Chapter

1
Chapter 3
Stoichiometry
2
Mole - Mass Relationships in Chemical Systems
3.1 The Mole
3.2 Determining the Formula of an Unknown Compound
3.3 Writing and Balancing Chemical Equations
3.4 Calculating the Amounts of Reactant and
Product
3.5 Fundamentals of Solution Stoichiometry
3
Moles

• To count sugar crystals, you could weigh the
  sugar and divide by weight of a crystal
• mass and number of particles are proportional
• Atoms are too small too count individually, so we
  weigh them and divide by atomic weight to get a
  count.
• We need a bigger number to count with
• count eggs by the dozen, count bottle rockets by
  the gross, count atoms by the mole.

4
Moles

• 1 dozen 12
• 1 gross 144
• 1 mole 6.022 x 1023
• Avogadros number 6.022 x 1023 anythings/mole
• 12 eggs 1 dozen eggs
• 12 carbon atoms 1 dozen carbon atoms
• 6.022 x 1023 eggs 1mole eggs
• 6.022 x 1023 C atoms 1mole of carbon

5
Moles

• Why use Avogadros number to count atoms?
• Avogadros number is the relationship between a
  gram and an amu
• 6.022 x 1023 amu 1 g
• Consider atomic weights
• C atom 6 protons and 6 electrons 12 amu/C atom
• 12 amu 1g ? C atoms 12 g/mole C
• C atom 6.022 x 1023 amu 1 mole C
• Note Masses on periodic table can be amu/atom or
  g/mole atoms

6
Oxygen 32.00 g
One Mole of Common Substances
Water 18.02 g
CaCO3 100.09 g
Figure 3.2
Copper 63.55 g
7
Molar Mass and atomic weight and molecular weight
and formula weight

• Molar Mass - the mass of a mole of any substance
  in grams (g/mole)
• Molar Mass grams/mole
• To find the molar mass of an element or compound
  add the atomic weights of all the elements.
• Ex.
• a H atom weighs 1 amu, and an O atom weighs 16
  amu, so a water molecule (2H, and 1 O) weighs 18
  amu.
• 1 Mole of H weighs 1 g, and 1 mole of O weighs 16
  g, so a mole of water (2 mole H, and 1 mole O)
  weighs 18 g.

8
Interconverting Moles, Mass, and Number of
Chemical Entities

• Molar Mass grams/mole
• 6.022 x 1023 how many/mole
• mass lt molar mass gt of moles lt Avagadros
  number gt of particles
• how much does 1 mole of water weigh?
• 1 mole 18g 18 g
• mole
• how many molecules in 1 mole of water?
• 1 mole water 6.022 x 1023 molecule water
  6.022 x 1023 molecules
• mole water
• how many moles is 1 million water molecules?
• 1,000,000 molecules mole
  water 1.66 x 10-18 moles water
• 6.022 x 1023 molecules

9
Interconverting Moles, Mass, and Number of
Chemical Entities

• Molar Mass grams/mole
• 6.022 x 1023 how many/mole
• mass lt molar mass gt of moles lt Avagadros
  number gt of particles
• how many hydrogen atoms in 5 mole of water?
• 5 mole H2O 2 mole H 6.022 x 1023 atoms H
  6.022 x 1024
• 1 mole H2O mole H
• how much does .75 mole of NaCl weigh?
• how many moles is 5.067g of NaCl?
• how many water molecules in 1 ml of water?
• how many sodium atoms in .37 mol of sodium
  carbonate?
• Very important type of Problems!!!! Try
  Examples 3.1, 3.2

10
Calculating the Mass Percents and Masses of
Elements in a Sample of Compound

• Anything part / total X 100
• mass element total mass of element in
  cmpd/mass of cmpd x 100
• Find mass of Na in sodium carbonate
• Find mass of Na in 23.34g of sodium carbonate
• Formula Na2CO3
• Molar Mass 2x22.99 1x12.01 3x16.00 105.99
  g/mole
• Na (2x22.99 g Na/ 105.99 g Na2CO3) x 100
  43.38 Na
• Mass of Na in 23.34g Na2CO3 23.34g Na2CO3 x
  .4338 10.12 g Na
• Or
• 23.34g Na2CO3 x (2x22.99 g Na/ 105.99 g Na2CO3)
  10.12 g Na

