2.c – conserving matter. objectives state and apply the law of conservation of matter. learn how...
TRANSCRIPT
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2.C – Conserving Matter
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Objectives
State and apply the Law of Conservation of Matter.
Learn how to balance chemical equations. Explain the mole concept. Calculate molar mass of a compound. Calculate percent composition of elements. Distinguish between renewable and
nonrenewable resources.
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C.1: Keeping Track of Atoms
Complete the reading guide handout
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In some chemical reactions, matter seems to be created. When a nail rusts.
In other reactions, matter seems to disappear. When paper burns and apparently vanishes
However, neither creation or destruction of matter occurs.
Matter may undergo physical or chemical changes.
C.1: Keeping Track of Atoms
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C.1: Keeping Track of Atoms
In a car engine gasoline is burned. What happens to the molecules of gasoline?
Gasoline is made up of carbon and hydrogen atoms (C and H atoms)
When gasoline burns these atoms react with oxygen atoms in air to form carbon dioxide (CO2), carbon monoxide (CO) and water (H2O).
The original atoms of gasoline are not destroyed but become rearranged.
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Basic chemical equation:
CxHx + O2 CO2 + CO + H2O
Molecules can be converted or decompose by chemical reactions; but the atoms remain.
reactants products
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C.1: Keeping Track of Atoms
Law of Conservation of Matter: Matter is neither created nor destroyed.
Since chemical reactions cannot create or destroy atoms, chemical equations representing the reactions must always be BALANCED.
This means that the number of atoms of each element is the same on the reactant and product sides of a chemical equation.
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Example:
The burning of coal is a reaction where carbon reacts with oxygen to produce carbon dioxide. The number of carbon and oxygen atoms is the same on both sides of the equation.
C + O2
CO2
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Atomic Perspective:
C O2 CO2
1 Carbon atom 1 oxygen molecule 1 carbon dioxide molecule
1. What are the reactants in this chemical equation?
2. What are the products in this chemical equation?
3. Are there the same number of atoms on both sides of the equation?
a. Were any atoms destroyed or created?
b. Was the Law of Conservation of Matter maintained?
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Atomic Perspective:
Cu (s) + O2 (g) CuO (s)
From the burning of copper lab:
Is this reaction balanced?
Does it obey the Law of Conservation of Matter?
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2 Cu (s) + O2 (g) 2 CuO (s)
COEFFICIENTS - indicates the number of units of each substance involved.
1. Does the oxygen molecule have a coefficient?
2. What do the subscripts represent?
3. Can subscripts be removed from chemical equations?
The above reaction reads: 2 copper atoms react with 1 oxygen to produce 2 molecules of copper oxide.
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C.2: Accounting for Atoms
To obey the Law of Conservation of Matter atoms can not be created or destroyed.
All atoms must be accounted for on both sides of the reaction arrow.
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Homework: Complete C1 Supplement worksheet
Due:
Accounting for Atoms
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C.2: Accounting for Atoms
Classwork: Page 155-157 Questions 1-5, parts A, C and D only
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C.3: Balancing Chemical Equations
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Balancing Equations
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Balancing EquationsLaw of Conservation of Atoms:
The number of atoms of each type of element must be the same on each side of the equation.
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Balancing Equations
Balancing hints:Balance the metals first.Balance the non metals next.Save the oxygen and hydrogen
atoms until the end.
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How do we Balance Equations?
H2 + O2 H2O
Below is the chemical equation for the reaction of hydrogen gas with oxygen gas to produce water.
Is this reaction balanced?
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Balancing Equations
Hydrogen and oxygen are diatomic elements. Their subscripts cannot be changed. The subscripts on water cannot be changed.
Hydrogen + oxygen water
H2 + O2 H2O
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Balancing Equation
Count the atoms on each side.Reactant side: 2 atoms H and 2 atoms OProduct side: 2 atoms H and 1 atom O
H2 + O2 H2O
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Balancing Equations
H2 + O2 H2O
If the subscripts cannot be altered, how can the atoms be made equal?
Adjust the number of molecules by changing the coefficients.
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Balancing Equations
Balance one atom at a time: Let’s start with balancing oxygen!
Reactants: 2 atoms of H and 2 atoms of O Products: 4 atoms of H and 2 atoms of O H is no longer balanced!
