Cpt 4. CHEMICAL STOICHIOMETRY

1
Cpt 4. CHEMICAL STOICHIOMETRY

• Definition Study of proportions of
• atoms inside substances
• substances in solutions or chemical reactions
• Objectives
• Calculate
• the composition formulas of chemical
  compounds
• the concentrations of solutions
• the quantitative outcome of chemical reactions

2
4.1. Mole and Molar mass

• Thought teaser
• Weight Species Number
• 1 g H 6.022E23 atoms
• 16 g O , , , ,
• 18 g H2O , , molecules
• 23 g Na() , , cations
• 35.5 g Cl(-) , , anions
• What trend is uncovered by the information
• from the table? See RQ2-19

3
RQ2-19

• What trend is uncovered by the information from
  the table on the previous slide?
• Any amount of substance with a density equal to
  its atomic or molecular density contains the same
  number of components 6.022E23
• Any amount of substance with a weight equal to
  its atomic or molecular weight contains the same
  number of components 6.022E23
• Substances with the same weight contain the same
  number of components 6.022E23

4
The Mole

• mole 6.022E23 Avogadro's number
• of components (atoms in an element,
• molecules in a compound) of a substance.
• Number of components of a substance in a weight
  in grams equal to the atomic or the molecular
  weight of the substance.
• Molar mass (MW) mass of a mole same magnitude
  but different units as the atomic or molecular
  mass.
• unit of molar mass g/mol
• Mass - Molar mass relation
• of moles (mass/molar mass)

5
RQ2-19B Previous Material Review

• On the chemical standpoint a mole is a a, rodent
  that lives in underground burrows b, measure of
  amount of substance c, measure of weight of
  substance
• The number of moles of a substance shows how much
  of a substance there is by showing the a, weight
  of substance b, volume of substance c, the
  number of components (atoms or molecules) of the
  substance.

6
WORD OF WISDOMLearning Steps

• Resolve, Dare to make the first step in the
  learning process REVIEW
• Reviewing leads to UNDERSTANDING
• Dont understand? Ask quickly for help to keep
  the momentum

7
WORD OF WISDOMBefore the Gain

• Unconfortable when studying? Expect the
  unconfort.
• Welcome the unconfort. It is beneficial to your
  learning
• Youre changing your thoughts
• From UNFOCUSED TO FOCUSED
• Say in your mind Im happy to do it. You feel
  a smile forming in your mind

8
WORD OF WISDOM

• Not enough time?
• MAKE the time!
• Sacrifice something.
• Your better future worth the sacrifice
• Tolerate Hardship on the way to Prosperity

9
WORD OF WISDOMHard Work

• Work is too hard?
• Keep trying. Itll get easier and easier
• Ask for help
• When the going gets tough, the tough gets going

10
Concept Illustration

• A 60 kg person has in average 36 kg of
• water. How many moles of water does
• that represent?
• mol wt / MW MW(H2O) 18.02 g/mol
• mol ?

11
Illustration 2

• Assume you have 0.123 mol of C14H18O4, benzoyl
  peroxide. What mass does that correspond to?
• Wt mol x MW
• MW (C14H18O4) (1412.01 18 416) g/mol ?
• Extra exercise 81, 82, pg 80

12
RQ2-20

• The adult human body has about 1014 cells. Does
  the body of an adult human have more or less than
  a mole of DNA?
• a. More, because the average adult has about 2000
  moles of water and DNA is larger than water.
• b. Less, because the human body can have no more
  than 1014 DNA molecules, which is less than a
  mole
• c. Less, because the average adult has about 2000
  moles of water and DNA is smaller than water.

13
4.2. Percent Composition

• Restriction wt/wt
• Definition by weight
• (wt of atom / wt of molecule) x 100
• (wt of component / wt of whole item)x100
• Law of Definite Composition the proportion of
  each element in a sample of pure compound is
  constant.