11
Empirical and Molecular Formulas
Empirical Formula -
- the lowest whole ratio of the elements in a
compound - can be determined experimentally
Molecular Formula -
- the actual number of atoms of each element in a
molecule. - always an integer multiple of the
empirical formula
benzene MF C6H6 EF CH water MF H2O
EF H2O
12
Determination of empirical formulas

• - empirical formulas can be found from masses or
  percentages of the elements in a compound
• Steps
• 1) 1) Determine the mass of each element in the
  compound.
• If are given can assume 100g of compound to get
  masses.
• 2) 2) Use molar masses to convert each mass to
  moles.
• 3) 3) Divide the number of moles of each element
  by the smallest mole value.
• 4) 4) If moles are in fractional values, .5, .33,
  .25, .20, etc. multiply by an integer to get
  values in whole numbers.

13
Sample Problem 3.4
Determining the Empirical Formula from Masses of
Elements
PROBLEM
Elemental analysis of a sample of an ionic
compound gave the following results 2.82g of Na,
4.35g of Cl, and 7.83g of O. What are the
empirical formula and name of the compound?
PLAN
Once we find the relative number of moles of each
element, we can divide by the lowest mol amount
to find the relative mol ratios (empirical
formula).
SOLUTION
2.82g Na
0.123 mol Na
mass(g) of each element
divide by M(g/mol)
4.35g Cl
0.123 mol Cl
amount(mol) of each element
7.83g O
0.489 mol O
use of moles as subscripts
preliminary formula
Na1 Cl1 O3.98
NaClO4
Na1 Cl1 O3.98
NaClO4
change to integer subscripts
empirical formula
NaClO4 is sodium perchlorate.
14
Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
PROBLEM
During physical activity. lactic acid
(M90.08g/mol) forms in muscle tissue and is
responsible for muscle soreness. Elemental
analysis shows that it contains 40.0 mass C,
6.71 mass H, and 53.3 mass O.
(a) Determine the empirical formula of lactic
acid.
(b) Determine the molecular formula.
PLAN
assume 100g lactic acid and find the mass of each
element
divide each mass by mol mass(M)
amount(mol) of each element
molecular formula
use mols as subscripts
divide mol mass by mass of empirical formula to
get a multiplier
preliminary formula
convert to integer subscripts
empirical formula
15
Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
continued
SOLUTION
Assuming there are 100.g of lactic acid, the
constituents are
40.0g C
6.71g H
53.3g O
3.33mol C
6.66mol H
3.33mol O
C3.33
H6.66
O3.33
CH2O
empirical formula
C3H6O3 is the molecular formula
3
16
Figure 3.4
Combustion Train for the Determination of the
Chemical Composition of Organic Compounds.
17
Sample Problem 3.6
Determining a Molecular Formula from Combustion
Analysis
difference (after-before) mass of oxidized
element
PLAN
find the mass of each element in its combustion
product
molecular formula
preliminary formula
empirical formula
find the mols
18
Chemical Formulas
Empirical Formula - Shows the relative number of
atoms of each element in the compound.
It is the simplest formula, and is
derived from masses of the elements. Note ionic
compounds only have E.F. Ex Benzene
CH Molecular Formula - Shows the actual number
of atoms of each element in the
molecule of the compound. Ex Benzene
C6H6 Structural Formula - Shows the actual
number of atoms, and the bonds between
them , that is, the arrangement of
atoms in the molecule. Note As molecular
complexity increases many structural formulas can
exist for a given molecular formula. Therefore
only a name, or structural formula can
specifically identify a substance.
19
Table 3.4 Two Compounds with Molecular Formula
C2H6O
Property
Ethanol
Dimethyl Ether
46.07
M(g/mol)
46.07
Colorless
Color
Colorless
-138.50C
Melting Point
-1170C
Boiling Point
78.50C
-250C
Density at 200C
0.789g/mL(liquid)
0.00195g/mL(gas)
Use
intoxicant in alcoholic beverages
in refrigeration
Structural formulas and space-filling model
Structural isomers different compounds with
same MF
20
3.3 Chemical Equations