H2 + O2 2H2O
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Balancing Equations
Now try to balance the hydrogens by placing a 2 in front of the reactant H2
Reactant side: 4 atoms of H and 2 atoms of O Product side: 4 atoms of H and 2 atoms of O It’s Balanced!
2H2 + O2 2H2O
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How do we Balance Equations?
2 H2 + O2 2 H2O
Subscripts
Coefficients
# of atoms in a compound
Number of compounds in the reaction
Subscripts can not be changed.
Coefficients balance atoms in an equation
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Make a table of elements
_____ __________
How to Balance By Inspection:
1
Reactants Products
H
O
C
CH4 + O2 H2 O CO2+_____
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2 Count the number of each element or ion on the reactants and products side.
_____ __________
How to Balance By Inspection:
Reactants Products
H
O
4
2 3
2
C 1 1
CH4 + O2 H2 O CO2+_____
Don’t forget to add all the atoms of the same element together—even if it appears in more than one compound!
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3Each time you add a coefficient, update your table with the new quantities of each atom.
Add coefficients to balance the numbers
_____ __________2
How to Balance By Inspection:
Reactants Products
H
O
4
2 3
2
C 1 1
2
4
44
CH4 + O2 _____ H2 O CO2+
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4
Filling each coefficient location lets you and the grader know that you finished the problem rather than you left some blank because you weren’t done!
Place a “1” in any empty coefficient location
_____ __________2
How to Balance By Inspection:
Reactants Products
H
O
4
2 3
2
C 1 1
2
4
44
1 1CH4 + O2 _____ H2 O CO2+
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What do Coefficients Really Mean?
CH4 + 2 O2 CO2 + 2 H2O
Total:1 C4 H4 O
Total:1 C4 H4 O
The equation is balanced.
H
C
H
HH
O O
O O
CO O HO
H
HO
H
H
CC
H
HH
O O
O O
CCO O HO
H
HO
H
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Balancing Equations
Count atoms.Reactants: 2 atoms N and 2 atoms HProducts: 1 atom N and 3 atoms of H
N2 + H2 NH3
Nitrogen + hydrogen ammonia
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Balancing EquationsNothing is balanced.Balance the nitrogen first by placing a
coefficient of 2 in front of the NH3.
N2 + H2 2NH3
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Balancing Equations Hydrogen is not balanced. Place a 3 in front of H2.
Reactant side: 2 atoms N, 6 atoms H Product side: 2 atoms N, 6 atoms H It’s balanced!
N2 + 3H2 2NH3
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Balancing Equations
Count Atoms: Reactants: Cu – 1, H – 2, S – 1, O – 4 Products: Cu – 1, H – 2, S - 2, O - 7
Cu + H2SO4 CuSO4 + H2O + SO2
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Balancing EquationsSulfur is not balanced.Place a two in front of sulfuric acid.Count Atoms: Reactants: Cu – 1, H – 4, S – 1, O – 8 Products: Cu – 1, H – 2, S - 2, O - 7
Cu + 2H2SO4 CuSO4 + H2O + SO2
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Balancing EquationsHydrogen needs to be balanced so
place a 2 in front of the H2O. Count the number of atoms.
Cu + 2H2SO4 CuSO4 + 2H2O + SO2
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Balancing Equations
Reactants: Cu – 1, H – 4, S – 2, O – 8 Products: Cu – 1, H – 4, S – 2, O – 8
It’s balanced!
Cu + 2H2SO4 CuSO4 + 2H2O + SO2
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Balancing Equations
Count atoms.Reactants: Ca – 3 atoms, P – 2 atoms,
O – 12 atoms, H – 2 atoms, S – 1 atom
Ca3(PO4)2 + H2SO4 CaSO4 + H3PO4
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Balancing Equations
Side note on Ca3(PO4)2
The subscript after the phosphate indicates two phosphate groups.
This means two P and eight O atoms.
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Balancing Equations
Ca3(PO4)2 + H2SO4 CaSO4 + H3PO4
Count atoms in the product.Ca atoms – 1, S atom – 1, O atoms – 8;
H atoms – 3, P atom – 1,
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Balancing Equations
Balance the metal first by placing a coefficient of 3 in front of CaSO4.
Ca3(PO4)2 + H2SO4 3CaSO4 + H3PO4
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Balancing Equations
Now balance the S atoms followed by the P atoms.
