 1023 carbon–12 atoms? 12.04 x 1023 is twice 6.02 x 1023 therefore the mass will be twice 12 g = 24 g

Questions 2-8 5. What is the mass of 24.01 x 1023 carbon–12 atoms? 24.01 x 1023 is four times 6.02 x 1023 therefore the mass will be four time 12 g = 48g

Questions 2-8 5. What is the mass of 24.01 x 1023 carbon–12 atoms? 24.01 x 1023 is four times 6.02 x 1023 therefore the mass will be four time 12 g = 48g

Questions 2-8 5. What is the mass of 24.01 x 1023 carbon–12 atoms? 24.01 x 1023 is four times 6.02 x 1023 therefore the mass will be four time 12 g = 48g

Questions 2-8 6. How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x1023 C = 18.06x1023C = 1.806x1024 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x1023 C = 2.01x1023C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x1023 C = 5.01x1022 C

Questions 2-8 6. How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x1023 C = 18.06x1023C = 1.806x1024 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x1023 C = 2.01x1023C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x1023 C = 5.01x1022 C

Questions 2-8 6. How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x1023 C = 18.06x1023C = 1.806x1024 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x1023 C = 2.01x1023C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x1023 C = 5.01x1022 C

Questions 2-8 6. How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x1023 C = 18.06x1023C = 1.806x1024 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x1023 C = 2.01x1023C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x1023 C = 5.01x1022 C

Questions 2-8 6. How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x1023 C = 18.06x1023C = 1.806x1024 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x1023 C = 2.01x1023C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x1023 C = 5.01x1022 C

Questions 2-8 6. How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x1023 C = 18.06x1023C = 1.806x1024 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x1023 C = 2.01x1023C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x1023 C = 5.01x1022 C

Questions 2-8 6. How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x1023 C = 18.06x1023C = 1.806x1024 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x1023 C = 2.01x1023C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x1023 C = 5.01x1022 C

Questions 2-8 6. How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x1023 C = 18.06x1023C = 1.806x1024 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x1023 C = 2.01x1023C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x1023 C = 5.01x1022 C

Questions 2-8 6. How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x1023 C = 18.06x1023C = 1.806x1024 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x1023 C = 2.01x1023C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x1023 C = 5.01x1022 C

Bonus Question 6.02x1023 C = 12.0 g C 1.0 x106 C = x Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C = 1.99x10-17g C

Bonus Question 6.02x1023 C = 12.0 g C 1.0 x106 C = x Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C = 1.99x10-17g C

Bonus Question 6.02x1023 C = 12.0 g C 1.0 x106 C = x Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C = 1.99x10-17g C

Bonus Question 6.02x1023 C = 12.0 g C 1.0 x106 C = x Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C = 1.99x10-17g C

Bonus Question 6.02x1023 C = 12.0 g C 1.0 x106 C = x Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C = 1.99x10-17g C

Bonus Question 6.02x1023 C = 12.0 g C 1.0 x106 C = x Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C = 1.99x10-17g C

Bonus Question 6.02x1023 C = 12.0 g C 1.0 x106 C = x Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C = 1.99x10-17g C

Bonus Question Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C 1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C (6.02x1023 C )

Bonus Question Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C 1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C (6.02x1023 C )

Bonus Question Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C 1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C (6.02x1023 C )

Bonus Question Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C 1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C (6.02x1023 C )

Bonus Question Cross multiply 1.0 x106 C multiplied by 12.0 g then divided by 6.0x1023 C 1.0 x106 C ( 12.0 g C ) = 1.99x10-17g C (6.02x1023 C )

6.02 x 1023 = P.T. grams 9. How many silver atoms are in 25.0 grams of

6.02 x 1023 = P.T. grams 9. How many silver atoms are in 25.0 grams of 25.0g Ag

6.02 x 1023 = P.T. grams 9. How many silver atoms are in 25.0 grams of 25.0g Ag 6.02 x 1023 Ag =

6.02 x 1023 = P.T. grams 9. How many silver atoms are in 25.0 grams of 25.0g Ag 6.02 x 1023 Ag = 107.9g

6.02 x 1023 = P.T. grams 9. How many silver atoms are in 25.0 grams of 25.0g Ag 6.02 x 1023 Ag =1.39x1023 Ag 107.9Ag

6.02 x 1023 = P.T. grams 10. What is the mass of 1.00 x 109 silver atoms? 1.00 x 109 Ag