• Reaction - when one or more substances are
  converted to other substances
• reactants starting materials that are used up
  in a reaction
• products substances produced in a chemical
  reaction
• chemical equations represent reactions with
  chemical formulas
• reactants ? products
• a complete chemical equation
• 1) is balanced - must have same number of each
  type of atom on each side of equation
• Why? Atomic theory says a reaction is just a
  rearrangement of atoms. Conservation of mass
  says you must end with the same number and type
  of atoms you start with.
• 2) includes phases (s), (l), (g), or (aq)

21
The formation of HF gas on the macroscopic and
molecular levels
Figure 3.6
22
3.3 Chemical Equations

• Magnesium burns in oxygen to produce ?
• Metal and nonmetal form ionic compound
• Write formulas for reactants and products using
  subscripts based on charge, remember which
  elements are diatomic.
• Mg O2 ? MgO
• balance rx using coefficients only
• 2Mg O2 ? 2MgO
• add phases
• 2Mg(s) O2(g) ? 2MgO(s)
• Coefficients in reactions show ratios, either
  atom/molecule or moles.

23
3.3 Chemical Equations

• Balancing Tips
• Balance the simplest substance last.
• Start Balancing the most complicated substance
  first.
• Ions that move from reactant to product whole can
  be balanced as if it were an atom.
• If you can balance by using fractions, do so. Go
  back and correct to integers when done.
• The smallest set of whole number coefficients is
  preferred.
• i.e. coefficients of 4Mg(s) 2O2(g) ? 4MgO(s)
  should be reduced.

24
3.3 Chemical Equations

• Ethanol burns in oxygen to produce carbon dioxide
  and water.
• Write formulas.
• C2H6O O2 ? CO2 H2O
• Start Balancing ethanol, C and H before O
• C2H6O O2 ? 2CO2 H2O
• C2H6O O2 ? 2CO2 3H2O
• C2H6O 3O2 ? 2CO2 3H2O
• Add states.
• C2H6O(l) 3O2(g) ? 2CO2(g) 3H2O(g)

25
Sample Problem 3.7
Balancing Chemical Equations
PROBLEM
Within the cylinders of a cars engine, the
hydrocarbon octane (C8H18), one of many
components of gasoline, mixes with oxygen from
the air and burns to form carbon dioxide and
water vapor. Write a balanced equation for this
reaction.
translate the statement
8
25/2
9
26
Stoichiometry

• coefficients in balanced chemical equations are
  used to relate amounts of each substance to each
  other.
• i.e. Can find products produced by given amount
  of starting material, can find amount reactant
  needed to produce a given amount of product, can
  find amount of one reactant needed to react with
  another,  
• Fe2O3(s) 3 CO(g) ? 2 Fe(s) 3 CO2(g)
• moles CO needed to react with 3 moles Fe2O3 ?
• moles CO2 produced from 1.5 moles CO ?
• mass Fe2O3 needed to produce 3 kg Fe ?
•  
• mm1 mole ratio
  mm2
• g1 ? moles1 ? mole 2
  ? g2

27
Stoichiometry

• Fe2O3(s) 3 CO(g) ? 2 Fe(s) 3 CO2(g)
• moles CO needed to react with 3 moles Fe2O3 ?
• moles CO2 produced from 1.5 moles CO ?
• mass Fe2O3 needed to produce 3 kg Fe ?
• mm1 mole ratio
  mm2
• g1 ? moles1 ? mole 2
  ? g2
• 3 moles Fe2O3 3 mole CO
• 1 mole Fe2O3 9 mole CO
• 1.5 mole CO 3 mole CO2
• 3 mole CO 1.5 mole CO2
• 3 Kg Fe 1000g Fe 1 mole Fe 1 mole Fe2O3 159.7
  g Fe2O3
• 1 Kg Fe 55.85 g Fe 2 mole Fe
  1 mole Fe2O3 4289 g Fe2O3