Ca3(PO4)2 + 3H2SO4 3CaSO4 + H3PO4
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Balancing EquationsA coefficient of 2 placed in front of
H3PO4 which balances both hydrogen and phosphate.
Ca3(PO4)2 + 3H2SO4 3CaSO4 + 2H3PO4
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Balancing Equations
This method of balancing equations is trial and error.
Practice.
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Let’s Practice #1
Example:Balance the
following equation
__ HCl + __ Ca(OH)2 __ CaCl2 + __ H2O
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Let’s Practice #2
Example:Balance the
following equation
__ H2 + __ O2 __ H2O
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Let’s Practice #3
Example:Balance the
following equation
__ Fe + __ O2 ___ Fe2O3
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C.4: Balancing Equations
Classwork: Complete questions 1-6 on page 160-
161
Due:
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Balancing Equations
Homework: Complete C3 Supplement worksheet
Due:
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Balancing Equations
Additional Worksheets
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C.5 The Mole Concept
Definition:
Mole – Unit for counting for chemists.
Counting atoms is impractical because atoms areso small and are not visible to the naked eye.
You can not weigh individual atoms on laboratory balance.
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Relating Mass to Numbers of Atoms
Introduction of three very important concepts:
1)The mole
2)Avogadro’s number
3)Molar mass
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What is a counting unit?
You’re already familiar with one counting unit…a “dozen”
“Dozen” 12
A dozen doughnuts 12 doughnuts
A dozen books
A dozen cars
A dozen people
12 books
12 cars
12 people
A dozen = 12
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A Mole of ParticlesA Mole of Particles Contains 6.02 x 1023 particles
Avogadro’s Number
1 mole C = 6.02 x 1023 C atoms
1 mole H2O = 6.02 x 1023 H2O molecules
1 mole NaCl= 6.02 x 1023 NaCl “molecules”
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How big is a mole? Enough soft drink cans to cover the
surface of the earth to a depth of over 200 miles.
If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.
If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.
One mole of paper clips would wrap around the earth 400 trillion times!
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What does a “mole” count in?
A mole = 6.02 1023 (called Avogadro’s number)
“mole” 6.02 1023
1 mole of doughnuts 6.02 1023 doughnuts
1 mole of atoms
1 mole of molecules
6.02 1023 atoms
6.02 1023 molecules
6.02 1023 = 602,000,000,000,000,000,000,000
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1. Mole of a substance = grams of substance/MW of substance
2. The mole enables chemists to move from the microscopic world of atoms and molecules to the real world of grams and kilograms.
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Molar Mass
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Definition
Molar Mass – The mass for one mole of an atom or molecule.
Other terms commonly used for the same meaning:
Molecular WeightAtomic Weight (used for atoms)
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Mass for 1 mole of atomsThe average atomic mass = grams for 1 mole
Element Mass
1 mole of carbon atoms 12.01 g
1 mole of oxygen atoms
1 mole of hydrogen atoms
16.00 g
1.01 g
Unit for molar mass: g/mole or g/mol
Average atomic mass is found on the periodic table
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One mole of carbon (12 grams) and one mole of copper (63.5 grams)
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Molar Mass
Examples:
Molar mass of lithium (Li) = 6.94 g/mol
Molar mass of helium (He) = 4.00 g/mol
Molar mass of mercury (Hg) = 200.6 g/mol
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A molar mass of an element contains one mole of atoms.
4.00 g helium = 1mole = 6.02 x 1023 atoms.
6.94 g lithium = 1mole = 6.02 x 1023 atoms.
200.6 g mercury = 1mole = 6.02 x 1023 atoms.
