copyright © houghton mifflin company.all rights reserved. presentation of lecture outlines, 2–1...

Copyright © Houghton Mifflin Company.All rights reserved. Presentation of Lecture Outlines, 2–1

Welcome !

• Review : Atomic structure and the periodic table.• Today: the mol, the law of conservation of mass, stoichiometry

2

Learning Objectives

Today we will:· Gain an understanding of where

major Chemistry strands spiral throughout the curriculum.

· Explore the states of matter, the mol, and the Law of Conservation of Mass.

· Measure and analyze percent composition of a substance in a compound.

· Explore where an understanding of stoichiometry can be useful in all content areas.

Definition of Matter

The two properties of matter are:

•Mass – amount of matter

•Volume – amount of space occupied

•Therefore…matter is anything that has mass and takes up space.

States of Matter

• Solid – definite shape and volume

• Liquid – definite volume, and takes the shape of the container

• Gas – takes the shape and volume of the container

• Plasma – a collection of high energy ions and electrons.

• Go to lecture on the ideal gas law…..


Crushed Can Expt

• When the can is heated, what occurs?• When the water vaporizes in the can, what

happens to the volume of the can?• What phase change occurs to the water

vapor in the can when the can is immersed in the cold water? The water ____________.

• What happens to the pressure of the water vapor in the can when it condenses?

• Would the same result occur if the can were immersed into the water with the open side upward?


Now we can introduce

the Mole

A mole is a unit of quantity.

A mole is 6.02 x 1023 things.

÷ 6.02 x 1023

6.02 x 1023 is known as Avogadro’s constant (NA)

Number of atoms, molecules or

fundamental unitsNumber of moles

(mol)

× 6.02 x 1023

Why the Mole?

Consider one molecule of water

How many molecules in 2000 mL of water?

6.7 x 1025 molecules

We count eggs by the dozen

We measure long periods of time in centuries.

6.7 x 1025 molecules is not a manageable number. Consider:

We measure long distances in our universe using light years.

The Mole

The mole is the SI unit for chemical quantity used to count the particles in a sample of pure substance.

There are many ways of measuring large quantities that utilize large units. The mole is one such unit.

One mole = 6.02x1023 particles.“One mole of anything = 6.02x1023 units of that thing”

The Mole

How many molecules of water in 2000 mL?

6.7 x 1025 molecules

Or 111 mol meaning 111 moles of water molecules. This is a much more manageable number.

Molar Mass

Are you sitting comfortably?

Mass and the Mole: Why 6.02×1023?

Consider1 atom of sulfurAR = 32.07 = 5.326×10-23 grams

1 mole = 6.02×1023 was chosen because this was the number of carbon-12 atoms that has a mass of 12 grams.

Pure carbon (a mix of isotopes) has a mass of 12.01 g per mole.

We call this value MOLAR MASS

Consider1 mole of sulfur = 6.02×1023 atoms = 32.07 grams

6.02×1023 is a REALLY BIG number

Consider 6.02×1023 sheets of paper stacked.

How many round trips to the Moon would this stack of paper be equivalent to?

8×1010 (eighty billion ) roundtrips. Create, pick, find a mole analogy of your own. Show the math.

Aluminum AR = 26.98 gram atomic mass = 26.98 g mol-1

Language issue

gram atomic massgram molecular massgram formula mass

Carbon dioxide (CO2) MR = 44.00 gram molecular mass = 44.00 g mol-1

Sodium chloride (NaCl) MR = 58.44 gram formula mass = 58.44 g mol-1

MOLARMASS

We weigh chemical quantities in grams. The molar mass value for a substance allows us to determine the number of moles from a measured mass.

Consider 5.68 g of MgCl2

Molar mass of MgCl2 = 95.21 g mol-1

-1

massnumber of moles =

molar mass5.68 g

=0.0597 mol92.51 g mol

n

÷ 6.02 x 1023Number of atoms,

molecules or fundamental

units

MOLES

× 6.02 x 1023

MASS

× molar mass

÷ molar mass

Molarity

• Used to describe solutions• M = moles/L



Activity – Determine the Molar Mass of Butane

1. Identify at least 4 possible sources of experimental error 2. The accepted molar mass of butane is 58.0 g/mole. Determine percentage error for your data. 3. Use your data, assume a molar mass of 58.0 g/mol, and derive the value of R, the gas constant. 4. The molar mass of the gas is 58.0 g/mol. Determine the volume of one mole of

butane at STP based on your data. What is the percentage error between your value and 22.4 L (the accepted value).