6.02 x 1023 = P.T. grams 10. What is the mass of 1.00 x 109 silver atoms? 1.00 x 109 Ag ( 107.9 g Ag ) =

6.02 x 1023 = P.T. grams 10. What is the mass of 1.00 x 109 silver atoms? 1.00 x 109 Ag ( 107.9 g Ag ) = (6.02 x 1023 Ag )

6.02 x 1023 = P.T. grams 10. What is the mass of 1.00 x 109 silver atoms? 1.00 x 109 Ag ( 107.9 g Ag ) = 1.79 x 10-13 g Ag (6.02 x 1023 Ag )

6.02 x 1023 = P.T. grams 11. What is the mass of 2.00 x 1023 gold atoms? 2.00 x 1023 Au ( 197.0 g Au ) = 65.4 g Au (6.02 x 1023 Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x 1023 Au = 3.06 X1023 Au 197.0 g Au

6.02 x 1023 = P.T. grams 11. What is the mass of 2.00 x 1023 gold atoms? 2.00 x 1023 Au ( 197.0 g Au ) = 65.4 g Au (6.02 x 1023 Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x 1023 Au = 3.06 X1023 Au 197.0 g Au

6.02 x 1023 = P.T. grams 11. What is the mass of 2.00 x 1023 gold atoms? 2.00 x 1023 Au ( 197.0 g Au ) = 65.4 g Au (6.02 x 1023 Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x 1023 Au = 3.06 X1023 Au 197.0 g Au

6.02 x 1023 = P.T. grams 11. What is the mass of 2.00 x 1023 gold atoms? 2.00 x 1023 Au ( 197.0 g Au ) = 65.4 g Au (6.02 x 1023 Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x 1023 Au = 3.06 X1023 Au 197.0 g Au

6.02 x 1023 = P.T. grams 11. What is the mass of 2.00 x 1023 gold atoms? 2.00 x 1023 Au ( 197.0 g Au ) = 65.4 g Au (6.02 x 1023 Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x 1023 Au = 3.06 X1023 Au 197.0 g Au

6.02 x 1023 = P.T. grams 11. What is the mass of 2.00 x 1023 gold atoms? 2.00 x 1023 Au ( 197.0 g Au ) = 65.4 g Au (6.02 x 1023 Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x 1023 Au = 3.06 X1023 Au 197.0 g Au

6.02 x 1023 = P.T. grams 11. What is the mass of 2.00 x 1023 gold atoms? 2.00 x 1023 Au ( 197.0 g Au ) = 65.4 g Au (6.02 x 1023 Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x 1023 Au = 3.06 X1023 Au 197.0 g Au

6.02 x 1023 = P.T. grams 11. What is the mass of 2.00 x 1023 gold atoms? 2.00 x 1023 Au ( 197.0 g Au ) = 65.4 g Au (6.02 x 1023 Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x 1023 Au = 3.06 X1023 Au 197.0 g Au

6.02 x 1023 = P.T. grams 13. What is the mass 2.00 x 1024 platinum atoms? 2.00 x 1024 Pt ( 195.1 g Pt ) = 648 g Pt (6.02 x 1023 Pt ) 14. What is the mass of 2.00 x 1022 silicon atoms? 2.00 x 1022 Si ( 28.1g Si ) = .93 g Si (6.02 x 1023 Si )

6.02 x 1023 = P.T. grams 13. What is the mass 2.00 x 1024 platinum atoms? 2.00 x 1024 Pt ( 195.1 g Pt ) = 648 g Pt (6.02 x 1023 Pt ) 14. What is the mass of 2.00 x 1022 silicon atoms? 2.00 x 1022 Si ( 28.1g Si ) = .93 g Si (6.02 x 1023 Si )

6.02 x 1023 = P.T. grams 13. What is the mass 2.00 x 1024 platinum atoms? 2.00 x 1024 Pt ( 195.1 g Pt ) = 648 g Pt (6.02 x 1023 Pt ) 14. What is the mass of 2.00 x 1022 silicon atoms? 2.00 x 1022 Si ( 28.1g Si ) = .93 g Si (6.02 x 1023 Si )

6.02 x 1023 = P.T. grams 13. What is the mass 2.00 x 1024 platinum atoms? 2.00 x 1024 Pt ( 195.1 g Pt ) = 648 g Pt (6.02 x 1023 Pt ) 14. What is the mass of 2.00 x 1022 silicon atoms? 2.00 x 1022 Si ( 28.1g Si ) = .93 g Si (6.02 x 1023 Si )