14
Illustration

• Calculate the weight of iron in Fe2O3. What
  mass of iron is in 25.0 g of Fe2O3?
• Extra Exercise Find the of C, H and O in
  glucose C6H12O6.
• Extra exercise 69,pg 126

15
Illustration (Continued)

• Information provided
• Formula Fe2O3
• Information requested of Fe in Fe2O3.
• Way to answer use MWs
• of Fe 100 x (2 x MW(Fe) / MW(Fe2O3)) ?
• Extra example 65, pg 126

16
Illustration (Continued 2)

• Information provided b
• Weight of Fe2O3 25. 0 g
• Information requested mass of Fe in the Fe2O3
  sample
• Way to answer
• Wt(Fe) x Wt(Fe2O3) / 100 ?
• Extra exercise 77, pg 127

17
RQ2-20B Previous Material Review

• The number of moles of a substance indicates the
  a, weight b, number of components c, density of
  the substance
• The percentage by weight of salt in a water
  solution indicates for example the number of
  grams of salt in a, 100g b, 100 mL c,10 g of
  solution

18
RQ2-20B2 Previous Material Review

• If you are given the mol of a substance for
  which you know the molar weight, you can find the
  weight of the substance by a, multiplying b,
  dividing c, adding mol and MW.
• If you are given the composition of solute in a
  solution for which you know the weight, you can
  find the weight of solute by using this
  expression
• a, wt(solute) / wt(solution) / 100
• b, wt(solute) x wt(solution) x 100
• c, wt(solute) x wt(solution) / 100

19
WANT
WISH
20
WORD OF WISDOMHard Work

• Work is too hard?
• Keep trying. Itll get easier and easier
• Ask for help
• When the going gets tough, the tough gets going

21
WORD OF WISDOMBefore the Gain

• Unconfortable when studying? Expect the
  unconfort.
• Welcome the unconfort. It is beneficial to your
  learning
• Youre changing your thoughts
• From UNFOCUSED TO FOCUSED
• Say in your mind Im happy to do it. You feel
  a smile forming in your mind

22
WORD OF WISDOM

• Not enough time?
• MAKE the time!
• Sacrifice something.
• Your better future worth the sacrifice
• Tolerate Hardship on the way to Prosperity

23
4.3. Derivation of Formulas

• Types of formulas
• Empirical formula shows the simplest ratios of
  atoms (groups of atoms) in a molecule
• Molecular Formula shows the actual numbers of
  atoms in a molecule
• a. Derivation of Empirical Formulas
• General procedure
• Weights of components ? Moles of components ? Raw
  Formula ? Intermediate formula ? Empirical formula

24
Empirical Formula Determination Procedure
Raw Formula
25
Detailed Procedure

• Raw formula obtained using (calculated) of
  moles for subscripts
• Intermediate formula obtained dividing all
  subscripts by the smallest subscript
• Empirical formula obtained after operation to
  round all fractional subscripts into the
  smallest whole numbers.

26
Special Fractional Subscripts

• Unrounded Subscripts that end with .33, .66 .25,
  .5 and .75
• Procedure
• Multiply the fractional subscript in the
  intermediate formula by the smallest number
  needed to produce the smallest whole number (or a
  number close to it).

27
Illustration

• Information provided
• composition of mandelic acid (an alpha hydroxy
  acid) C(63.15), H(5.30), and O(31.55)
• Information requested EF of mandelic acid.
• How do you proceed to solve the problem? See
  RQ2-20b

28
RQ2-20b

• How do you proceed to solve the problem in the
  previous question using the KNU method. Justify
  your answer.
• a. Find RF from IF, itself from EF, itself found
  using the mol of components of mandelic acid.
• b. Find EF from IF, itself from RF, itself found
  using the mol of components of mandelic acid.
• c. Find IF from EF, itself from RF, itself found
  using the mol of components of mandelic acid.

29
RQ2-20C Previous Material Review

• The empirical formula of a compound is determined
  first by using as subscripts the a, weight b,
  volume c, number of moles of the atoms that make
  up the compound
• If one of the subscripts in the intermediate
  formula ends with .67, the subscript should be
  a, rounded to the next whole number b,
  multiplied by 3 c, divided by 3 to get the
  empirical formula

30
Illustration Solution

• Empirical Formula lt- Intermediate Formula lt- Raw
  Formula CxHyOz
• x, y, z mol of C, H, and O
• mol wt / MW
• Wt unknown. Use values
• Wt of C ?
• MW 12.0 g/mol
• Wt of H ?
• MW 1.01 g/ mol
• Wt of O ?
• MW 16.0 g/mol

31
Illustration Solution (Continued)

• Plug info of Wt and MW in the expression of mol
• Plug info of mol in the expression of raw
  formula
• Use the raw formula to find the intermediate
  formula
• Use the intermediate formula to find the
  empirical formula
• Extra exercise 81, pg 127

32
Molecular Formulas

• Empirical Formula (EF)-gt Empirical weight (EW)
• Molecular Formula (MF)-gt Molecular weight (MW)
• MW/EW Molecular/Empirical Ratio (MER)
• MF/EF MER -gt MF EF x MER

