Chapter 3 Stoichiometry

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Chapter 3Stoichiometry
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Preview

• Concepts of atomic mass, moles, molar mass, and
  percent compositions.
• Introduction to chemical reactions.
• Balancing chemical equations.
• Stoichiometric calculations for reactants and
  products in a chemical reaction.
• Limiting reactants and percent yields.

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Counting by Weighing
Chapter 3 Section 1

• Can you measure an atom and tell what mass it
  has??

• To find the average mass for one atom you use

• Using the same procedure we can count the number
  of atoms in a sample.

Sample of atoms () has all identical atoms.
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Atomic Masses
Chapter 3 Section 2

• Atomic mass is the mass of one atom in atomic
  mass unit (amu).
• Atomic Mass Unit
• Dalton used the hydrogen to be the reference for
  the masses of the other elements and he assigned
  it a mass of 1.
• Scientists then adopted the oxygen as a reference
  and assign it a mass of 16.0000.
• By definition used by modern systems,
  carbon-twelve 12C is assigned a mass of exactly
  12 amu.
• Masses of all other atoms are given relative to
  this standard mass.
• Carbon exists in the form of three isotopes
  12C (98.89), 13C (1.11), and 14C (lt
  0.001).

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The Atumium exhibit in Brussels
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Mass Spectrometers
Chapter 3 Section 2

• How are isotopes detected experimentally?

• You can determine
• Mass ratios.
• Relative abundance.

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Mass Spectrum of Carbon
Chapter 3 Section 2
Three isotopes of carbon are present in nature
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Mass Spectroscopic Analysis of 12C and 13C
Isotopes
Chapter 3 Section 2
Exact number by definition

• Mass 13C/ Mass12C 1.0836129.
• Mass of 13C (1.0836129)(12 amu)
• 13.003355 amu
• Masses of other elements could be determined in
  the same way.
• Question
• So why the carbon in the periodic table has a
  mass of 12.01 amu and not 12 amu??

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Average Atomic Weight
Chapter 3 Section 2

• Answer
• Carbon exists naturally as a mixture of three
  isotopes, and thus the atomic mass unit used for
  the carbon atom in the periodic table is the
  average value of the masses of those isotopes.
• 98.89 of 12 amu (12C) 1.11 of 13.0034 amu
  (13C)
• (0.9889)(12 amu) (0.0111)(13.0034 amu)
• 12.010 amu (the average atomic mass or just
  the atomic mass of the carbon atom).
• That is applied on all the elements of the
  periodic table.

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Average Atomic Weight
Chapter 3 Section 2

• Remember that, there is no single carbon atom has
  the mass of 12.01 amu. This is the average mass
  per carbon atom.
• We can apply counting by weighing concept here.
  
• To obtain 1000 atoms of carbon, you will have to
  weigh 12010 amu of carbon atoms.

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Mass Spectra
Chapter 3 Section 2

• Mass spectrometer is also used to determine the
  isotopic compositions of an element (Relative
  abundance).

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Exercise 29 on Page 117
Chapter 3 Section 2

• 187Re is 62.60 with a mass of 186.956 amu.
• Mass of Re 186.207 amu
• (0.6260)(186.956amu)
  (0.3740)(?? amu)

Answer is Mass of 185Re 184.9533 amu 185.0
amu
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The Mole
Chapter 3 Section 3

• 1 mole of a substance is equal to 6.022141023
  (Avogadros number) entities of that substance.
• 6.0221023 602,200,000,000,000,000,000,000
• Can you imagine it??
• Scientific (SI) definition of the mole
• The number equal to the number of carbon atoms in
  exactly 12-g sample of pure 12C.
• But this can be extended to any type of elements.

1 mole of seconds 1 mole of marbles 1 mole of
paper sheets
? Try this website http//www2.ucdsb.on.ca/tiss/s
tretton/ChemFilm/Mole_Concept/sld001.html
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How Big is the Mole??
Chapter 3 Section 3
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The Mole
Chapter 3 Section 3

• 12 g of 12C has 1 mole of 12C atoms. (By
  definition)
• 12.01 g of C has 1 mole of C atoms.
• Because
• Then both samples of 12C and natural C contain
  the same no. of components (1 mole).
• This is applied on all other elements when their
  masses are determined with respect to the mass of
  the 12C atom.