6.02 x 1023 = P.T. grams 13. What is the mass 2.00 x 1024 platinum atoms? 2.00 x 1024 Pt ( 195.1 g Pt ) = 648 g Pt (6.02 x 1023 Pt ) 14. What is the mass of 2.00 x 1022 silicon atoms? 2.00 x 1022 Si ( 28.1g Si ) = .93 g Si (6.02 x 1023 Si )

6.02 x 1023 = P.T. grams 13. What is the mass 2.00 x 1024 platinum atoms? 2.00 x 1024 Pt ( 195.1 g Pt ) = 648 g Pt (6.02 x 1023 Pt ) 14. What is the mass of 2.00 x 1022 silicon atoms? 2.00 x 1022 Si ( 28.1g Si ) = .93 g Si (6.02 x 1023 Si )

6.02 x 1023 = P.T. grams 13. What is the mass 2.00 x 1024 platinum atoms? 2.00 x 1024 Pt ( 195.1 g Pt ) = 648 g Pt (6.02 x 1023 Pt ) 14. What is the mass of 2.00 x 1022 silicon atoms? 2.00 x 1022 Si ( 28.1g Si ) = .93 g Si (6.02 x 1023 Si )

6.02 x 1023 = P.T. grams 13. What is the mass 2.00 x 1024 platinum atoms? 2.00 x 1024 Pt ( 195.1 g Pt ) = 648 g Pt (6.02 x 1023 Pt ) 14. What is the mass of 2.00 x 1022 silicon atoms? 2.00 x 1022 Si ( 28.1g Si ) = .93 g Si (6.02 x 1023 Si )

6.02 x 1023 = P.T. grams 13. What is the mass 2.00 x 1024 platinum atoms? 2.00 x 1024 Pt ( 195.1 g Pt ) = 648 g Pt (6.02 x 1023 Pt ) 14. What is the mass of 2.00 x 1022 silicon atoms? 2.00 x 1022 Si ( 28.1g Si ) = .93 g Si (6.02 x 1023 Si )

6.02 x 1023 = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x 1023 Si = 2.57X1023 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x 1023 Pb = 3.49 X1022 Pb 207.2 g Pb

6.02 x 1023 = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x 1023 Si = 2.57X1023 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x 1023 Pb = 3.49 X1022 Pb 207.2 g Pb

6.02 x 1023 = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x 1023 Si = 2.57X1023 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x 1023 Pb = 3.49 X1022 Pb 207.2 g Pb

6.02 x 1023 = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x 1023 Si = 2.57X1023 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x 1023 Pb = 3.49 X1022 Pb 207.2 g Pb

6.02 x 1023 = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x 1023 Si = 2.57X1023 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x 1023 Pb = 3.49 X1022 Pb 207.2 g Pb

6.02 x 1023 = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x 1023 Si = 2.57X1023 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x 1023 Pb = 3.49 X1022 Pb 207.2 g Pb

6.02 x 1023 = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x 1023 Si = 2.57X1023 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x 1023 Pb = 3.49 X1022 Pb 207.2 g Pb

6.02 x 1023 = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x 1023 Si = 2.57X1023 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x 1023 Pb = 3.49 X1022 Pb 207.2 g Pb

6.02 x 1023 = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x 1023 Si = 2.57X1023 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x 1023 Pb = 3.49 X1022 Pb 207.2 g Pb

