1:08 am unit 1: stoichiometry chemistry 2202 1. 1:08 am stoichiometry stoichiometry deals with...

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Unit 1: Stoichiometry

Chemistry 2202

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Stoichiometry

Stoichiometry deals with quantities used in OR produced by a chemical reaction

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3 Parts

Mole Calculations (Chp. 2 & 3) Stoichiometry and Chemical

Equations (Chp. 4) Solution Stoichiometry (Chp. 6)

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PART 1 - Mole Calculations

Isotopes and Atomic Mass (pp. 43 - 46)

Avogadro’s number (pp. 47 – 49) Mole Conversions (pp. 50 - 74)

M, MV, NA, n, m, v, N

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Questions

p. 45 #’s 1 – 4p. 46 #’s 1 – 6p. 75 #’s 9 – 12p. 51-53 #’s 5 – 15p. 57 #’s 16 – 19p. 59,60 #’s 20 –

27p. 63,64 #’s 28 - 37

p. 54 #’s 5 – 8p. 65 #’s 2, 4, 5p. 75 #’s 13, 14p. 76 #’s 15, 17–

19,

21-23p. 73 #’s 38 – 43p. 74 #’s 1 – 4p. 76 #’s 26, 27

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PART 1 - Mole Calculations

Percent composition: - given mass (p. 79 - 82) - given the chemical formula (p. 83 - 86)

Empirical Formulas (pp. 87 - 94) Molecular Formulas (pp. 95 - 98) Lab: Formula of a Hydrate

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Questions

p. 82 #’s 1 – 4p. 85 #’s 5 – 8p. 89 #’s 9 – 12p. 91 #’s 13 – 16p. 97 #’s 17 - 20

p. 103 #’s 23 – 24p. 86 #’s 1, 3 – 6p. 94 #’s 1 - 7p. 106 #’s 1 - 3,

6, 7p. 107 – 109

#’s 5 – 23, 25

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Isotopes and Atomic Mass

atomic number - the number of protons in an atom or ion

mass number - the sum of the protons and neutrons in an atom

isotope - atoms which have the same number of protons and electrons but different numbers of neutrons

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Isotopes and Atomic Mass eg.

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Isotopes and Atomic Mass

1735

1737C l C l

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Isotopes and Atomic Mass

1224

1225

1226M g M g M g

not all isotopes are created equal

79 % 10 % 11 %

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Isotopes and Atomic Mass

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Isotopes and Atomic Mass

atomic mass unit (AMU - p.43)- a unit used to describe the mass of

individual atoms- the symbol for the AMU is u- 1 u is 1/12 of the mass of a carbon-12

atom

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Isotopes and Atomic Mass

average atomic mass (AAM)- the AAM is the weighted average of

all the isotopes of an element (p. 45)

p. 14 # 5p. 45 #’s 1 – 4p. 46 #’s 1 – 6 p. 75 #’s 9 - 12 14

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Finding % Abundance

eg. Br has two naturally occurring isotopes. Br-79 has a mass of 78.92 u and Br-81 has a mass of 80.92 u. If the AAM of Br is 79.90 u, determine the percentage abundance of each isotope.

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Let x = fraction of Br-79Let y = fraction of Br-81

x + y = 178.92x + 80.92y = 79.90

Finding % Abundance

y = 1 - x

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x + y = 178.92x + 80.92y = 79.90

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Avogadro’s Number (p. 47)

p. 48

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Avogadro’s Number The MOLE is a number used by chemists to

count atoms The MOLE is the number of atoms contained in

exactly 12 g of carbon-12. In honor of Amedeo Avogadro, the number of

particles in 1 mol has been called Avogadro’s number.

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How big is Avogadro's number?

An Avogadro's number of soft drink cans would cover the surface of the earth to a depth of over 200 miles.

Avogadro's number of unpopped popcorn kernels spread across the USA, would cover the entire country to a depth of over 9 miles.

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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.

How big is Avogadro's number?

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Avogadro’s Number

1 mole = 6.02214199 x 1023 particles

1 mol = 6.022 x 1023 particles

NA = 6.022 x 1023 particles/mol

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Avogadro’s Number

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Avogadro’s Number

Number of moles Number of atoms

5 mol

0.01 mol

4.65 x 1024 atoms

8.01 x 1021 atoms

7.72 mol

0.0133 mol

6.022 x 1021 atoms

3.011 x 1024 atoms

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Avogadro’s Number

Formulas:

AN

Nn n = # of moles

N = # of particles (atoms, ions, molecules, or formula units)

NA = Avogadro’s #

N = n x NA

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How many moles are contained in the following?

a) 2.56 x 1028 Pb atoms

b) 7.19 x 1021 CO2 molecules

Avogadro’s Number

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Avogadro’s Number

eg. Calculate the number of moles in 4.98 x 1025 atoms of Al.

eg. How many formula units of Na2SO4 are in 5.69 mol of Na2SO4?