Relative masses of a single atom of 12C and
natural C
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The Mole
Chapter 3 Section 3
Average weight 0.5 kg 10 pieces of orange 5 kg
Average weight 1.0 kg 10 pieces of grapefruit
10 kg
A sample of a natural element with a mass equal
to the elements atomic mass expressed in grams
contains 1 mole of atoms
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The Mole
Chapter 3 Section 3

• 1 mole of Li atoms 6.0221023 Li atoms 6.941
  g of Li.
• Al???
• 1 mole of Al atoms 6.0221023 Al atoms 26.98
  g of Al.
• Mercury??
• 1 mole of Hg atoms 6.0221023 Hg atoms 200.6
  g of Hg.

A sample of a natural element with a mass equal
to the elements atomic mass expressed in grams
contains 1 mole of atoms
For any substance (numerically) Atomic mass
(amu) Mass for 1 mole (g/mol)
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Relationship between the Atomic Mass Unit and the
Gram
Chapter 3 Section 3

• For carbon-twelve atoms
• (6.0221023 atoms) 12 g
• 6.0221023 amu 1 g

Exact
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The Mole
Chapter 3 Section 3

• Some types of problems
• Given the of atoms and asked to find their
  mass.
• Given the mass of a sample of atoms and asked to
  find out how many atoms are there.
• Asked to compute the number of moles of a sample
  of atoms given its mass.
• Study Sample Exercises 3.3 through 3.5.

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The Mole
Chapter 3 Section 3

• moles of C 5.01021 atoms of C
• mass in g 5.01021 atoms of C
  

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Molar Mass
Chapter 3 Section 4

• Molar mass is the mass of 1 mole of a compound
• Molar mass of a compound is obtained by adding up
  the atomic masses of the atoms composing the
  compound.
• Examples (MM molar mass)
• Atomic mass for O 16.00 g/mol.
• Atomic mass for C 12.01 g/mol.
• MM for CO (12.0116.00) g/mol 28.01 g/mol.
• MM for CaCO3 (40.0812.01316.00) g/mol.
• 100.09 g/mol.

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Summary for Sections 1-4

• Mass of a 12C atom 12 amu.
• Mass of 1 mole of 12C 12.00 g
• Mass of 1 mole of element X in grams is
  numerically equal to the atomic mass of the same
  element in atomic mass unit (amu).
• 1 mole 6.0221023 Avogadros number
• Molar mass (MM) the mass of 1 mole of a
  substance usually expressed in g/mol.

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Molar Mass
Chapter 3 Section 4

• Sample Exercise 3.6
• a. Calculate the molar mass of juglone
  (C10H6O3).
• b. How many moles of juglone are in a 1.5610-2g
  sample?

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Molar Mass
Chapter 3 Section 4

• Sample Exercise 3.8

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Exercise 49 on Page 118
Chapter 3 Section 4

• MM of N2H4 (14.012 1.014) g/mol 32.06
  g/mol
• of N2H4 molecules In 1 g of N2H4
• 1g N2H4
• of N atoms in 1g of N2H4 molecules
• of N2H4 molecules

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Percent Composition of Compounds
Chapter 3 Section 5

• Mass can be calculated by comparing the molar
  mass of the atom to the molar mass of the
  molecule.
• Example is ethanol (C2H5OH)
• Mass O 100
• (100) 34.73
• Mass H 100
• (100) 13.13

1 (MM of O) MM of C2H5OH
1 mole of C2H5OH
16.00 g/mol 46.07 g/mol
6 moles of H
2 moles of C
6 (MM of H) MM of C2H5OH
1 mole of O
6(1.01) g/mol 46.07 g/mol
Mass s must be added up to 100
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Determining the Formula of a Compound
Chapter 3 Section 6

• Combustion of the sample is one on of the
  techniques used to analyze for carbon and
  hydrogen.
• It is done by reacting the sample with O2 to
  produce CO2, H2O, and N2.