33
Molecular Formulas (Example)

• Information provided
• MW of cacodyl 210 g/mol
• C 22.88
• H 5.76
• As 71.36
• Information requested Empirical and molecular
  formulas of cacodyl

34
Molecular Formulas (Example Continued)

• EF Intermediate formula lt- Raw formula CxHyAsz
• x, y, and z mol of C, H, and As
• mol wt / MW
• Find wt using s.
• What do you use the EF information for?
• See RQ2-20C

35
EF and MF of Cacodyl (Continued)

• EF -gt EW ?
• MW / EW MER ?
• MF EF x MER ?
• Extra exercise 117, pg 128

36
Molecular Formulas (Example 2)

• Info provided
• wt of S 1.256 g
• wt of SFx 5.722 g
• Info requested
• Value of x in SFx.

37
Molecular Formulas (Example 2 Solution)

• EF lt- Raw formula SmFn
• n, m mol of S and F
• mol wt / MW
• mol of S ?
• Find wt of F using wt of SFx wt of S
• mol of F ?
• Extra example 121, pg 128

38
EF/MF Example 3 Using Results of reactions120,
pg 128

• Information provided
• Weight of estrone (made of C, H, and O) 1.893
  g
• Weight of CO2 from combustion of estrone 5.545
  g
• Weight of H2O from combustion of estrone
  1.388 g
• MW of the estrone 270.36 g/mol
• Information requested MF of estrone

39
EF/MF Example 3 (Solution)

• MF EF x MER
• EF lt- IF lt- Raw formula, CxHyOz.
• x, y, and z mol of C, H and O.
• mol of C mol of CO2.
• mol of H 2 x mol of H2O
• mol of O wt/MW
• Wt of O total wt (sample) wt of C and H
• Continued on next slide

40
EF/MF Example 3 (Solution, Continued)

• Plug resulting weight into the expessions of mol
  of O
• Use mol of C, H, and O to get the RF.
• Use the RF to get the IF
• Use the IF to get the EF
• Use the EF to calculate the EW
• Use the EW and the MW to get the MER
• Use the MER and the EF to get the MF
• Extra exercise 119, pg 128.

41
RQ2-21

• What type of information about moles does the
  molecular formula provide?
• a. The actual number of moles of component atoms
  per mole of compound
• b. The simplest ratio of moles of component atoms
  per mole of compound
• c. It does not give any information about moles.
  Only the actual numbers of component atoms per
  molecule of compound

42
4.4. Solutions

• Solution homogeneous mixture of two or more
  substances.
• Solvent most abundant component.
• Solute least abundant
• Concentration amount of solute in a given amount
  of solvent
• Molarity (M) Measure of the concentration of a
  solution
• moles of solute / volume (liters) of solution.
  M mol/V
• Units mol/L
• Commercial unit molar (M)

43
Molarity (Illustration)

• Info provided
• Wt of K2Cr2O7 2.335 g
• Volume of final solution 500 mL 0.500 L
• Info requested
• Molar concentration of K2Cr2O7
• Molar concentrations of K() and Cr2O7(2-)
• Extra ex 53, pg 178

44
Dilution/Concentration of Solutions

• Dilution -gt decreased final molarity
• Concentration -gt increased final molarity.
• Total mol of solute constant same before and
  after dilution/concentration
• Initial mol(solute) Final mol (solute)
• Mi x Vi Mf x Vf
• I initial f final

45
Dilution/Concentration of Solutions (Illustration)

• Info provided
• Volume of original CuSO4 solution 4.00 mL
• Molarity of original CuSO4 solution 0.0250 M
• Final volume of CuSO4 solution 10.0 mL
• Info requested final molarity of CuSO4 solution.
• Extra ex 55, 178

46
RQ2-21B Previous Material Review

• If you divide the mol of solute by the volume
  (in Ls) of solution, you are determining the a
  density b, percent composition c, molarity of
  the solution
• If you want to determine the mol of solute in a
  solution, all you have to do is a, multiply b,
  divide c, add the molarity by the volume of the
  solution

47
RQ2-22

• Compare the stability of composition by weight
  and molarity in function of temperature
• a. by weight is stable and molarity is not.
  Unlike volumes, weights change with temperature
• b. molarity is stable and by weight is not.
  Unlike weights, volumes do not change with
  temperature
• c. by weight is stable and molarity is not.
  Unlike volumes, weights do not change with
  temperature

