Stoichiometry

1
Stoichiometry

• Mass Relationships in Chemical Reactions
• Limiting Reactants
• Theoretical, Actual and Percent Yields
• Empirical Formulas Via Chemical Analysis
• Solution Stoichiometry
• Gas Stoichiometry

2
Mass relationshipsin chemical reactions

• Stoichiometry - The calculation of quantities of
  reactants and products in a chemical reaction.

3
Mole calculations

• The balanced equation shows the reacting ratio
  between reactants and products.
• 2C2H6 7O2 4CO2 6H2O
• For each chemical, you can determine the
• moles of each reactant consumed
• moles of each product made
• There are numerous ways of determining the number
  of moles of a species.

4
Moles, Moles and More Moles

• There are many types of stoichiometry problems.
• All involve mole-mole relationships.
• Mass-Mass
• Solve using mass molar mass
• Solution Stoichiometry
• Solve using concentration volume
• Gas Stoichiometry (Mass-Volume Volume-Volume)
• Solve using PVnRT
• Important relationship
• For the reaction aA bB ? cC dD

5
Limiting reactant

• Limiting reactant - the material that is in the
  shortest supply based on a balanced chemical
  equation.

6
Limiting reactant example
7
Example

• For the following reaction, which is limiting if
  you have 5.0 g of hydrogen and 10 g oxygen?
• Balanced Chemical Reaction
• 2H2 O2 ________gt 2H2O
• You need 2 moles of H2 for each mole of O2.
• Moles of H2 5 g 2.5 mol
• Moles of O2 10g 0.31 mol

8
Example

• Balanced Chemical Reaction
• 2H2 O2 2H2O
• You need 2 moles of H2 for each mol of O2
• You have 2.5 moles of H2 and 0.31 mol of O2
• Need a ratio of 21
• but we have a ratio of 2.5 0.31 or 8.3 1.
• Hydrogen is in excess and oxygen is the
• limiting reactant.

9
Theoretical, actual and percent yields

• Theoretical yield
• The amount of product that should be formed
  according to the chemical reaction.
• Actual yield
• The amount of product actually formed.
• Percent yield
• Ratio of actual to theoretical yield, as a .

10
Yield

• Less product is almost always produced than
  expected.
• Possible reasons
• A reactant may be impure.
• Some product is lost mechanically since the
  product must be handled to be measured.
• The reactants may undergo unexpected reactions -
  side reactions.
• No reaction truly has a 100 yield due to the
  limitations of equilibrium.
• This will also make more sense when we discuss
  rates of reaction.

11
Percent yield

• The amount of product actually formed divided by
  the amount of product calculated to be formed,
  times 100.
• yield
  x 100
• In order to determine yield, you must be able
  to recover and measure all of the product in a
  pure form.

12
Yield example

• Example. The final step in the production of
  aspirin is the reaction of salicylic acid with
  acetic anhydride.
• 48.6 g of aspirin is produced when 50.0 g of
  salicylic acid and an excess of acetic anhydride
  are reacted. What is the yield?

13
Structural Formulae
C A T A L Y S T
14
Empirical formula

• This type of formula shows the ratios of the
  number of atoms of each kind in a compound.
• For organic compounds, the empirical formula can
  be determined by combustion analysis.
• Elemental analyzer
• An instrument in which an organic compound is
  quantitatively converted to carbon dioxide and
  water -- both of which are then measured.

15
Elemental analyzer
A sample is burned, completely converting it to
CO2 and H2O. Each is collected and measured as a
weight gain. By adding other traps elements like
oxygen, nitrogen, sulfur and halogens can also
be determined.
16
Elemental analysis

• Example
• A compound known to contain only carbon,
  hydrogen and nitrogen is assayed by elemental
  analysis. The following information is obtained.
• Original sample mass 0.1156 g
• Mass of CO2 collected 0.1638 g
• Mass of H2O collected 0.1676 g
• Determine the of each element in the compound.

17
Empirical formula

• Empirical formula
• The simplest formula that shows the ratios of
  the number of atoms of each element in a
  compound.
• Example - the empirical formula for hydrogen
  peroxide (H2O2) is HO.
• We can use our percent composition information
  from the earlier example to determine an
  empirical formula.