Molar Mass
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Molar mass for molecules
The molar mass for a molecule = the sum of the molar masses of all the atoms
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Calculating a Molecule’s Mass
Count the number of each type of atom
Find the molar mass of each atom on the periodic table
Multiple the # of atoms molar mass for each atom
Find the sum of all the masses
1
2
3
4
To find the molar mass of a molecule:
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Example: Molar Mass
Example:Find the
molar mass for
CaBr2
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Example: Molar Mass
Count the number of each type of atom1
Ca
Br
1
2
Example:Find the
molar mass for
CaBr2
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Example: Molar Mass
Find the molar mass of each atom on the periodic table2
Ca
Br
1
2
40.08 g/mole
79.91 g/mole
Example:Find the
molar mass for
CaBr2
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Example: Molar Mass
Multiple the # of atoms molar mass for each atom3
Ca
Br
1
2
40.08 g/mole
79.91 g/mole
Example:Find the
molar mass for
CaBr2
= 40.08 g/mole
= 159.82 g/mole
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Example: Molar Mass
Find the sum of all the masses4
Ca
Br
1
2
40.08 g/mole
79.91 g/mole
= 40.08 g/mole
= 159.82 g/mole+
199.90 g/mole
1 mole of CaBr2 molecules would have a mass of 199.90 g
Example:Find the
molar mass for
CaBr2
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Molar Mass
A molar mass of a compound is the sum of the molar masses of the elements.
Example: Water, H2O:
2 H = 2 x 1 g/mole = 2 g/mole1 O = 1 x 16 g/mole = 16 g/mole
molar mass of H20 =18 g/mol
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Molar Mass
A molar mass of a compound is the sum of the molar masses of the elements.
Example: methane, CH4:
4 H = 4 x 1 g = 4 g/mole1 C = 1 x 12g = 12 g/mole
molar mass of CH4 =16 g/mol
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Example: Molar Mass & Parenthesis
Be sure to distribute the subscript outside the parenthesis to each element inside the parenthesis.
Example:Find the
molar mass for Sr(NO3)2
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Let’s Practice #1
Example:Find the
molar mass for CH2Cl2
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Let’s Practice #2
Example:Find the
molar mass for Al(OH)3
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Classwork: C.6: Molar Masses
Page 163; questions1-9
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Homework: C.5 SupplementWorksheet
Due:
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C.8: Molar Relationships
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Molar mass for molecules
The molar mass for a molecule = the sum of the molar masses of all the atoms in the molecule.
Example:
1 mole of H2O = 18 grams
1 mole of H2O weighs 18 grams.
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What happens when you do not have 18 grams of water?
How do you calculate the number of moles of water?
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Conversion Factors
Conversion factor – a way that can be used to convert from one unit to the other.
Example: the conversion between quarters and dollars:
4 quarters 1 dollar 1 dollar or 4 quarters
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Conversion Factors
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Lets return to the term dozen.
12 roses = 1 dozen roses (conversion factor)6 roses = ½ dozen
Calculations:
6 roses____ = 0.5 dozen12 roses/dozen
Conversion Factors
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Molar masses can be used as a conversion factor in chemical calculations.
Molar masses allow the chemist to convert from grams to moles of a compound and from moles to grams of a compound.
Grams Molesmolar mass
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Gram/Mole Conversions
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Example: Moles to Grams
Example:How many grams are
in 1.25 moles of water?
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Example: Moles to Grams
1.25 mol H2O = _______ g H2Omol H2O
g H2O18.02
1
22.53
When converting between grams and moles, the molar mass is needed
1 mole H2O molecules = 18.02 g
HO
21
1.01 g/mole16.00 g/mole
= 2.02 g/mole= 16.00 g/mole+
18.02 g/mole
Example:How many grams are
in 1.25 moles of water?
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Problem
How many grams are in 2.25 moles of iron, Fe?
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Problem
How many grams are in 2.25 moles of iron, Fe?
Answer: 126 grams Fe
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Problem
How many grams are in 0.375 moles of
potassium, K?
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Problem
How many grams are in 0.375 moles of
potassium, K?
Answer: 14.7 grams
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Problem
How many grams are in 20 moles of methane, CH4?
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Problem
How many grams are in 20 moles of methane, CH4?
Answer: 320 grams
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Grams to Moles
If you are given a quantity in grams you can determine how moles you have by using the molar mass.
molar massGrams Moles
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Let’s Practice
Example:How many moles are in 25.5 g NaCl?
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Let’s Practice
25.5 g NaCl = _______ mole NaClg NaCl
mole NaCl1
58.44
0.44
1 mole NaCl molecules = 58.44 g
NaCl
11
22.99 g/mole35.45 g/mole
= 22.99 g/mole= 35.45 g/mole+
58.44 g/mole
Example:How many moles are in 25.5 g NaCl?
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Example: Grams to Moles
A chemist produced 11.9 g of aluminum, Al. How many moles of aluminum were produced?
mass of Al in grams amount of Al in moles
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Problem
How many moles of calcium, Ca, are in 5.00 grams of calcium?