# of Na ions? # of Oxygen atoms?

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Avogadro’s Number

1. How many molecules of glucose are in 0.435 mol of C6H12O6?

How many carbon atoms?2. Calculate the number of moles in

a sample of glucose that has 3.56 x 1022 hydrogen atoms.

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Avogadro’s Number

pp. 51 – 53: #’s 5 – 15p. 54:  #’s 4 - 8

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Molar Mass

The mass of one mole of a substance is called the molar mass of the substance

eg. 1 mole of Pb has a mass of 207.19 g1 mole of Ag has a mass of 107.87 g

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Molar Mass

The symbol for molar mass is M and the unit is g/mol

eg. MPb = 207.19 g/mol

MAg = 107.87 g/mol

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Molar Mass

The molar mass of a compound is the sum of the molar masses of the elements in the compound

eg. Calculate the molar mass of:a) H2O b) C6H12O6 c) Ca(OH)2

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Molar Mass

H2O has 2 hydrogens and 1 oxygen

2 x 1.01 = 2.021 X 16.00 = 16.00

18.02 g/mol

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Molar Mass

C6H12O6

6 x 12.01 = 72.0612 x 1.01 = 12.126 x 16.00 = 96.00

180.18 g/mol

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Molar Mass

Ca(OH)2

1 x 40.08 = 40.082 x 16.00 = 32.002 x 1.01 = 2.02

74.10 g/mol

Your calculator may not show the zeroes.There should be 2 digits after the decimal when adding molar masses

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Molar Massp. 57: #’s 16 – 19 & Molar Masses Handout

1. 151.92 g/mol 7. 58.44 g/mol 2. 120.38 g/mol 8. 100.09 g/mol 3. 105.99 g/mol 9. 44.02 g/mol 4. 100.40 g/mol 10. 248.22 g/mol 5. 74.44 g/mol 11. 115.04 g/mol 6. 78.01 g/mol

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Molar Mass Calculations

AN

Nn

M

mn

N = n x NA

mass

molar

mass

m = n x M

Avogadro’s #

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Molar Mass Calculations

AN

Nn

M

mn

N = n x NA

m = n x M

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Molar Mass Calculations

eg. How many moles are in 25.3 g of NO2?

m = 25.3 gMNO2 = 46.01 g/mol

M

mn

mol/g.

g.

0146

325

mol.5500

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Molar Mass Calculations

eg. What is the mass of 4.69 mol of water?

n = 4.69 molMwater = 18.02 g/mol

m = n x M = (4.69 mol)(18.02 g/mol) = 84.5 g

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Molar Mass Calculations

Practice: p. 59 #’s 20 - 23p. 60 #’s 24 - 27

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The Mole #4answers

1.a) 0.038 mol 2.a) 17.4 g 3.a) 5631 g b) 3.75 mol b) 1560. g b) 3.73 g c) 0.276 mol c) 528 g c) 10.02 g d) 23 mol d) 97513 g d) 1662.9 g e) 0.0575 mol e) 9328 g e) 981.2 g

4.a) 1.82 mol c) 0.02133 mol e) 0.0000573 mol b) 13 mol d) 4.3423 mol

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eg. How many molecules are in 26.9 g of water?m = 26.9 gMwater= 18.02 g/mol

NA = 6.022 x 1023 molecules/mol

Find N

Particle–Mole-Mass Conversions

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M

mn

02.18

9.26

= 1.493 mol H2O

N = n x NA

= 1.493 X 6.022 x 1023

= 8.99 x 1023 molecules

Particle–Mole-Mass Conversions

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eg. A sample of Sn contains 4.69 x 1028 atoms. Calculate its mass.

N = 4.69 x 1028 NA = 6.022 x 1023 molecules/mol

MSn = 118.69 g/mol

Find m

Particle–Mole-Mass Conversions

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AN

Nn

23

28

10x022.6

10x69.4

= 77,881 mol

m = n x M

= 77881 mol x 118.69 g/mol

= 9.24 x 10 6 g

Particle–Mole-Mass Conversions

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1. Calculate the mass of 4.80 x 1024 water molecules.