18
Empirical formula

• From our earlier example, we found that our
  compound had a composition of
• If we assume that we have a 100.0 gram sample,
  then we can divide each percentage by the
  elements atomic mass and determine the number of
  moles of each.

C 38.67 H 16.22 N 45.11
19
Empirical formula
20
Empirical formula

• The empirical formula is then found by looking
  for the smallest whole number ratio.
• C 3.220 / 3.220 1.000
• H 16.09 / 3.220 4.997
• N 3.220 / 3.220 1.000
• The empirical formula is CH5N

21
Molecular formula

• Molecular formula - shows the actual number of
  each type of atom in a molecule.
• They are multiples of the empirical formula.
• If you know the molecular mass, then the
  molecular formula can be found.
• For our earlier example, what would be the
  molecular formula if you knew that the molecular
  mass was 62.12?

22
Molecular formula

• Empirical formula CH5N
• Empirical formula mass 31.06 g/mol
• Actual molecular mass 62.12 g/mol
• Ratio 62.12 / 31.06 2
• The molecular formula is C2H10N2
• Note This does not tell you have the atoms are
  arranged in the compound!

23
We Now Digress

• Let us predict reactions!!
• Then we can do more stoichiometry with them!!
• Reactions in aqueous solution are mostly review
  from last year!
• We now address REDOX and COORDINATION CHEMISTRY

24
Solution Stoichiometry

• Remember the relationship
• Moles can be obtained via molarity and volume.
• Molarity moles/L
• Volume L
• (Molarity)(Volume) ? (mol/L)(L) mol

25
Concentration of solutions

• We need a way to tell how much solute is in a
  solution - concentration.
• There are many systems - we will cover four.
• Weight / volume percent
• Volume / volume percent
• Weight / weight percent
• Molarity

26
Weight/Volume
Mass solute Total Volume
Weight/Volume
x 100
use g and ml
If 5 grams of NaCl is dissolved in water to make
200 ml of solution, what is the
concentration? 5 g / 200 ml 100 2.5
wt/v Saline is a 0.9 wt/v solution of NaCl in
water.
27
Volume/Volume
Volume Solute Total Volume
x 100
Volume/Volume
Use the same units for both
If 10 ml of alcohol is dissolved in water to make
200 ml of solution, what is the
concentration? 10 ml / 200 ml 100 5
V/V Alcohol in wine is measured as a V/V.
28
Weight/Weight
Mass Solute Total Mass
Weight/Weight
x 100
Use the same units for both
If a ham contained 5 grams of fat in 200 g of
ham, what is the wt/wt? 5 g / 200g
100 2.5 wt/wt On the label, it would say
97.5 fat free.
29
Very low concentrations

• Pollutants in air and water are typically found
  at very low concentrations. Two common units are
  used to express these trace amounts.
• Parts per million - ppm
• Parts per billion - ppb
• Both are modifications of the system which
  could be viewed as parts per hundred - pph.
• Both mass and volume systems are used.

30
Low concentrations in air

• Trace amounts in are are expressed as
  volume/volume ratios.
• ppm x 106
• ppb x 109
• Example. One cm3 of SO2 in one m3 of air would
  be expressed as 1 ppm or 1000 ppb.

31
Low concentrations in water

• Mass percentages are used for water pollutants.
• ppm x 106
• ppb x 109
• Example. One ppm of a toxin in water is the same
  as 1 mg / liter since one liter of water has a
  mass of approximately 106 mg.

32
Molarity

• Molarity
• Recognizes that compounds have different
• formula weights.
• A 1 M solution of glucose contains the
• same number of molecules as 1 M ethanol.
• - special symbol which means molar
• ( mol/L )

33
Molarity

• Calculate the molarity of a 2.0 L solution that
  contains 10 moles of NaOH.
• MNaOH 10 molNaOH / 2.0 L
• 5.0 M

34
Molarity

• Whats the molarity of a solution that has 18.23
  g HCl in 2.0 liters?
• First, you need the FM of HCl.
• FMHCl 1.008 x 1 H 35.45 x 1 Cl
• 36.46 g/mol
• Next, find the number of moles.
• molesHCl 18.23 gHCl / 36.46 g/mol
• 0.50 mol
• Finally, divide by the volume.
• MHCl 0.50 mol / 2.0 L
• 0.25 M

35
Solution preparation

• Solutions are typically prepared by
• Dissolving the proper amount of solute and
  diluting to volume.
• Dilution of a concentrated solution.
• Lets look at an example of the calculations
  required to prepare known molar solutions using
  both approaches.