28
Sample Problem 3.8
Calculating Amounts of Reactants and Products
PROBLEM
In a lifetime, the average American uses
1750lb(794g) of copper in coins, plumbing, and
wiring. Copper is obtained from sulfide ores,
such as chalcocite, or copper(I) sulfide, by a
multistage process. After an initial grinding
step, the first stage is to roast the ore (heat
it strongly with oxygen gas) to form powdered
copper(I) oxide and gaseous sulfur dioxide. (a)
How many moles of oxygen are required to roast
10.0mol of copper(I) sulfide? (b) How many grams
of sulfur dioxide are formed when 10.0mol of
copper(I) sulfide is roasted? (c) How many
kilograms of oxygen are required to form 2.86Kg
of copper(I) oxide?
PLAN
write and balance equation
find mols O2
find mols SO2
find mols Cu2O
find g SO2
find mols O2
find kg O2
29
Sample Problem 3.8
Calculating Amounts of Reactants and Products
continued
SOLUTION
10.0mol Cu2S
15.0mol O2
(a)
10.0mol Cu2S
(b)
641g SO2
2.86kg Cu2O
(c)
20.0mol Cu2O
20.0mol Cu2O
0.960kg O2
30
Sample Problem 3.9
Calculating Amounts of Reactants and Products in
a Reaction Sequence
PLAN
SOLUTION
write balanced equations for each step
cancel reactants and products common to both
sides of the equations
sum the equations
31
Summary of the Mass-Mole-Number Relationships in
a Chemical Reaction
Figure 3.8
32
Limiting Reactants

• When one of the reactants in a chemical reaction
  runs out, the reaction stops.
• The reactant that runs out is called the limiting
  reactant, because it limits how much product can
  be made.
• The other reactant(s) are said to be in excess.
• In any problem in which amounts of 2 or more
  reactants are given, the limiting reactant must
  be found. Only it can determine amounts of other
  reactants used, and products made.
• To determine the limiting reactant, find which
  reactant makes the least amount of product.

33
An Ice Cream Sundae Analogy for Limiting Reactions
Figure 3.9
34
Sample Problem 3.10
Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
PROBLEM
A fuel mixture used in the early days of rocketry
is composed of two liquids, hydrazine(N2H4) and
dinitrogen tetraoxide(N2O4), which ignite on
contact to form nitrogen gas and water vapor.
How many grams of nitrogen gas form when
1.00x102g of N2H4 and 2.00x102g of N2O4 are mixed?
PLAN
We always start with a balanced chemical equation
and find the number of mols of reactants and
products which have been given.
In this case one of the reactants is in molar
excess and the other will limit the extent of the
reaction.
divide by M
molar ratio
mol of N2
mol of N2
35
Sample Problem 3.10
Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
continued
SOLUTION
2
4
3
1.00x102g N2H4
3.12mol N2H4
N2H4 is the limiting reactant because it produces
less product, N2, than does N2O4.
3.12mol N2H4
4.68mol N2
4.68mol N2
131g N2
2.00x102g N2O4
2.17mol N2O4
2.17mol N2O4
6.51mol N2
36
Percent Yield

• Theoretical yield - The amount of product that
  can be produced in a reaction based on
  calculations.
• Actual yield - The amount of product actually
  produced in the reaction based on experimental
  result.
• yield actual yield/theoretical yield x 100

37
Sample Problem 3.11
Calculating Percent Yield
PROBLEM
Silicon carbide (SiC) is an important ceramic
material that is made by allowing sand(silicon
dioxide, SiO2) to react with powdered carbon at
high temperature. Carbon monoxide is also
formed. When 100.0kg of sand are processed,
51.4kg of SiC are recovered. What is the percent
yield of SiC in this process?
PLAN
SOLUTION
write balanced equation
100.0kg SiO2
1664 mol SiO2
find mol reactant product
mol SiO2 mol SiC 1664
find g product predicted
1664mol SiC
66.73kg
actual yield/theoretical yield x 100
percent yield
x100
77.0
38
Molarity

• Solute the component of smaller quantity in a
  solution
• Solvent the component of larger quantity in a
  solution
• Concentration the amount of solute dissolved in
  a given amount of solution
• - most useful measure of concentration is
  molarity
• Molarity (M) moles of solute/ L of solution
• ex 1 mole of NaCl dissolved in enough water to
  make 1L of solution is a 1 M 1 molar NaCl
  solution.
• 2 mols of NaCl dissolved to make 1 L is 2 M NaCl
  solution
• 2 mols of NaCl dissolved to make .5 L is 4 M NaCl
  solution