6.02 x 1023 = P.T. grams 17. What is the mass of 2.0 x109 Pb atoms? 2.0 x109 Pb 207.2g Pb =6.88x10-13g Pb 6.02 x 1023 Pb 18. What is the mass of 1.51 x1023 C6H12O6? 1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6 (6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams 17. What is the mass of 2.0 x109 Pb atoms? 2.0 x109 Pb 207.2g Pb =6.88x10-13g Pb 6.02 x 1023 Pb 18. What is the mass of 1.51 x1023 C6H12O6? 1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6 (6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams 17. What is the mass of 2.0 x109 Pb atoms? 2.0 x109 Pb 207.2g Pb =6.88x10-13g Pb 6.02 x 1023 Pb 18. What is the mass of 1.51 x1023 C6H12O6? 1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6 (6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams 17. What is the mass of 2.0 x109 Pb atoms? 2.0 x109 Pb 207.2g Pb =6.88x10-13g Pb 6.02 x 1023 Pb 18. What is the mass of 1.51 x1023 C6H12O6? 1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6 (6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams 17. What is the mass of 2.0 x109 Pb atoms? 2.0 x109 Pb 207.2g Pb =6.88x10-13g Pb 6.02 x 1023 Pb 18. What is the mass of 1.51 x1023 C6H12O6? 1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6 (6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams 17. What is the mass of 2.0 x109 Pb atoms? 2.0 x109 Pb 207.2g Pb =6.88x10-13g Pb 6.02 x 1023 Pb 18. What is the mass of 1.51 x1023 C6H12O6? 1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6 (6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams 17. What is the mass of 2.0 x109 Pb atoms? 2.0 x109 Pb 207.2g Pb =6.88x10-13g Pb 6.02 x 1023 Pb 18. What is the mass of 1.51 x1023 C6H12O6? 1.51 x1023 C6H12O6 (180 g C6H12O6 ) )= 45.0g C6H12O6 (6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams 17. What is the mass of 2.0 x109 Pb atoms? 2.0 x109 Pb 207.2g Pb =6.88x10-13g Pb 6.02 x 1023 Pb 18. What is the mass of 1.51 x1023 C6H12O6? 1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6 (6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams 17. What is the mass of 2.0 x109 Pb atoms? 2.0 x109 Pb 207.2g Pb =6.88x10-13g Pb 6.02 x 1023 Pb 18. What is the mass of 1.51 x1023 C6H12O6? 1.51 x1023 C6H12O6 (180 g C6H12O6 )= 45.0g C6H12O6 (6.02 x 1023 C6H12O6)

6.02 x 1023 = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO3)2 454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO3)2 187.5g Cu(NO3)2

6.02 x 1023 = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO3)2 454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO3)2 187.5g Cu(NO3)2

6.02 x 1023 = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO3)2 454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO3)2 187.5g Cu(NO3)2

6.02 x 1023 = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO3)2 454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO3)2 187.5g Cu(NO3)2

6.02 x 1023 = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO3)2 454g Cu(NO3)2 6.02 x 1023 Cu(NO3)2 =1.46x1024 Cu(NO3)2 187.5g Cu(NO3)2

Mole Concept 6.02 x 1023 is known as _____________________ number. 6.02 x 1023 particles is known as __________of particles just as 12 particles is known as a _______ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x 1023 of those particles or ___________ of those particle. __________ g = 1 mole of particles = 6.02 x 1023 particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms 6

Mole Concept 6.02 x 1023 is known as ___Avogadro’s___ number. 6.02 x 1023 particles is known as __________of particles just as 12 particles is known as a _______ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x 1023 of those particles or ___________ of those particle. __________ g = 1 mole of particles = 6.02 x 1023 particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms 6

Mole Concept 6.02 x 1023 is known as ___Avogadro’s___ number. 6.02 x 1023 particles is known as _1mole__of particles just as 12 particles is known as a _______ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x 1023 of those particles or ___________ of those particle. __________ g = 1 mole of particles = 6.02 x 1023 particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms 6

Mole Concept 6.02 x 1023 is known as ___Avogadro’s___ number. 6.02 x 1023 particles is known as _1mole__of particles just as 12 particles is known as a _dozen__ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x 1023 of those particles or ___________ of those particle. __________ g = 1 mole of particles = 6.02 x 1023 particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms 6

Mole Concept 6.02 x 1023 is known as ___Avogadro’s___ number. 6.02 x 1023 particles is known as _1mole__of particles just as 12 particles is known as a _dozen__ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x 1023 of those particles or __1mole__ of those particle. __________ g = 1 mole of particles = 6.02 x 1023 particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms 6

Mole Concept 6.02 x 1023 is known as ___Avogadro’s___ number. 6.02 x 1023 particles is known as _1mole__of particles just as 12 particles is known as a _dozen__ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x 1023 of those particles or __1mole__ of those particle. ___P.T.____ g = 1 mole of particles = 6.02 x 1023 particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x 1023 C atoms 6

Mole Concept 6.02 x 1023 is known as ___Avogadro’s___ number. 6.02 x 1023 particles is known as _1mole__of particles just as 12 particles is known as a _dozen__ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x 1023 of those particles or __1mole__ of those particle. ___P.T.____ g = 1 mole of particles = 6.02 x 1023 particles atomic mass / molecular mass / formula mass 12 __12.0____ g C = 1.00 moles of C = 6.02 x 1023 C atoms 6