48
4.5. Calculations based on Equations and Reactions

• Study case C3H8 5O2 -gt 3CO2 4H2O
• Info from Equations -gt Identity theoretical
  proportions between reactants products of a
  reaction.
• Proportions -gt mole ratios in equations
• Info from Actual reactions amounts of reactants
  actually used / products actually formed
• Weights -gt of moles -gt mole ratios in actual
  reactions.
• General Principle
• mole ratios from actual reaction mole ratios
  from equation

49
a. Target 1 amounts of reactants or products

• Condition Reactants proportions are as shown by
  the mole ratios from equations
• General Principle
• mole ratios from equation mole ratios from
  actual reaction
• Mol ratio mol of target substance / mol of
  known substance
• mol of target substance mol ratio of
  (Target/Known) x mol of known substance
• How do you determine the weight of target
  substance? See RQ2-23

50
RQ2-23

• Use the expression of mol of target substance to
  determine the weight of target substance
• a. wt of target substance mol ratio of
  (Target/Known) x mol of known x MW of target
  substance
• b. wt of target substance mol ratio of
  (Known/Target) x mol of known x MW of known
  substance
• c. wt of target substance mol ratio of
  (Target/Known) x mol of known / MW of target
  substance

51
Finding amounts of reactants or products
(Illustration)

• Info provided
• CH4 2O2 -gt CO2 2H2O
• Wt of CH4 25.5 g
• Info requested wt of O2 needed to burn the CH4
  completely
• What relation do you use to solve the problem?
  See RQ2-24

52
RQ2-24

• What relation do you use to solve the problem in
  the previous question? Which parameter is known
  and which one should be determined next?
• a. Wt(O2) mol (O2) / MW. MW is known and
  mol(O2) should be determined next.
• b. Wt(O2) MW / mol (O2). mol(O2) is known and
  MW should be determined next.
• c. Wt(O2) mol (O2) x MW. MW is known and
  mol(O2) should be determined next.

53
Illustration (Solution)

• Wt(O2) mol (O2) x MW
• mol(O2) mole ratio of (O2/CH4) x mol(CH4)
• mol(CH4) ?
• Use mol(O2) to find wt(O2).
• Extra exercise 33, pg 177

54
Finding amounts of reactants or products
(Illustration 2)

• Info provided
• Volume of HNO3 solution 50.0 mL
• Molarity of the HNO3 solution 0.125 M
• Reaction equation
• Na2CO3 2HNO3 -gt 2NaNO3 CO2 H2O
• Info requested wt of Na2CO3 needed for the
  complete reaction.

55
Illustration 2 (Solution)

• Wt(Na2CO3) mol x MW (Na2CO3)
• mol (Na2CO3) mole ratio of (Na2CO3)/ HNO3) x
  mol(HNO3)
• What parameter is not known in the previous
  expression? How do you find it?
• See RQ2-25

56
RQ2-25

• What parameter is not known in the previous
  expression mol (Na2CO3) mole ratio of
  (Na2CO3)/ HNO3) x mol(HNO3) ? How do you find
  it?
• a. Unknown mol(HNO3), found using the relation
  mol wt/MW
• b. Unknown mol(HNO3), found using the relation
  mol M x V
• c. Unknown mole ratio, found using the relation
  wt mol x MW

57
Illustration 2 (Solution, Continued)

• mol(HNO3) M x V ?
• Use mol(Na2CO3) to find wt(Na2CO3).
• Extra exercise 59, pg 179

58
RQ2-25B Previous Material Review

• If you know the mol of reagent A in the reaction
  A 3B -gt AB3, you can find the mol of B using
  the a, weight b, mole c, volume ratio of B to
  A.
• In order to find the weight of B in the previous
  reaction, the mole ratio of B to A is multiplied
  by the mol of A and the molar weight of a, B
  b, AB3 c, A

59
b. Target 2 Limiting Reagent (LR)

• LR reactant that is used in less than enough
  amount to complete the reaction with the other
  reactant.
• LR limits the reaction. It stops when LR is used
  up.
• General procedure
• Find mol of reactants
• Find the simplest reactant/reactant mole ratio
• from actual reaction in decimal form
• Compare reactant/reactant mole ratios from
• actual reaction and from equation (also in
  decimal form) the insufficient reactant
  limiting reagent

60
Illustration

• Info provided
• Reaction equation 2Al 3Cl2 -gt 2AlCl3
• Wt of Al 2.70 g
• Wt of Cl2 4.05 g
• Info requested limiting reactant ?

61
Way to the Solution

• LR insufficient reactant through comparison of
  reactant/reactant mole ratios from equation and
  actual reaction.
• Al/Cl2 mole ratio from equation 2/3
• Al/Cl2 mole ratio from actual reaction ?
• mol (Al) ?
• mol (Cl2) ?