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Problem
How many moles of calcium, Ca, are in 5.00 grams of calcium?
Answer: 0.125 moles
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Problem
How many moles of sodium chloride, NaCl, are in 75 grams of NaCl?
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Problem
How many moles of sodium chloride, NaCl, are in 75 grams of NaCl?
Answer: 1.28 moles
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C.8: Molar Relationships
Classwork: C.8 Problems
Page 166 questions 1-3 (hold on 4)
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Homework: C.7-1 SupplementWorksheet
Due:
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C.7:Equations and Molar Relationships
How are chemical equations and molar masses related?
2 CuO + C 2 Cu + CO2
2 moles 1 mole 2 moles 1mole
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C.7:Equations and Molar Relationships
2 CuO + C 2 Cu + CO2
2 moles 1 mole 2 moles 1mole
159.1 g + 12.01 g 127.1 g + 44 g
171.1 g total 171.1 g total
Same total mass on the reactant and product side
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C.7:Equations and Molar Relationships
StoichiometryPowerPoint
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Example: Grams to Molecules
Example:How many molecules
are in 25.5 g NaCl?
Skip
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25.5 g NaCl
Example: Grams to Molecules
= _________ molecules NaCl
g NaCl
mol NaCl1
58.44
2.63 1023
1 mol = 6.021023 molecules
1 moles NaCl molecules = 58.44 g
NaCl
11
22.99 g/mole35.45 g/mole
= 22.99 g/mole= 35.45 g/mole+
58.44 g/mole
mol NaCl
molecules NaCl6.021023
1
Example:How many molecules
are in 25.5 g NaCl?
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Let’s Practice #4
Example:How many grams is a sample of 2.75 × 1024
molecules of SrCl2?
Skip
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Homework: C.7-2 SupplementWorksheet
Due:
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C.9: Percent Composition
Percent Composition - The percent mass of each element found in a mixture
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Sample problem 1: A post-1982 penny with a mass of 2.50 g is composed of 2.44 g of zinc and 0.06 g copper. What is the percent composition of each element?
Mixtures
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Zinc: 2.44 g zinc x 100% = 97.5% 2.50 g total
Copper: 0.06 g Cu x 100% = 2.5% 2.50 g total
Total percent must equal 100%
The percent composition of the penny can be found by dividing the mass of each element by the total mass of the penny and multiplying by 100%.
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Sample problem 2: The formula for the copper containing mineral chalcocite is Cu2S.
1) What percentage of copper (Cu) is in this mineral?
2) What percentage of sulfur (S) is in this mineral?
Relating molar mass and percent composition
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% Cu = mass of Cu x 100% mass of Cu2S
% Cu: 127.1 g Cu x 100% = 79.9% 159.2 g Cu2S
The formula for chalcocite indicates one mole of Cu2S contains two molecules of Cu (127.1 g) and one mole of S (32 g). The molar mass of Cu2S = 159.2 g.
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% S = mass of S x 100% mass of Cu2S
% Cu: 32 g Cu x 100% = 20.1% 159.2 g Cu2S
The formula for chalcocite indicates one mole of Cu2S contains two molecules of Cu (127.1 g) and one mole of S (32 g). The molar mass of Cu2S = 159.2 g.
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C.10: Percent Composition
Classwork: page 168, problems 2 and 4
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C.10: Percent Composition
Homework: C.10: Worksheetpercent composition
Due:
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C.11:
Lab: Retrieving Copper
Read over the lab procedure
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C.12: Conservation
Renewable resources: resources that can be replenished.
Examples include fresh water, air, fertile soil, plants and animals.
As long as natural cycles are not disturbed too much, supplies of renewable resources can be maintained indefinitely.
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Nonrenewable resources: resources that can not be readily replenished.
Examples include metals, natural gas, coal and petroleum.
The length of time to replenish these resources are very large.
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End of Unit
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Replacing: replace a resource by finding a substitute, preferably from renewable resources.
Reusing: refurbish or repair an item to use again. Examples include car parts and printer cartridges.
Recycle: reprocess to use again. Examples include aluminum cans, newspaper and glass bottles.
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C.12: Conservation
C12: Complete worksheet.
These terms will appear again on a test or final exam!