chapters 8-10

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Chapters 8-10Chapters 8-10

The mole, equations and stoichiometry

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The Structure of AtomsThe Structure of Atoms

Atomic Mass Unit

1 amu = 1/12 of the mass of one atom of Carbon-12

1 amu = 1.6605 x 10-24 g

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Mass: proton = 1.00728 amuneutron = 1.0086 amuelectron = 0.0005486 amu12C atom = 12.00000 amu13C atom = 13.00335 amu

Atomic and Molecular MassAtomic and Molecular Mass

1 amu mass of carbon 12 atom12

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• The atomic masses as tabulated in the periodic table are the averages of the naturally occurring isotopes.

• Mass of C = average of 12C and 13C

= 0.9889 x 12 amu + 0.0111 x 13.0034 amu

= 12.011 amu

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The mass of a molecule is just the sum of the masses of the atoms making up the molecule.

m(C2H4O2) = 2·mC + 4·mH + 2·mO

= 2·(12.01) + 4·(1.01) + 2·(16.00)

= 60.06 amu

http://www.wfu.edu/chemistry/courses/jonesbt/111/Chapter3/MW.xls

http://www.wfu.edu/chemistry/courses/jonesbt/111/Chapter3/MW.xls

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Avogadro and the MoleAvogadro and the Mole

• One mole of a substance is the gram mass value equal to the amu mass of the substance.

• One mole of any substance contains 6.02 x 1023 units of that substance.

• Avogadro’s Number (NA, 6.022 x 1023) is the

numerical value assigned to the unit, 1 mole.

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• Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)

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• The Mole: Allows us to

make comparisons

between substances

that have

different

masses.

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Counting atomsCounting atoms

Amount in moles can be converted to number of atoms:

the conversion is similar to changing 20 dozen eggs into the equivalent number of individual eggs:

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Convert 2.66 mol of an element to its equivalent number of atoms:

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How many moles in 2.54 x 1024 Fe atoms? 2.54 x 1024 atoms x

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The moleThe mole

# particles# particles

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Avogadro’s number X /

The moleThe mole

# Particles# Particles

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Practice problems p 278Practice problems p 278

1) How many atoms are present in 3.7 mol of sodium?

2) How many atoms are present in 155 mol of arsenic?

3) How many moles of xenon is 5.66 x 1026 atoms?4) How many moles of silver is 2.888 x 1015 atoms?

5) Recall that 1 mol = 6.02 x 1023 atoms

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Balancing Chemical EquationsBalancing Chemical Equations

• A balanced chemical equation represents the conversion of the reactants to products such that the number of atoms of each element is conserved.

reactants products

limestone quicklime + gas

Calcium carbonate calcium oxide + carbon dioxide

CaCO3(s) CaO(s) + CO2(g)

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Molar mass relates moles to gramsMolar mass relates moles to grams

Molar mass- is numerically equal to the element’s atomic mass and has the unit grams per mole (g/mol)

- molar masses are used as conversion factors in the relationship between moles and the mass in grams of a substance.

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Avogadro’s number X / (x 1mol/6.02EE23)6.02EE23

x mass x 1 mol 1 mol mass

The MoleThe Mole

# of Particles # of Particles

Mass (g)Mass (g)

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CaCO3(s) CaO(s) + CO2(g)

The letters in parentheses following each substance are

called State Symbols

(g) → gas (l) → liquid (s) → solid (aq) → aqueous

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A balanced equation MUST have the same number of

atoms of each element on both sides of the equation.

H2 + O2 → H2O Not Balanced

H2 + ½O2 → H2O Balanced

2H2 + O2 → 2H2O Balanced

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The numbers multiplying chemical formulas

in a chemical equation are called:

Stoichiometric Coefficients (S.C.)

2H2 + O2 → 2H2O Balanced

Here 2, 1, and 2 are stoichiometric coefficients.

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Hints for Balancing Chemical Equations:

1) Save single element molecules for last.

2) Try not to change the S.C. of a molecule containing

an element that is already balanced.