62
Way to Solution (Continued)

• Mole ratio of Al/Cl2 from actual reaction ?
• Put the ratio in same form as 2/3 0.67 from the
  equation. Divide numerator and denominator by
  same number that produces 2 at the numerator (for
  example).
• Compare the two mole ratios. The insufficient
  reactant LR ?
• Extra ex 37, pg 177

63
Limiting Reagent and Co-reactant leftover

• What amount of unreacted other reactant is left?

64
c. Target 3 Reaction Yield

• Yield Quantitative result of an actual reaction
• 1. Theoretical yield (TY)
• TY weight of product expected from a reaction
  that runs completely and perfectly
• TY mol x MW of target product(TP).
• mol(TP) Mol ratio of (TP / LR) x Mol of LR

65
Reaction Yield (Continued)

• 2. Actual yield (AY)
• AY Weight of material actually collected from
  the reaction
• 3. () Yield
• () Yield (AY / TY) x 100

66
Illustration

• Under certain conditions the formation of ammonia
  from nitrogen and hydrogen has a 7.82 yield.
  Under these conditions, how many grams of NH3
  will be produced from the reaction 25.0 g N2 with
  2.00 g H2?
• The reaction occurs according to the following
  equation
• N2(g) 3 H2(g) -gt 2 NH3(g)

67
Way to the Solution

• Info provided
• Reaction N2 3H2 -gt 2NH3
• Wt of N2 25.0 g
• Wt of H2 2.0 g
• yield of the reaction 7.82
• Info requested
• Wt of NH3 produced (AY)

68
Way to Solution (2)

• Expression of requested wt of NH3 is AY Y x TY
  / 100
• Expression of TY mole ratio of (NH3/LR) x mol
  of LR x MW of NH3
• Limiting reagent insufficient reactant between
  N2 and H2.
• How do you determine the limiting reagent?
• See RQ2-26

69
RQ2-26

• How do you determine the limiting reagent?
• a. Compare mole ratios from actual reaction to
  those from equation
• b. Compare moles from actual reaction to those
  from equation
• c. Compare weight ratios from actual reaction to
  those from equation

70
Way to Solution (Continued)

• Compare mole ratios from actual reaction to those
  from equation
• Mole ratio from equation ?
• Mole ratio from actual reaction mol of H2 used
  / mol of N2 used
• mol of N2 used ?
• mol of H2 used ?

71
Way to Solution (4)

• TY (mol ratio of NH3/H2) x mol of H2 x MW of
  NH3 (2/3 x 0.990 mol) x 17.0 g/mol 11.22 g
• Requested wt of NH3 is AY ( x TY)/ 100 7.82
  x 11.22 g / 100 0.877 g
• Extra Exercise 45, pg 179

72
Calculations on Reactions (Additional
Illustration)

• A mass of 2.052 g of a metal carbonate, MCO3, is
  heated to give the metal oxide and 0.4576 g CO2.
• MCO3(s) -gt MO(s) CO2(g)
• What is the identity of the metal?

73
Empirical Formula (Additional Illustration)

• A 4.236 g sample of a hydrocarbon is combusted to
  give 3.810 g of H2O and 13.96 g of CO2. What is
  the empirical formula of the compound?

74
Calculations on Reactions (Additional
Illustration 2)

• A 1 g of chewable vitamin C requires 27.85 mL of
  0.102 M Br2 solution for titration to the
  equivalence point. What is the mass of vitamin C
  in the tablet?

75
Calculations on Reactions (Additional
Illustration 2b)

• Info provided
• Reaction of Vit C
• Wt of Vit C Tablet
• Volume of Br2 solution
• Molarity of , , , ,
• Info requested wt of Vit C in tablet

76
Empirical Formulas (Example 2)

• A compound contains only C, H, and N. Combustion
  of 35.0 mg of the compound produces 33.5 mg of
  CO2 and 41.1 mg of H2O. What is the empirical
  formula of the compound?

77
Empirical Formula Problem2 (Analysis)

• Information provided
• Weight of the compound 35.0 mg
• Weight of CO2 33.5 mg
• Weight of H2O 41.1 mg
• Information requested
• Empirical formula of the compound

78
Empirical Formula Problem 2 (Solution)

• Determination of EF requires raw formula CxHyNz.
  x, y, z mol of C, H, and N.
• mol of C same as mol of CO2
• mol of H 2 x mol of H2O
• mol of N wt/MW. Wt of N Total wt wt of C
  and N.