36
Making a solution

• You are assigned the task of preparing 100.0 ml
  of a 0.5000 M solution of sodium chloride.
• What do you do?
• First, you need to know how many moles of NaCl
  are in 100.0 ml of a 0.5 M solution.
• mol M x V (in liters)
• 0.5000 M x 0.1000 liters
• 0.05000 moles NaCl

37
Making a solution

• Next, we need to know how many grams of NaCl to
  weigh out.
• gNaCl mol x Molar MassNaCl
• 0.05000 mol x 58.44 g/mol
• 2.922 grams

38
Making a solution

• Finally, youre ready to make the solution.
• Weigh out exactly 2.922 grams of dry, pure NaCl
  and transfer it to a volumetric flask.
• Fill the flask about 1/3 of the way with pure
  water and gently swirl until the salt dissolves.
• Now, dilute exactly to the mark, cap and mix.

39
Dilution

• Once you have a solution, it can be diluted by
  adding more solvent. This is also important for
  materials only available as solutions
• M1V1 M2V2
• 1 initial
• 2 final
• Any volume or concentration unit can be used as
  long as you use the same units on both sides of
  the equation.

40
Dilution

• What is the concentration of a solution produced
  by diluting 100.0 ml of 1.5 MNaOH to 2.000
  liters?
• M1V1 M2V2
• M1 1.5 M M2 ???
• V1 100.0 ml V2 2000 ml
• M2 M1V1 / V2
• M2 (1.5 M) (100.0 ml) 0.075 M
• (2000. ml)

41
Solution stoichiometry

• Extension of earlier stoichiometry problems.
• First step is to determine the number of moles
  based on solution concentration and volume.
• Final step is to convert back to volume or
  concentration as required by the problem.
• You still need a balanced equation and must use
  the coefficients for working the problem.

42
Solution stoichiometry example

• Determine the volume of 0.100 M HCl that must be
  added to completely react with 250 ml of 2.50 M
  NaOH
• Balanced chemical equation
• HCl(aq) NaOH(aq) NaCl(aq) H2O (l)
• The first step is to determine how many moles of
  NaOH we have.

43
Solution stoichiometry example

• We have 250 ml of a 2.50 M solution.
• molNaOH 0.250 L x 2.50 mol/L
• 0.625 molNaOH
• From the balanced chemical equation, we know that
  we need one mole of HCl for each mole of NaOH.
• That means we need 0.625 molHCl.

44
Solution stoichiometry example

• Now we can determine what volume of our 0.100 M
  HCl solution is required.
• L molHCl / MHCl
• 0.625 mol
• 6.26 L

45
Titration

• Method based on measurement of volume.
• You must have a solution of known concentration -
  standard solution.
• Preparing a standard/standardizing a solution is
  an important skill
• It is added to an unknown solution while the
  volume is measured.
• The process is continued until the end point is
  reached - a change that we can measure.
• Acids and bases are commonly measured using
  titrations.

46
Neutralization

• The reaction of an acid with a base to produce a
  salt and water.
• HCl (aq) NaOH (aq) NaCl (aq) H2O (l)
• If we prepare a standard solution of NaOH, we can
  then use it to determine the concentration of HCl
  in a sample.

47
Titrations
Buret - volumetric glassware used for
titrations. It allows you to add a known
amount of your titrant to the solution you are
testing. An indicator will give you the
endpoint.
48
Titrations
49
Titrations

• Note the color change which indicates that the
  endpoint has been reached.
• It is VERY faint!!!

50
Indicator examples

• Acid-base indicators are weak acids that undergo
  a color change at a known pH.

pH
phenolphthalein
51
Indicator examples
bromthymol blue commonly used in aquarium
water test kits
methyl red
52
Gas Stoichiometry

• Remember the relationship
• Moles can be obtained via PVnRT
• Moles PV/RT

53
The Gas Laws A Brief Review

• Since gases are highly compressible and will
  expand when heated, these properties have been
  studied extensively.
• The relationships between volume, pressure,
  temperature and moles are referred to as the gas
  laws.
• To understand gas stoichiometry, we must review a
  few concepts.