Review safety concerns and protocols !!!!!

21

An Introduction…

• C6H12O6 + O2 H20 + CO2 + Energy

• What is wrong with this equation?• How did you learn to balance an

equation?


– The reactants are starting substances in a chemical reaction. The arrow means “yields.” The formulas on the right side of the arrow represent the products.

– A chemical equation is the symbolic representation of a chemical reaction in terms of chemical formulas.

Chemical Reactions: Equations

• Writing chemical equations

NaCl2ClNa2 2

– For example, the burning of sodium and chlorine to produce sodium chloride is written


– In many cases, it is useful to indicate the states of the substances in the equation.

– When you use these labels, the previous equation becomes

Chemical Reactions: Equations

• Writing chemical equations

)s(NaCl2)g(Cl)s(Na2 2

Balancing an equation activity…..

Balancing chemical reactions video link

24

Law of Conservation of Mass Activity

25

Law of Conservation of Mass:

• This experiment verifies the Law of Conservation of Matter: Matter is neither created or destroyed as a result of chemical changes but may be changed in form. The balanced equations are as follows:

• 2NaOH (aq) + CuSO4 (aq) -----> Na2SO4 (aq) + Cu(OH)2 (s)

• 4NH3 (aq) + CuSO4 (aq) -------> Cu(NH3)4SO4 (s)

• Na2CO3 (aq) + CuSO4 (aq) -------> Na2SO4 (aq) + CuCO3 (s)

26

Law of Conservation of Mass: Content Blast

• Matter is neither created nor destroyed in any process.

• Changes in matter are always accompanied by changes in energy.

• Energy changes operate under the Law of Conservation of Energy.

• Atoms are neither created nor destroyed in a chemical reaction.

• For mass to remain constant during a chemical reaction, the number of atoms of each element must be the same before and after a chemical reaction.

• Equations are balanced using coefficients before the formulas for reactants and products.

27

Bubble Gum ActivityAn Introduction to

Stoichiometry

• Complete Task 1 and Procedures 1-3.– Take a 15 minute break while chewing

the gum.• Return to your groups and

complete the remainder of the activity.

• NOTE the review hints.• Track and record content,

pedagogical issues/examples, or diversity issues that may appear in the classroom.

28

Bubble Gum Activity Report Out Summary

The manufacturer of one common type of bubble gum indicates on the package that each piece of bubble gum has a mass of 5 grams. The sugar content for each piece is 4 grams. Determine the percent by mass of sugar in this type of gum. How do your results compare to the manufacturer’s stated content? Using your data collected calculate the total mass of one piece of gum and the mass of sugar in one piece of gum. How many moles of sugar are present in one piece of gum (assume that the sugar is sucrose)? How many molecules of sugar are present in one piece of gum if the sugar is sucrose? Were any of the questions you or your group had answered by this activity? If so, which ones?

29

Bubble Gum Activity Report Out Summary

• What questions did your group have prior to the activity?

• What challenges did you experience while completing the activity?

• Were your questions answered? If not, what adjustments could be made to the activity to help it better address your questions?

30

Bubble Gum Activity Content Blast

• Percent composition is the mass of each component relative to the total mass of the substance.

• Molar mass is the mass in grams of one mole of a substance.

• The mole establishes a relationship between the atomic mass unit and the gram.

• The mass in grams of 1 mole of a substance is numerically equal to its atomic mass or formulas mass in atomic mass units.

• The number of molecules in a mole of any molecular compound is 6.02 x 1023.

Stoichiometry

• Using a balanced chemical equation to determine how much reactant or product is consumed or produced in a chemical reaction.

• 2 H2 (g) + O2 (g) 2 H2O (l)– How many moles of H2O are produced from the

reaction of 2 moles of H2?

– How many moles of O2 are required to produce 4 moles of H2O?