2. How many grams are in 2.53 x 1026 CH4 molecules?

3. How many atoms are in 68.0 g of Sn?4. Calculate the number of molecules

in 105 g of C6H12O6.

Particle–Mole-Mass Conversions

47

Particles (N) Moles (n) Mass (m)

4.80 x 1024 H2O molecules

2.53 x 1026 CH4 molecules

68.0 g of Sn

105 g of C6H12O6

Particle–Mole-Mass Conversions

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particles(N)

Moles(n)

Mass(m)

5.98 x 1026 Cu atoms

4.50 g H2O

6.15 mol O3

Particle–Mole-Mass Conversions

06:42 AM 49

particles(N)

Moles(n)

Mass(m)

4.18 x 1022 Al atoms

2.45 g C8H18

0.145 mol S2F4

Particle–Mole-Mass Conversions

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Practice: p. 63 #’s 28 - 33p. 64 #’s 34 – 37p. 76 # 15

moles (n)

mass (m)

particles(N)

x M

÷ M÷ NA

x NA

Particle–Mole-Mass Conversions

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eg. How many molecules are in 4.78 g of glucose?m = 4.78 gMwater= 180.18 g/mol

NA = 6.022 x 1023 molecules/mol

Find N

Particle–Mole-Mass Conversions

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M

mn

180.18

4.78

= 0.02653 mol glucose

N = n x NA

= 0.02653 X 6.022 x 1023

= 8.99 x 1023 molecules

Particle–Mole-Mass Conversions

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Molar Mass Calculations

Practice: p. 63 #’s 28 - 33p. 64 #’s 34 – 37p. 76 # 15

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Molar Mass Calculations

Practice:p. 54 #’s 5 - 8p. 65 #’s 2, 4, 5p. 75 #’s 13, 14, p. 76 #’s 15, 17 – 19, 21 -23

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Molar Volume

•The volume of a gas increases when temperature increases but decreases when pressure increases .

•The volume of gases is measured under conditions of Standard Temperature and Pressure (STP)

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Molar Volume

Standard Pressure – 101.3 kPa Standard Temperature – 0 °C Avogadro hypothesized that equal

volumes of gases at the same temperature and pressure contain equal numbers of molecules.

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Molar Volume

Experimental evidence shows the volume of one mole of ANY GAS at STP is 22.4 L/mol

OR VSTP = 22.4 L/mol

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Molar Mass Calculations

AN

Nn

M

mn

N = n x NA

m = n x M

STPV

vn

v = n x VSTP

given volume in Litres

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Molar Mass Calculations

moles (n)

mass (m)

particles(N)

x M

÷ M÷ NA

x NA

volume (v)

x VSTP ÷ VSTP

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Molar Volume

p. 73 #’s 38 – 43

p. 74 #’s 1 – 4

p. 76 #’s 26, 27

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Percent Composition (p. 79)

The mass percent of a compound is the mass of each element in a compound expressed as a percent of the total mass of the compound.

mass percen tm ass o f elem en t

to ta l m ass o f com poundX 100%

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Percent Composition

eg. 8.50 g of a compound was analyzed and found to contain 6.00 g of hydrogen and 2.50 g of carbon. Calculate the mass percent for each element.

p. 82 #’s 1 - 464

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Percent Composition

mass percent may be found from the molar mass of a compound.

eg. Find the percentage composition for CH4

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Percent Composition

M = 12.01 g/mol + 4(1.01 g/mol) = 12.01 g/mol + 4.04 g/mol

= 16.05 g/mol

%25.2

100x16.05

4.04H%

%74.8

100x16.05

12.01C%

p. 85 #’s 5 - 8

66

p. 85 #’s 5 - 8p. 86 #’s 1, 3 – 6p. 107 #’s 6 – 10

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Empirical Formulas

An empirical formula gives the simplest ratio of elements in a compound.A molecular formula shows the actual number of atoms in a molecule of a compound.Ionic compounds are always written as empirical formulas

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Empirical Formulas

Compound Molecular Formula

Empirical Formula

butane C4H10

glucose C6H12O6

water H2O

benzene C6H6

C2H5

CH2O

H2O

CH69

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Empirical Formulas (p.87)

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Empirical Formulas

The empirical formula of a compound may be determined by using the % composition of a given compound.