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x1020C? 7.5x1020C ( 12.0g C ) = .015 g C ( 6.02x1023C)

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x1020C? 7.5x1020C ( 12.0g C ) = .015 g C ( 6.02x1023C)

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x1020C? 7.5x1020C ( 12.0g C ) = .015 g C ( 6.02x1023C)

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x1020C? 7.5x1020C ( 12.0g C ) = .015 g C ( 6.02x1023C)

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x1020C? 7.5x1020C ( 12.0g C ) = .015 g C ( 6.02x1023C)

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x1020C? 7.5x1020C ( 12.0g C ) = .015 g C ( 6.02x1023C)

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x1020C? 7.5x1020C ( 12.0g C ) = .015 g C ( 6.02x1023C)

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x1025 C atom sample? 1.806x1025 C(1mole C ) = 30 moles C (6.02x1023 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x1023C) = 5.04x1024 C atoms ( 12.0g C )

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x1025 C atom sample? 1.806x1025 C(1mole C ) = 30 moles C (6.02x1023 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x1023C) = 5.04x1024 C atoms ( 12.0g C )

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x1025 C atom sample? 1.806x1025 C(1mole C ) = 30 moles C (6.02x1023 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x1023C) = 5.04x1024 C atoms ( 12.0g C )

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x1025 C atom sample? 1.806x1025 C(1mole C ) = 30 moles C (6.02x1023 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x1023C) = 5.04x1024 C atoms ( 12.0g C )

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x1025 C atom sample? 1.806x1025 C(1mole C ) = 30 moles C (6.02x1023 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x1023C) = 5.04x1024 C atoms ( 12.0g C )

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x1025 C atom sample? 1.806x1025 C(1mole C ) = 30 moles C (6.02x1023 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x1023C) = 5.04x1024 C atoms ( 12.0g C )

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x1025 C atom sample? 1.806x1025 C(1mole C ) = 30 moles C (6.02x1023 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x1023C) = 5.04x1024 C atoms ( 12.0g C )

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x1025 C atom sample? 1.806x1025 C(1mole C ) = 30 moles C (6.02x1023 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x1023C) = 5.04x1024 C atoms ( 12.0g C )

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 25. How many atoms of carbon are in .750 mole sample? .750 mole C ( 6.02 x1023 C) = 4.52 x1023 C 1mole C

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 25. How many atoms of carbon are in .750 mole sample? .750 mole C ( 6.02 x1023 C) = 4.52 x1023 C 1mole C

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 25. How many atoms of carbon are in .750 mole sample? .750 mole C ( 6.02 x1023 C) = 4.52 x1023 C 1mole C

Mole Concept II 6.02 x 1023 = P.T. grams = 1.0 mole 25. How many atoms of carbon are in .750 mole sample? .750 mole C ( 6.02 x1023 C) = 4.52 x1023 C 1mole C

6.02 x 1023 = P.T. grams 26. What is the mass of 3.50 moles of C2H5OH? 3.50 mole C2H5OH 46 g C2H5OH=161 g C2H5OH 1mole C2H5OH 27. How many molecules are in .250 mole of C2H5OH? .250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023

6.02 x 1023 = P.T. grams 26. What is the mass of 3.50 moles of C2H5OH? 3.50 mole C2H5OH 46 g C2H5OH=161 g C2H5OH 1mole C2H5OH 27. How many molecules are in .250 mole of C2H5OH? .250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023

6.02 x 1023 = P.T. grams 26. What is the mass of 3.50 moles of C2H5OH? 3.50 mole C2H5OH 46 g C2H5OH=161 g C2H5OH 1mole C2H5OH 27. How many molecules are in .250 mole of C2H5OH? .250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023

6.02 x 1023 = P.T. grams 26. What is the mass of 3.50 moles of C2H5OH? 3.50 mole C2H5OH 46 g C2H5OH=161 g C2H5OH 1mole C2H5OH 27. How many molecules are in .250 mole of C2H5OH? .250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023

6.02 x 1023 = P.T. grams 26. What is the mass of 3.50 moles of C2H5OH? 3.50 mole C2H5OH 46 g C2H5OH=161 g C2H5OH 1mole C2H5OH 27. How many molecules are in .250 mole of C2H5OH? .250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023