Chemical equation = recipe

Chemical Equations

• 2 H2 (g) + O2 (g) H2O (l)

http://www.chemistry.ohio-state.edu/betha/nealChemBal/

4 moles molecules + 2 moles molecules 4 moles molecules

Coefficients in Chemical Equation represents mole ratios

Stoichiometry“Chemical Equation = recipe”

• 2 bread slices + 1 cheese slice 1 sandwich

– Suppose need to make as many cheese sandwiches as possible for a party and have 20 slices of bread...how many slices of cheese are needed?

• By inspection of recipe (chemical equation)• Or use coefficient ratio (mole ratio) from equation:

2 bread slices = 1 cheese slice• cheese slices = 20 bread slices x = 10 cheese slices needed

slices_bread2

slice_cheese1

Image source: http://www.fotosearch.com/clip-art/sandwich.html

Stoichiometry

• 4 Fe + 3 O2 2 Fe2O3

• How many moles of Fe2O3 will be produced from the reaction of 1.50 mol of iron?

• ….mol Fe2O3 = 1.50 mol Fe x = 0.75 mol Fe2O3

molFe4

OmolFe2 32

Mole ratio is coefficient ratio

Stoichiometry – using grams

• Suppose the given is grams, not moles…• Consider the unbalanced reaction:

Al (s) + Fe2O3 (s) Al2O3 (s) + Fe (l)

If you react 4.0 g of Al with excess Fe2O3, how many moles of Fe will be produced??

• First balance the equation:

2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (l)• Chemical equation speaks to us in moles, not grams! So to

use the mole ratio it gives us (2mol Al: 2 mol Fe), – first need to convert 4.0 g Al to moles of Al– Then take moles of Al and “convert” to moles of Fe using mole ratio

from balanced equation (coefficient ratio) as was done on previous slides.

• Road map:



• Suppose we wanted to solve for grams, not moles

2 Al (s) + Fe2O3 (s) Al2O3 (s) + 2 Fe (l)

• Question: If you react 4.0 g of Al with excess Fe2O3, how many grams of Fe will be produced?

• Road map is same as before, except one last step to convert mol Fe to g Fe.– Remember that to do any stoichiometry problem, you must

be in the language of moles since the chemical equation (source of relationship or conversion factor) only speaks to

us in moles, not grams!

• Road map: g → mol → mol → g


39

Mg + 2 HCl (aq) MgCl2(aq) + H2 (g)

Demonstration –Limiting Reagents

1. 100 mL of 1.0 M HCl would have how many mol HCl present?

2. From the balanced equation, we can do a calculation that will show 0.1 mole HCl requires exactly how many Mg to completely react?

0.1 mol HCl x 1 mole Mg x 24.3 g Mg = 1.2 g Mg 2 mol HCl 1 mole Mg

This is the amount in the Red balloon! Therefore, both the HCl and the Mg in the 2nd flask are completely consumed! Cool!

3. How much H2 should be produced in the Red balloon?

1.20 g Mg x 1 mole Mg x 1 mole H2 = 0.050 mol H2

24.3 g Mg 1 mole Mg



4. How much hydrogen should be produced by the Blue balloon flask?

Since there was 0.60 Mg in the blue balloon flask, the Mg will limit the amount of hydrogen produced. Therefore, use the amount of Mg to determine the amount of H2 produced.

0.60 g Mg x 1 mole Mg x 1 mole H2 = 0.025 mol H2 (Blue balloon) 24.3 g Mg 1 mole Mg This is half the amount of gas formed in the red balloon. This explains why the blue balloon is half the size of the red balloon.



5. How much HCl would be needed to react with the 0.60 g Mg?

0.60 g Mg x 1 mole Mg x 2 mole HCl x ___1 L___ = 50 mL! 24.3 g Mg 1 mole Mg 1 mole HCl That means 50 mL of HCl are left over un-reacted in the 1st flask!

6. What about the 3rd flask with the Yellow balloon? You can do similar calculations to those shown previously.

42

Content Specific Activities Report Out Summary

• What questions did your group have prior to the activity?

• What challenges did you experience while completing the activity?

• Were your questions answered? If not, what adjustments could be made to the activity to help it better address your questions?

43

Day Two Learning Objectives

ReviewToday we:· Gained an understanding of where major

Chemistry strands spiral throughout the curriculum.

· Explored the Law of Conservation of Mass.

· Measured and analyzed percent composition of a substance in a compound.

· Explored stoichiometry

copyright © houghton mifflin company.all rights reserved. presentation of lecture outlines, 2–1...

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