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Empirical Formulas

Method: assume you have 100.0 g of the

compound (ie. change % to g) calculate the moles (n) for each

element divide each n by the smallest n to get

the ratio for the empirical formula72

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Empirical Formulas

eg. A compound was analyzed and found to contain 87.4% N and 12.6 % H by mass. Determine the empirical formula of the compound.

p. 89 #’s 9 - 12

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Empirical Formulas

When finding the EF, the mole ratio may not be a whole number ratio.

eg. A compound contains 84.73% N and 15.27 % H by mass. Determine the empirical formula of the compound.

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p. 90

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Empirical Formulas

eg. A compound contains 89.91% C and 10.08 % H by mass. Determine the empirical formula of the compound.

p. 91 #’s 13 – 16p. 94 #’s 2-4, 6, 7

Answers on p. 109

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MgO Lab

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Molecular Formulas The molecular formula of a compound is

a multiple of the empirical formula.

See p. 95

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Molecular Formulas

To find the molecular formula we need the empirical formula and the molar mass of the compound

eg. The empirical formula of hydrazine is NH2. The molar mass of hydrazine is 32.06 g/mol. What is the molecular formula for hydrazine?

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Molecular Formulas

p. 97 #’s 17 – 20

p. 107, 108 #’s 11 - 14

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CHC analyzer (p. 99 – 101)

1. Describe the operation of a carbon hydrogen combustion analyzer.

2. 22.0 g of carbon dioxide and 10.8 g of water is collected in a CHC analysis. Determine the empirical formula of the hydrocarbon.

p. 101 #’s 21, 2281

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Formula of a hydrate

To determine the formula of a hydrate:

- calculate the moles of water

- calculate the moles of anhydrous compound

- determine the simplest ratio

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Formula of a hydrate

eg. Use the data below to determine the value of x in LiCl• xH2O.

mass of crucible = 26.35 g

crucible + hydrate = 42.15 g

crucible + anhydrous compound= 34.94 g

83

mwater = 42.15 – 34.94 = 7.21 g H2O

mLiCl = 34.94 – 26.35 = 8.59 g LiCl

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eg. Na2CO3 ● xH2O

crucible = 15.96 g

crucible + hydrate = 22.19 g

crucible + anhydrous compound = 19.67 g

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eg. CoCl2 • xH2O

crucible = 151.96 gcrucible + hydrate = 164.35 g crucible + anhydrous compound = 158.23 g

p. 103; # 24 Lab: pp. 104-105

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Formula of a hydrate

mass of empty beaker

mass of beaker & hydrate

mass of beaker & anhydrous compound

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Review – Chp. 3

p. 86 #’s 1, 3 – 6p. 94 #’s 1 – 7p. 103 # 23pp. 107–109 #’s 5 – 22

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Test

p. 45 #’s 1 – 4p. 46 #’s 1 – 6p. 75 #’s 9 – 12p. 51-53 #’s 5 – 15p. 57 #’s 16 – 19p. 59,60 #’s 20 –

27p. 63,64 #’s 28 - 37

p. 54 #’s 5 – 8p. 65 #’s 2, 4, 5p. 75 #’s 13, 14p. 76 #’s 15, 17–

19,

21-23p. 73 #’s 38 – 43p. 74 #’s 1 – 4p. 76 #’s 26, 27

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Test

p. 82 #’s 1 – 4p. 85 #’s 5 – 8p. 89 #’s 9 – 12p. 91 #’s 13 – 16p. 97 #’s 17 - 20

p. 103 #’s 23 – 24p. 86 #’s 1, 3 – 6p. 94 #’s 1 - 7p. 106 #’s 1 - 3,

6, 7p. 107 – 109

#’s 5 – 23, 25

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Test Monday – Dec. 7

1.Formula of a hydrate- lab activity- p. 103 # 24

2. % composition- given mass

p. 82 #’s 1 – 4 - given formula

p. 85 #’s 5 - 8p. 86 #’s 3 & 4

p. 103 # 23p. 107 # 5

3. EF and MF- What is an empirical formula- Finding EF from % composition pp. 89, 91 #’s 9 - 16 (watch for .5 or .333)- Finding MF from EF and molar mass p. 97 #’s 17 – 20 p. 108 #13

4. Mole calculations – Chp. 2

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Stoichiometry (Chp.4)

Stoichiometry is the determination of quantities needed for, or produced by, chemical reactions.

Ratios from balanced chemical equations are used to predict quantities.

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Stoichiometry – p. 111

Clubhouse sandwich recipe

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Clubhouse sandwich recipe

Slices of Toast

Slices of Turkey

Strips of Bacon

# of Sandwiche

s

12

27

66

100

Fill in the missing quantities:

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Mole Ratios

A mole ratio is a mathematical expression that shows the relative amounts of two species involved in a chemical change.