6.02 x 1023 = P.T. grams 26. What is the mass of 3.50 moles of C2H5OH? 3.50 mole C2H5OH 46 g C2H5OH=161 g C2H5OH 1mole C2H5OH 27. How many molecules are in .250 mole of C2H5OH? .250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023

6.02 x 1023 = P.T. grams 26. What is the mass of 3.50 moles of C2H5OH? 3.50 mole C2H5OH 46 g C2H5OH=161 g C2H5OH 1mole C2H5OH 27. How many molecules are in .250 mole of C2H5OH? .250 mole C2H5OH 6.02 x 1023 C2H5OH= 1.51 X1023

6.02 x 1023 = P.T. grams = 1.0 mole 28. How many moles of C2H5OH are in a 1.0x10 3 molecule sample of C2H5OH? 1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles 6.02 x 1023 C2H5OH 29. How many moles of C2H5OH are in 12.0 grams of C2H5OH? 12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 28. How many moles of C2H5OH are in a 1.0x10 3 molecule sample of C2H5OH? 1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles 6.02 x 1023 C2H5OH 29. How many moles of C2H5OH are in 12.0 grams of C2H5OH? 12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 28. How many moles of C2H5OH are in a 1.0x10 3 molecule sample of C2H5OH? 1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles 6.02 x 1023 C2H5OH 29. How many moles of C2H5OH are in 12.0 grams of C2H5OH? 12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 28. How many moles of C2H5OH are in a 1.0x10 3 molecule sample of C2H5OH? 1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles 6.02 x 1023 C2H5OH 29. How many moles of C2H5OH are in 12.0 grams of C2H5OH? 12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 28. How many moles of C2H5OH are in a 1.0x10 3 molecule sample of C2H5OH? 1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles 6.02 x 1023 C2H5OH 29. How many moles of C2H5OH are in 12.0 grams of C2H5OH? 12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 28. How many moles of C2H5OH are in a 1.0x10 3 molecule sample of C2H5OH? 1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles 6.02 x 1023 C2H5OH 29. How many moles of C2H5OH are in 12.0 grams of C2H5OH? 12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 28. How many moles of C2H5OH are in a 1.0x10 3 molecule sample of C2H5OH? 1.00 x 103 C2H5OH 1mole C2H5OH = 1.67x10-21moles 6.02 x 1023 C2H5OH 29. How many moles of C2H5OH are in 12.0 grams of C2H5OH? 12.0 gC2H5OH 1mole C2H5OH= .26 moles C2H5OH 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C2H5OH? 2.50 mole C2H5OH 46.0g C2H5OH = 115 g C2H5OH 1mole C2H5OH 31. How many molecules are found in 454 g C2H5OH 454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C2H5OH? 2.50 mole C2H5OH 46.0g C2H5OH = 115 g C2H5OH 1mole C2H5OH 31. How many molecules are found in 454 g C2H5OH 454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C2H5OH? 2.50 mole C2H5OH 46.0g C2H5OH = 115 g C2H5OH 1mole C2H5OH 31. How many molecules are found in 454 g C2H5OH 454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C2H5OH? 2.50 mole C2H5OH 46.0g C2H5OH = 115 g C2H5OH 1mole C2H5OH 31. How many molecules are found in 454 g C2H5OH 454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C2H5OH? 2.50 mole C2H5OH 46.0g C2H5OH = 115 g C2H5OH 1mole C2H5OH 31. How many molecules are found in 454 g C2H5OH 454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C2H5OH? 2.50 mole C2H5OH 46.0g C2H5OH = 115 g C2H5OH 1mole C2H5OH 31. How many molecules are found in 454 g C2H5OH 454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C2H5OH? 2.50 mole C2H5OH 46.0g C2H5OH = 115 g C2H5OH 1mole C2H5OH 31. How many molecules are found in 454 g C2H5OH 454 gC2H5OH 6.02 X 10 23 C2H5OH= 5.94X10 24 46.0 g C2H5OH