3) If possible, begin with the most complex molecule

that has no elements balanced.

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Hints for Balancing Chemical Equations:

4) Otherwise, trial and error!!

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Example 1: CH4 + O2 → CO2 + H2O

Balance O2 last

C is already balanced

Start by changing S.C. of H2O to balance H

CH4 + O2 → CO2 + 2H2O

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Example 1: CH4 + O2 → CO2 + 2H2O

Now C and H are balanced

Balance O by changing the S.C. of O2

CH4 + 2O2 → CO2 + 2H2O

BALANCED!

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Example 2: B2H6 + O2 → B2O3 + H2O

Balance O last

B is already balanced

Start by changing S.C. of H2O:

B2H6 + O2 → B2O3 + 3H2O

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Example 2: B2H6 + O2 → B2O3 + 3H2O

B and H are balanced

Balance O by changing S.C. of O2

B2H6 + 3O2 → B2O3 + 3H2O

BALANCED!

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Example 3: MnO2 + KOH + O2 → K2MnO4 +

H2O

Balance O last

Mn is already balanced

Change S.C. of KOH to balance K

MnO2 + 2KOH + O2 → K2MnO4 + H2O

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Example 3: MnO2 + 2KOH + O2 → K2MnO4 +

H2O

Mn, K, and H are balanced (H was balanced by chance)

Balance O

MnO2 + 2KOH + ½O2 → K2MnO4 + H2O

or

2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O

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Example 4: NaNO2 + H2SO4→

NO + HNO3 + H2O + Na2SO4

Hard one (no single element molecules)

S is balanced

Start with NaNO2 to balance Na

2NaNO2 + H2SO4→ NO + HNO3 + H2O + Na2SO4

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Example 4: 2NaNO2 + H2SO4→

NO + HNO3 + H2O + Na2SO4

S, Na, and N are balanced

Cannot balance H without changing S.C. for H2SO4!

Boo! Option 1: trial and error

Option 2: Go on to next problem!

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• Balance the following equations:

C6H12O6 → C2H6O + CO2

Fe + O2 → Fe2O3

NH3 + Cl2 → N2H4 + NH4Cl

KClO3 + C12H22O11 → KCl + CO2 + H2O

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C6H12O6 → 2C2H6O + 2CO2

Fe + O2 → Fe2O3



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C6H12O6 → 2C2H6O + 2CO2

4Fe + 3O2 → 2Fe2O3 (balance O first)



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C6H12O6 → 2C2H6O + 2CO2

4Fe + 3O2 → 2Fe2O3 (balance O first)


N:H is 1:3 on left, must get 1:3 on right!

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N:H is 1:3 on left, must get 1:3 on right!

4NH3 + Cl2 → N2H4 + 2NH4Cl

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C6H12O6 → 2C2H6O + 2CO2

4Fe + 3O2 → 2Fe2O3

4NH3 + Cl2 → N2H4 + 2NH4Cl

KClO3 + C12H22O11 → KCl + CO2 + H2O (tough!)

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balance C

KClO3 + C12H22O11 → KCl + 12CO2 + H2O

balance H

KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O

balance O

8KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O

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8KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O

balance K (and hope Cl is balanced)

8KClO3 + C12H22O11 → 8KCl + 12CO2 + 11H2O

Balanced!

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• Write a balanced equation for the reaction of element A (red spheres) with element B (green spheres) as represented below:

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StoichiometryStoichiometry

• Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities.

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Aqueous solutions of NaOCl (household bleach)

are prepared by the reaction of NaOH with Cl2:

2 NaOH(aq) + Cl2(g) NaOCl(aq) + NaCl(aq) + H2O(l)

How many grams of NaOH are needed to react

with 25.0 g of Cl2?