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Mole Ratios

A mole ratio Comes from a balanced chemical

equation Shows the relative amounts of the

reactants/products in moles Looks like

96

coeffic ien t

coeffic ien trequ ired

g iven

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N2(g) + 3 H2 (g) → 2 NH3 (g)

20

66

140

81

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C3H8 + 5 O2 → 3 CO2 + 4 H2O

How many moles of CO2 are produced when 31.5 mol of O2 react?

98

n n xcoeffic ien t

coeffic ien trequ ired g ivenrequ ired

g iven

06:42 AM

C3H8 + 5 O2 → 3 CO2 + 4 H2O

How many moles of H2O are produced when 1.35 mol of O2 react?

06:42 AM 99

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C3H8 + 5 O2 → 3 CO2 + 4 H2O

How many moles of C3H8 are needed to react with 0.369 mol of O2?

06:42 AM 100

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C5H12 + O2 → CO2 + H2O

How many moles of CO2 are produced when 6.35 mol of O2 react?

06:42 AM 101

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Al(s) + Br2(l) → AlBr3(s)

How many moles of Br2 are needed to produce 0.315 mol of AlBr3?

p. 115 #’s 4 – 7p. 117 #’s 8 - 10

06:42 AM 102

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Mole to Mole Stoichiometry

1. How many moles of copper would be produced if 20.5 mol of copper (II) oxide decomposes?

06:42 AM 103

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Mass to Mole Stoichiometry

1. How many moles of water are produced when 20.6 g of CH4 burns?

06:42 AM 104

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Mass to Mole Stoichiometry

2. How many moles of nitrogen gas are needed to produce 6.75 g of NH3 in a reaction with hydrogen gas?

105

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Mass to Mole Stoichiometry

3. How many moles of silver would be produced if 10.0 g of AgNO3 reacts with copper metal?

106

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Mass to Mole Stoichiometry

4. How many moles of CO2 are produced when 10.6 g of C3H8 burns?

107

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Mole to Mass Stoichiometry

1. What mass of CaCl2 is produced when

4.38 mol of Ca(NO3)2 reacts with NaCl?

108

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Mole to Mass Stoichiometry

2. Calculate the mass of copper produced if 20.5 mol of CuO decomposes?

06:42 AM 109

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1. What mass of water would be produced when 5.45 g of C3H8 burns?

06:42 AM 110

Mass to Mass Stoichiometry

06:42 AM

Four step stoichiometry

1. Write a balanced chemical equation

2. Calculate moles given

3. Mole ratio – find moles required

4. Calculate required quantity

m = n x M v = n x Vstp N = n x NA

111

M

mn

STPV

vn

AN

Nn

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Mass to Mass Stoichiometry

2. Calculate the mass of HCl needed to react with 3.56 g of Fe to produce FeCl2.

06:42 AM 112

06:42 AM 113

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Stoichiometry (Chp.4)

eg. What mass of CO2 gas is produced when 45.9 g of CH4 burns ?

Step #1 CH4 + 2 O2 → CO2 + 2 H2O

45.9 g ? g

114

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eg. What mass of HCl is needed to react with 3.56 g of Fe to produce FeCl2.

116

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Mole Calculations (p. 121 #13)

3.56 g ? g

= 0.06374 mol Fe

Fe + 2 HCl → FeCl2 + H2

55.85g/mol

g3.56nFe Step #2

Step #3

Fe mol 1

HCl mol 2 x Fe mol 0.06374nHCl

= 0.12748 mol HCl117

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Mole Calculations

p. 122 #15Given 32.0 g of sulfur (M = 256.56 g/mol) Find mass of ZnS

#2 n = 0.1247 mol S8

#3 n = 0.9976 mol ZnS(M = 97.45 g/mol)

#4 m = 97.2 g ZnS118

06:42 AM

Mole Calculations

p. 123 #18Given 33.5 g of H3PO4 (M = 98.00 g/mol) Find mass of MgO

#2 n = 0.3418 mol H3PO4

#3 n = 0.5128 mol MgO(M = 40.31 g/mol)

#4 m = 20.7 g MgO119

06:42 AM

Mole Calculations

p. 123 #17Given 25.0 g of Al4C3 (M = 143.95 g/mol) Find volume of CH4

#2 n = 0.174 mol Al4C3

#3 n = 0.522 mol CH4 (MV = 22.4 L/mol)

#4 m = 11.7 L CH406:42 AM 120

How many moles of aluminum chloride can be produced from the reaction of chlorine and 10.8 mol of aluminum ?