Excess
Increase in mass of absorbents determines the
mass of carbon and hydrogen
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Determining the Formula of a Compound
Chapter 3 Section 6

• CxHyNz O2 in excess ? CO2 H2O
  O2 N2
• What is the chemical formula of CxHyNz ??
• MM of CO2 12.01 2(16.00) 44.01 g/mol
• Mass of C in CO2 0.1638 g CO2 12.01 g C /
  44.01 g CO2
• 0.04470 g C.
• Mass of C in CxHyNz 0.04470 g C /0.1156 g
  CxHyNz 100 38.67 C

Limiting reactant
Excess
Unreacted
0.1156g
0.1638g
0.1676g
Fraction of C present by mass in CO2
(a)
Continue ?
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Determining the Formula of a Compound
Chapter 3 Section 6

• Similarly
• Mass of H in H2O ? Mass of H in CxHyNz ?
  16.22 H
• Then
• 100 - mass C - mass H mass N 45.11 N
• Assuming having 100 g of CxHyNz, there will be
  38.67g C , 16.22g H, and 45.11g N.
• mol of C 38.67g C 1 mol C / 12.01 g C
• In the same way we get 16.09 mol H and 3.219
  mol N.
• Finding the smallest whole number ratio by
  dividing by 3.2
• C ? 1.0 H ? 4.9 N?1.0
• The (empirical) formula is C1H5N1 or simply CH5N
• The molecular mass is needed to determine the
  molecular (actual) formula.
• If MM 31.06 then it is CH5N
• If MM 62.12 then it is C2H10N2
• and so on

(b) (c) (d) (e)
3.220 mol C
? Practice also the sample exercise 3.13
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Determining the Formula of a Compound
Chapter 3 Section 6

• Sample Exercise 3.12
• A sample was analyzed and found to contain
  43.64 phosphorous and 56.36 oxygen. If the MM
  283.88 g/mol, find the empirical and molecular
  formula.

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Chemical Equations
Chapter 3 Section 7

• A chemical reaction is the chemical change
  involving reorganization of the atoms in one or
  more substances by breaking bonds and forming
  other new bonds.
• CH4 O2 ? CO2 H2O
• Reactants Products

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Chemical Equations
Chapter 3 Section 7

• Chemical equations must be balanced so that the
  numbers of each type of atoms in the reactant and
  product sides are equal.
• CH4 2 O2 ? CO2 2 H2O
• Reactants Products
• Good web link http//richardbowles.tripod.com/che
  mistry/balance.htm

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Chemical Equations
Chapter 3 Section 7
How much information can balanced chemical
reactions give us??
Physical state of the compound
Reaction coefficients
1
1
(s) (l) (g) (aq)
Physical States of Products and Reactants
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Balancing Chemical Equations
Chapter 3 Section 8

• Chemical equations must be balanced in order to
  make sense. Unbalanced equations give incorrect
  results.
• To balance a chemical equation
• You can only change the reaction coefficients,
  not the atom subscripts or the molecular
  formulas.
• Use trial and error methods.
• Start balancing the atoms one by one in the
  reactant and product sides.
• Write the balanced equation in the final form and
  do a reality check.

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Exercises
Chapter 3 Section 8

• Sample Exercise 3.14 on page 100
• (NH4)2Cr2O7 ? Cr2O3 N2
  H2O
• 2N , 8H , 2Cr , 7O 2N , 2H , 2Cr , 4O
• Balancing the hydrogen and oxygen in one step
• (NH4)2Cr2O7 ? Cr2O3 N2
  4 H2O

Exercise 85(a) on page 120 C6H6 (l) O2
(g) ? CO2 (g) H2O (g)
6C , 6H , 2O
1C , 2H , 3O 6C , 6H , 13O
6C , 2H , 13O
6C , 6H , 15O
6C , 6H , 15O 2 C6H6 (l)
15 O2 (g) ? 12 CO2 (g) 6 H2O (g)
6
3
15/2
13/2
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Stoichiometric Calculations
Chapter 3 Section 9

• Stoichiometry is
• the accounting or math behind chemistry.
• using enough information to get more quantitative
  results, such as numbers of moles, masses,
  percent yields, etc., for a given chemical
  reaction.
• Useful web links about stoichiometry
• http//dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry
• /Stoichiometry.html.
• http//www.shodor.org/UNChem/basic/stoic/index.htm
  l.
• http//www.chem.vt.edu/RVGS/ACT/notes
• /Study_Guide-Moles_Problems.html.