6.02 x 1023 = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO3)2 ? 0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO3)2 164.1g Ca(NO3)2 33. What is the mass of .125 moles Ca(NO3)2? .125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2 1 mole Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO3)2 ? 0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO3)2 164.1g Ca(NO3)2 33. What is the mass of .125 moles Ca(NO3)2? .125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2 1 mole Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO3)2 ? 0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO3)2 164.1g Ca(NO3)2 33. What is the mass of .125 moles Ca(NO3)2? .125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2 1 mole Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO3)2 ? 0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO3)2 164.1g Ca(NO3)2 33. What is the mass of .125 moles Ca(NO3)2? .125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2 1 mole Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO3)2 ? 0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO3)2 164.1g Ca(NO3)2 33. What is the mass of .125 moles Ca(NO3)2? .125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2 1 mole Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO3)2 ? 0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO3)2 164.1g Ca(NO3)2 33. What is the mass of .125 moles Ca(NO3)2? .125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2 1 mole Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO3)2 ? 0.10 g Ca(NO3)2 1mole Ca(NO3)2 =.000609 mol Ca(NO3)2 164.1g Ca(NO3)2 33. What is the mass of .125 moles Ca(NO3)2? .125 moles Ca(NO3)2 164.1 g Ca(NO3)2 =20.5 g Ca(NO3)2 1 mole Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO3)2? 454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2 = 1.67x1024 Ca(NO3)2 164.1g Ca(NO3)2 35. What is the mass of .50 moles Ca(NO3)2? .50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2 (1mol Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO3)2? 454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2 = 1.67x1024 Ca(NO3)2 164.1g Ca(NO3)2 35. What is the mass of .50 moles Ca(NO3)2? .50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2 (1mol Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO3)2? 454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2 = 1.67x1024 Ca(NO3)2 164.1g Ca(NO3)2 35. What is the mass of .50 moles Ca(NO3)2? .50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2 (1mol Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO3)2? 454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2 = 1.67x1024 Ca(NO3)2 164.1g Ca(NO3)2 35. What is the mass of .50 moles Ca(NO3)2? .50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2 (1mol Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO3)2? 454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2 = 1.67x1024 Ca(NO3)2 164.1g Ca(NO3)2 35. What is the mass of .50 moles Ca(NO3)2? .50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2 (1mol Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO3)2? 454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2 = 1.67x1024 Ca(NO3)2 164.1g Ca(NO3)2 35. What is the mass of .50 moles Ca(NO3)2? .50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2 (1mol Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO3)2? 454g Ca(NO3)2 6.02 X 10 23 Ca(NO3)2 = 1.67x1024 Ca(NO3)2 164.1g Ca(NO3)2 35. What is the mass of .50 moles Ca(NO3)2? .50 moles Ca(NO3)2(164.1g Ca(NO3)2=82.1g Ca(NO3)2 (1mol Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 36. How many units are in .250 mol Ca(NO3)2? .250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2 1mole Ca(NO3)2 37. How many moles are in 1.0x109 Ca(NO3)2? 1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol 6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 36. How many units are in .250 mol Ca(NO3)2? .250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2 1mole Ca(NO3)2 37. How many moles are in 1.0x109 Ca(NO3)2? 1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol 6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 36. How many units are in .250 mol Ca(NO3)2? .250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2 1mole Ca(NO3)2 37. How many moles are in 1.0x109 Ca(NO3)2? 1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol 6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 36. How many units are in .250 mol Ca(NO3)2? .250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2 1mole Ca(NO3)2 37. How many moles are in 1.0x109 Ca(NO3)2? 1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol 6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 36. How many units are in .250 mol Ca(NO3)2? .250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2 1mole Ca(NO3)2 37. How many moles are in 1.0x109 Ca(NO3)2? 1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol 6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 36. How many units are in .250 mol Ca(NO3)2? .250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2 1mole Ca(NO3)2 37. How many moles are in 1.0x109 Ca(NO3)2? 1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol 6.02x1023 Ca(NO3)2

6.02 x 1023 = P.T. grams = 1.0 mole 36. How many units are in .250 mol Ca(NO3)2? .250 mole Ca(NO3)2 6.02x1023 Ca(NO3)2 =1.51x1023 Ca(NO3)2 1mole Ca(NO3)2 37. How many moles are in 1.0x109 Ca(NO3)2? 1.0x109 Ca(NO3)2 ( 1mole Ca(NO3)2= 1.66x10-15 mol 6.02x1023 Ca(NO3)2

Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S1.0 ( 32 g S ) 1.56 50 g O ( 1 mol O ) = 3.13 mol O2.0 ( 16 g O ) 1.56 SO2

Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S1.0 ( 32 g S ) 1.56 50 g O ( 1 mol O ) = 3.13 mol O2.0 ( 16 g O ) 1.56 SO2

Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S1.0 ( 32 g S ) 1.56 50 g O ( 1 mol O ) = 3.13 mol O2.0 ( 16 g O ) 1.56 SO2

Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S1.0 ( 32 g S ) 1.56 50 g O ( 1 mol O ) = 3.13 mol O2.0 ( 16 g O ) 1.56 SO2

Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S1.0 ( 32 g S ) 1.56 50 g O ( 1 mol O ) = 3.13 mol O2.0 ( 16 g O ) 1.56 SO2

Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S1.0 ( 32 g S ) 1.56 50 g O ( 1 mol O ) = 3.13 mol O2.0 ( 16 g O ) 1.56 SO2

Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S1.0 ( 32 g S ) 1.56 50 g O ( 1 mol O ) = 3.13 mol O2.0 ( 16 g O ) 1.56 SO2

Empirical formula Count the number of particles of each element by determining the number of moles of each element Divide by the smallest number of moles in order to determine the simplest whole number ratio that the elements combine in to form the compound If the ratios are not whole numbers multiply each ratio by a common factor until they are whole number ratios

Empirical formula Count the number of particles of each element by determining the number of moles of each element Divide by the smallest number of moles in order to determine the simplest whole number ratio that the elements combine in to form the compound If the ratios are not whole numbers multiply each ratio by a common factor until they are whole number ratios

Empirical formula Count the number of particles of each element by determining the number of moles of each element Divide by the smallest number of moles in order to determine the simplest whole number ratio that the elements combine in to form the compound If the ratios are not whole numbers multiply each ratio by a common factor until they are whole number ratios

Empirical formula Count the number of particles of each element by determining the number of moles of each element Divide by the smallest number of moles in order to determine the simplest whole number ratio that the elements combine in to form the compound If the ratios are not whole numbers multiply each ratio by a common factor until they are whole number ratios

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co1.0 (59.0gCo ) .603 64.4gCl ( 1mole Cl)=1.81 mol Cl3.0 ( 35.5 g Cl) .603 CoCl3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula B) CoCl2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al1.0 (27.0g Al ) 1.96 47.1 gO ( 1mole O)=2.94 mol O1.5 ( 16.0 g O) 1.96 Al1.0O1.5 x 2 = Al2O3

Empirical Formula D) MgO E) 2.83gK, .435g C, 1.735 g O 2.83 g K ( 1 mole K ) = .0724 mole K2 ( 39.1 g K ) .0363 .435 g C ( 1 mole C ) = .0363 mole C ( 12.0 g C ) .0363 1.735 g O ( 1 mole O ) = .108 mole O3 ( 16.0 g O ) .0363

Empirical Formula E) 2.83gK, .435g C, 1.735 g O 2.83 g K ( 1 mole K ) = .0724 mole K2 ( 39.1 g K ) .0363 .435 g C ( 1 mole C ) = .0363 mole C ( 12.0 g C ) .0363 1.735 g O ( 1 mole O ) = .108 mole O3 ( 16.0 g O ) .0363

Empirical Formula E) 2.83 g K ( 1 mole K ) = .0724 mole K2 .435 g C ( 1 mole C ) = .0363 mole C ( 12.0 g C ) .0363 1.735 g C ( 1 mole O ) = .108 mole O3 ( 16.0 g O ) .0363

Empirical Formula E) 2.83 g K ( 1 mole K ) = .0724 mole K2 .435 g C ( 1 mole C ) = .0363 mole C ( 12.0 g C ) .0363 1.735 g C ( 1 mole O ) = .108 mole O3 ( 16.0 g O ) .0363

Empirical Formula E) 2.83 g K ( 1 mole K ) = .0724 mole K2 .435 g C ( 1 mole C ) = .0363 mole C ( 12.0 g C ) .0363 1.735 g C ( 1 mole O ) = .108 mole O3 ( 16.0 g O ) .0363

Empirical Formula E) 2.83 g K ( 1 mole K ) = .0724 mole K2 .435 g C ( 1 mole C ) = .0363 mole C ( 12.0 g C ) .0363 1.735 g C ( 1 mole O ) = .108 mole O3 ( 16.0 g O ) .0363

Empirical Formula F) ZnSO3 G) CaC2O4 H) NaCN I) Al2S3O12  Al2(SO4)3 J) Fe2S3O12  Fe2(SO4)3 K) MgCl2O8  Mg(ClO4)2