Cl2(g) + Al(s) → AlCl3(s)

06:42 AM 121

How many moles of magnesium are needed to react with 27 g of iodine to form magnesium iodide?

06:42 AM 122

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Mole Calculations (p. 121 #14)

2.34 g ? L

#2 n = 0.05086 mol NO2

#3 n = 0.01272 mol O2

(MV = 22.4 L/mol)

#4 n = (0.01272)(22.4) = 0.285 L O2123

06:42 AM

Limiting Reactant (p. 128)

10.0 g of Li reacts with 15.0 g of Br2. Calculate the mass of LiBr produced.

124

Limiting Reactant (p. 128)

06:42 AM 125

2 Li Br2 2 LiBr

10.0

15.0

06:42 AM

Limiting Reactant (p. 128)

The Limiting Reactant (LR) OR Limiting Reagent (LR) is the substance that is completely used in a chemical reaction.

The Excess Reactant is the reactant that is left over after a reaction is complete.

126

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Limiting Reactant (p. 128)

eg. 2.00 g of NaI reacts with 2.00 g of Pb(NO3)2. Determine the LR and calculate the amount of PbI2 produced.

write a balanced equation find n for each reactant (Step #2) find moles produced by each

reactant (Step #3) 127

Pb(NO3)2 + 2 NaI → 2 NaNO3 + PbI2

nPb(NO)3 = 2.00 g 331.21 g/mol = 0.006038 mol

nNaI = 2.00 g 149.89 g/mol = 0.013343 mol

06:42 AM 128

nPbI2 = 0.006038 mol Pb(NO3)2 x 1 mol PbI2

1 mol Pb(NO3)2

= 0.006038 mol PbI2

nPbI2 = 0.013343 mol NaI x 1 mol PbI2

2 mol NaI

= 0.006672 mol PbI2

mPbI2 = 0.006038 mol x 460.99 g/mol

= 2.78 g PbI2

06:42 AM 129

What mass of calcium carbonate will be produced when 20.0 g of calcium phosphate reacts with 15.0 g of sodium carbonate?

(14.2 g)

3 Na2CO3 + Ca3(PO4)2 → 3 CaCO3 + 2 Na3PO4

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What mass of barium hydroxide will be produced when 10.0 g of barium nitrate reacts with 30.0 g of sodium hydroxide? (6.56 g)

Ba(NO3)2 + NaOH → Ba(OH)2 + NaNO3

= 0.03826 mol Ba(NO3)2 = 0.750 mol NaOH

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g/mol 40.00

g30.00nNaOH g/mol 261.35

g10.00n

23 )Ba(NO

Using Ba(NO3)2

= 0.03826 mol Ba(OH)2

Using NaOH

= 0.375 mol Ba(OH)2

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23

223Ba(OH) )Ba(NO mol 1

Ba(OH) mol 1 x )Ba(NO mol 0.03826n

2

NaOH mol 2

Ba(OH) mol 1 x NaOH mol 0.750n 2

Ba(OH)2

g 6.56

g/mol 171.35 x mol 0.03826m2Ba(OH)

What volume of hydrogen gas at STP will be produced when 10.0 g of zinc metal reacts with 20.0 g of hydrogen chloride?

Zn + 2 HCl → H2 + ZnCl2

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Law of Conservation of Mass (p. 118)

In a chemical reaction, the total mass of reactants always equals the total mass of products.

eg. 2 Na3N → 6 Na + N2

When 500.00 g of Na3N decomposes 323.20 g of N2 is produced. How much Na is produced in this decomposition?

136

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Law of Conservation of Mass( p. 118)

eg. To produce 90.1 g of water, what mass of hydrogen gas is needed to react with 80.0 g of oxygen?

eg. If 3.55 g of chlorine reacts with exactly2.29 g of sodium, what mass of NaCl willbe produced?

137

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The theoretical yield is the amount of product that we calculate using stoichiometry

The actual yield is the amount of product obtained from a chemical reaction

Percent yield (p. 137)

138

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Percent yield (p. 137)

percent y ie ld actua l y ie ld

theoretica l y ie ld x 100

139

p. 139 #’s 31, 32, & 33

DEMO: silver nitrate + copper

Equation:

Mass AgNO3 =

Mass Cu =

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DEMO: silver nitrate + copper

Mass of filter paper and precipitate =

Mass of empty filter paper =

Mass of precipitate =

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