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Stoichiometric Calculations
Chapter 3 Section 9

• Steps of stoichiometric calculations
• Balance the chemical equation.
• Always make the numbers of moles of the reactants
  and products (the stoichiometric ratios) to be
  the reference (the keys) for your calculations.
• C3H8 (g) 5O2 (g) ? 3CO2 (g)
  4H2O (g)
• What mass of O2 reacts with 96.1g of C3H8?

96.1 g C3H8
mol of C3H8
mol of O2
?? g of O2
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Sample Exercise 3.17 on Page 105
Chapter 3 Section 9

• Which is more effective antacid per gram?

NaHCO3
Mg(OH)2
In other words, which one will react with more
acid (HCl)?
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Sample Exercise 3.17 on Page 105
Chapter 3 Section 9

• Solution
• NaHCO3 (s) HCl (aq) ? NaCl (aq) H2O (l)
  CO2 (aq)
• Mg(OH)2 (s) 2HCl (aq) ? MgCl2 (aq) H2O (l)
• 1.00 g NaHCO3
  1.1910-2 mol NaHCO3
• 1.00 g Mg(OH)2
  3.4210-2 mol Mg(OH)2

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Limiting Reactants
Chapter 3 Section 10

• Limiting reactant
• is the reactant that is consumed fully before
  any other reactant. It is the reactant used for
  stoichiometric calculations.
• Excess reactant(s)
• is the reactant that is not fully consumed in a
  chemical reaction.

2Na Cl2 ? 2NaCl
Also visit http//www.science.uwaterloo.ca/cchie
h/cact/c120/limitn.html
Excess
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Limiting Reactants
Chapter 3 Section 10

• Exercise If you put equal weights of sodium
  metal (Na) and chlorine gas (Cl2) into a reaction
  vessel, which is the limiting reagent?
• Solution Consider 1 g from each.
• 2Na (s) Cl2 (g)
  ? 2NaCl (s)
• 1.00g Na
• 1.00g Cl2
• 1 mole of Cl2 needs 2 moles of Na gt 0.0141g of
  Cl2 need 0.0282g of Na. But we have 0.0435g of Na
  (excess).

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Limiting Reactants
Chapter 3 Section 10

• In stoichiometric calculations involving limiting
  reactants,
• follow these steps
• 1. Balance the chemical equation.
• 2. Use the key ( of moles of reactants and
  products) to make comparison between the amount
  of each substance involved in the calculations.
• 3. Determine which reactant is the limiting one.
• 4. Continue with your calculations based on the
  limiting reactant.
• 5. Convert from moles into grams if you need to.

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Limiting Reactants
Chapter 3 Section 10

• Ready for a good example??
• Exercise In one process, 124 g of Al are reacted
  with 601 g of Fe2O3 According to
• 2Al Fe2O3 ? Al2O3 2Fe
• Calculate the mass of Al2O3 formed.

• Solution
• Make sure the equation is balanced.
• Find out which one is limiting reactant.
• Use the limiting reactant to get the moles (then
  grams) for the product Al2O3.

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Limiting Reactants
Chapter 3 Section 10

• 2Al Fe2O3 ? Al2O3 2Fe
• 124g
  601g
• g Al we start with ? mol Al ? mol Fe2O3 ? g
  Fe2O3 we need.
• 124 g Al
  367g
  Fe2O3
• 124 g Al needs 367 g Fe2O3, but we have 601 g
  Fe2O3 (excess) and Al is the limiting reactant.
• Then, g Al ? mol Al ? mol Al2O3 ? g Al2O3
• 124 g Al
  234g
  Al2O3

1 mol Al 27.0g Al
1 mol Fe2O3 2 mol Al
160.0 g Fe2O3 1 mol Fe2O3
1 mol Al 27.0g Al
1 mol Al2O3 2 mol Al
102.0 g Al2O3 1 mol Al2O3
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Calculating Percent Yield
Chapter 3 Section 10

• Theoretical Yield is the amount of product that
  would result if all the limiting reagent reacted
  (very seldom!).
• Actual Yield is the amount of product actually
  (experimentally) obtained from the reaction.
• Example For the previous reaction, if one is
  getting 198g of Al2O3, what is then the percent
  yield for Al2O3?
• Solution
• Yield 100 84.6

198g Al2O3 234g Al2O3