39
Sample Problem 3.12
Calculating the Molarity of a Solution
PROBLEM
Hydrobromic acid(HBr) is a solution of hydrogen
bromide gas in water. Calculate the molarity of
hydrobromic acid solution if 455mL contains
1.80mol of hydrogen bromide.
mol of HBr
SOLUTION
divide by volume
3.96M
concentration(mol/mL) HBr
103mL 1L
molarity(mol/L) HBr
40
Sample Problem 3.13
Calculating Mass of Solute in a Given Volume of
Solution
PROBLEM
How many grams of solute are in 1.75L of 0.460M
sodium hydrogen phosphate?
PLAN
Molarity is the number of moles of solute per
liter of solution. Knowing the molarity and
volume leaves us to find the moles and then the
of grams of solute. The formula for the solute
is Na2HPO4.
volume of soln
SOLUTION
multiply by M
1.75L
moles of solute
0.805mol Na2HPO4
multiply by M
0.805mol Na2HPO4
grams of solute
114g Na2HPO4
41
Sample Problem 3.14
Preparing a Dilute Solution from a Concentrated
Solution
PROBLEM
Isotonic saline is a 0.15M aqueous solution of
NaCl that simulates the total concentration of
ions found in many cellular fluids. Its uses
range from a cleaning rinse for contact lenses to
a washing medium for red blood cells. How would
you prepare 0.80L of isotonic saline from a 6.0M
stock solution?
PLAN
It is important to realize the number of moles of
solute does not change during the dilution but
the volume does. The new volume will be the sum
of the two volumes, that is, the total final
volume.
MdxVd mol solute McxVc
volume of dilute soln
SOLUTION
multiply by M of dilute solution
moles of NaCl in dilute soln mol NaCl in
concentrated soln
0.80L soln
0.12mol NaCl
divide by M of concentrated soln
0.12mol NaCl
0.020L soln
L of concentrated soln
42
Converting a Concentrated Solution to a Dilute
Solution
Figure 3.13
43
Fig. 3.11
44
Sample Problem 3.15
Calculating Amounts of Reactants and Products for
a Reaction in Solution
PROBLEM
Specialized cells in the stomach release HCl to
aid digestion. If they release too much, the
excess can be neutralized with antacids. A
common antacid contains magnesium hydroxide,
which reacts with the acid to form water and
magnesium chloride solution. As a government
chemist testing commercial antacids, you use
0.10M HCl to simulate the acid concentration in
the stomach. How many liters of stomach acid
react with a tablet containing 0.10g of magnesium
hydroxide?
PLAN
Write a balanced equation for the reaction find
the grams of Mg(OH)2 determine the mol ratio of
reactants and products use mols to convert to
molarity.
45
Sample Problem 3.15
Calculating Amounts of Reactants and Products for
a Reaction in Solution
continued
SOLUTION
0.10g Mg(OH)2
1.7x10-3 mol Mg(OH)2
1.7x10-3 mol Mg(OH)2
3.4x10-3 mol HCl
3.4x10-3 mol HCl
3.4x10-2 L HCl
46
Sample Problem 3.16
Solving Limiting-Reactant Problems for Reactions
in Solution
PROBLEM
Mercury and its compounds have many uses, from
filling teeth (as an alloy with silver, copper,
and tin) to the industrial production of
chlorine. Because of their toxicity, however,
soluble mercury compounds, such mercury(II)
nitrate, must be removed from industrial
wastewater. One removal method reacts the
wastewater with sodium sulfide solution to
produce solid mercury(II) sulfide and sodium
nitrate solution. In a laboratory simulation,
0.050L of 0.010M mercury(II) nitrate reacts with
0.020L of 0.10M sodium sulfide. How many grams
of mercury(II) sulfide form?
PLAN
As usual, write a balanced chemical reaction.
Since this is a problem concerning a limiting
reactant, we proceed as in Sample Problem 3.10
and find the amount of product which would be
made from each reactant. We then chose the
reactant which gives the lesser amount of product.
47
Sample Problem 3.16
Solving Limiting-Reactant Problems for Reactions
in Solution
continued
SOLUTION
0.050L Hg(NO3)2
0.020L Hg(NO3)2
x 0.010 mol/L
x 0. 10 mol/L
5.0x10-4 mol HgS
2.0x10-3 mol HgS
Hg(NO3)2 is the limiting reagent.
5.0x10-4 mol HgS
0.12g HgS
48
Laboratory Preparation of Molar Solutions
Figure 3.12
49
Figure 3.14