54
Units we will be using

• Volume liters, although other units could be
  used
• (The value of R would need to modified)
• Temperature Must use an absolute scale.
• K - Kelvin is most often used.
• Pressure Atm, although others could be used
• (The value of R would need to be modified)

55
Gas laws

• Laws that show the relationship between volume
  and various properties of gases
• Boyles Law
• Charles Law
• Gay-Lussacs Law
• Avogadros Law
• The Ideal Gas Equation combines several of these
  laws into a single relationship.

56
Boyles law

• The volume of a gas is inversely proportional to
  its pressure.
• PV k
• or
• P1 V1 P2 V2
• Temperature and number of moles must be held
  constant!

57
Boyles law
Increasing the pressure on a sample on
gas decreases it volume at constant
temperature. Note the effect here as weight is
added.
58
Charles law
The volume of a gas is directly proportional
to the absolute temperature (K).
V T
k
V1 V2 T1 T2
or

Pressure and number of moles must be held
constant!
59
Charles law
When you heat a sample of a gas, its volume
increases.
The pressure and number of moles must be held
constant.
60
Charles Law
Placing an air filled balloon near liquid
nitrogen (77 K) will cause the volume to be
reduced. Pressure and the number of moles are
constant.
61
Gay-Lussacs Law

• Law of of Combining Volumes.
• At constant temperature and pressure, the
  volumes of gases involved in a chemical reaction
  are in the ratios of small whole numbers.
• Studies by Joseph Gay-Lussac led to a better
  understanding of molecules and their reactions.

62
Gay-Lussacs Law

• Example.
• Reaction of hydrogen and oxygen gases.
• Two volumes of hydrogen will combine with one
  volume of oxygen to produce two volumes of
  water.
• We now know that the equation is
• 2 H2 (g) O2 (g) 2
  H2O (g)

63
Avogadros law

• Equal volumes of gas at the same temperature and
  pressure contain equal numbers of molecules.
• V k n
• V1 V2
• n1 n2

64
Avogadros law
If you have more moles of a gas, it takes up
more space at the same temperature and pressure.
65
Standard conditions (STP)

• Remember the following standard conditions.
• Standard temperature 273.15 K
• Standard pressure 1 atm
• At these conditions
• One mole of a gas has
• a volume of 22.4 liters.

66
The ideal gas law

• A combination of Boyles, Charles and Avogadros
  Laws
• PV nRT
• P pressure, atm
• V volume, L
• n moles
• T temperature, K
• R 0.082 06 L atm/K mol
• (gas law constant)

67
Ideal gas law

• When you only allow volume and one other factor
  to vary, you end up with one of the other gas
  laws.
• Just remember
• Boyle Pressure
• Charles Temperature
• Avogadro Moles

68
Ideal gas law
P2V2 n2T2
P1V1 n1T1
R
Any value that is the same on both sides (i.e.
held constant) can be cancelled, thus showing the
relationships described by the individual gas
laws.
69
Ideal gas law
Example - if n and T are held constant
P2V2 n2T2
P1V1 n1T1

This leaves us P1V1 P2V2
Boyles Law
70
Example
If a gas has a volume of 3.0 liters at 250
K, what volume will it have at 450 K ?
P1V1 n1T1
P2V2 n2T2

V1 T1
V2 T2

71
Daltons law of partial pressures

• The total pressure of a gaseous mixture is the
  sum of the partial pressure of all the gases.
• PT P1 P2 P3 .....
• Air is a mixture of gases - each adds it own
  pressure to the total.
• Pair PN2 PO2 PAr PCO2 PH2O

72
Partial pressure example

• Mixtures of helium and oxygen are used in scuba
  diving tanks to help prevent the bends.
• For a particular dive, 46 liters of O2 and 12
  liters of He were pumped in to a 5 liter tank.
  Both gases were added at 1.0 atm pressure at
  25oC.
• Determine the partial pressure for both gases in
  the scuba tank at 25oC.

73
Partial pressure example

• First calculate the number of moles of each gas
  using PV nRT.
• nO2 1.9 mol
• nHe 0.49 mol

74
Partial pressure example

• Now calculate the partial pressures of each.
• PO2 9.3 atm
• PO2
• 2.4 atm
• Total pressure in the tank is 11.7 atm.