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2 NaOH + Cl2 → NaOCl + NaCl + H2O

25.0 g Cl2 reacts with ? g NaOH

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22 353.0

90.70

10.25 Clmoles

Clg

ClmoleClg

NaOHgNaOHmole

NaOHg

Clmole

NaOHmoles

Clg

ClmoleClg 2.28

1

0.40

1

2

90.70

10.25

22

22

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• Calculate the molar mass of the following:

Fe2O3 (Rust)

C6H8O7 (Citric acid)

C16H18N2O4 (Penicillin G)

• Balance the following, and determine how many

moles of CO will react with 0.500 moles of Fe2O3.

Fe2O3(s) + CO(g) Fe(s) + CO2(g)

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Fe2O3 + CO → Fe + CO2

Balance (not a simple one)Save Fe for last

C is balanced, but can’t balance OIn the products the ratio C:O is 1:2 and can’t change

Make the ratio C:O in reactants 1:2

Fe2O3 + 3CO → 2Fe + 3CO2

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Fe2O3 + 3CO → 2Fe + 3CO2

COmolesOFemole

COmoleOFemoles 50.1

1

3500.0

3232

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• Aspirin is prepared by reaction of salicylic acid (C7H6O3)

with acetic anhydride (C4H6O3) to form aspirin (C9H8O4)

and acetic acid (CH3CO2H). Use this information to

determine the mass of acetic anhydride required to

react with 4.50 g of salicylic acid. How many grams of

aspirin will result? How many grams of acetic acid will

be produced as a by-product?

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Salicylic acid + Acetic anhydride →

Aspirin + acetic acid

C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H

C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2

Balanced!

Equal # moles for all

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4.50 g Salicylic acid (C7H6O3) = ? moles

MW C7H6O3 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00

= 138.12 g/mole

..0326.0..12.138

..1..50.4 ASmoles

ASg

ASmoleASg

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Since all compounds have the same S.C., there must be

0.0326 moles of all 4 of them involved in the reaction.

g Aspirin (C9H8O4) = 0.0326 moles x MW Aspirin

= .0326 x [9x12.01 + 8x1.008 + 4x16.00]

=.0326 mole x 180.15 g/mole

5.87 g Aspirin

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• Yields of Chemical Reactions: If the actual amount

of product formed in a reaction is less than the

theoretical amount, we can calculate a percentage

yield.

100% yieldproduct lTheoretica

yieldproduct Actual yield%

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• Dichloromethane (CH2Cl2) is prepared by reaction

of methane (CH4) with chlorine (Cl2) giving

hydrogen chloride as a by-product. How many

grams of dichloromethane result from the reaction

of 1.85 kg of methane if the yield is 43.1%?

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CH4 + Cl2 → CH2Cl2 + HCl

Balance

CH4 + 2Cl2 → CH2Cl2 + 2HCl

1.85 kg CH4 = ? moles CH4

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1.85 kg CH4 = ? moles CH4

MW CH4 = 1x12.01 + 4x1.008 = 16.04 g/mole

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44 115

4.16

1100085.1 CHmoles

CHg

CHmole

kg

gCHkg

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115 moles CH4

in theory we should produce:

115 moles of CH2Cl2 and 230 moles of HCl

And use up 230 moles of Cl2

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115 moles of CH2Cl2 = ? g

MW CH2Cl2 = 12.01 + 2x1.008 + 2x35.45 = 84.93

115 moles x (84.03 g/mole) = 9770 g

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Expect 9770 g CH2Cl2

but the yield is 43.1%

So we produced just 0.431 x 9770 g

4.21 kg CH2Cl2

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Suppose the reaction went to completion

(100% yield)

Is mass conserved?

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Start with 115 moles CH4 and 230 moles Cl2

total mass = 115x16.04 + 230x70.90

= 1850 + 16300 = 18150

only 3 sig. figs. → 18.2 kg

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End with 115 moles CH2Cl2 and 230 moles HCl

total mass = 115x84.93 + 230x36.46

= 9770 + 8390 = 18160

only 3 sig. figs → 18.2 kg

chapters 8-10

Documents

02ee23 x mass x

amu mass

mass of c

avogadros number x x

particles14avogadros

gramsmolar mass

atoms x13the mole

equivalent number of