1. Chapter 4 Chemical Equations and Stoichiometry

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3. STOICHIOMETRY
- the study of the quantitative aspects of chemical reactions.

4. STOICHIOMETRY It rests on the principle of the conservation of matter.
2 Al(s) Br2(liq) f Al2Br6(s)

5. Write the balanced chemical equation NH4NO3(s) f N2O(g) + 2 H2O(g)
PROBLEM: If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed? What is the theoretical yield of products?
STEP 1
Write the balanced chemical equation
NH4NO3(s) f N2O(g) H2O(g)

6. 454 g of NH4NO3 f N2O + 2 H2O
STEP 2 Convert mass of reactant (454 g) to amount (mol)
STEP 3 Convert amount of reactant (5.68 mol) to amount (mol) of product.

7. 454 g of NH4NO3 f N2O + 2 H2O
STEP 3 Convert moles reactant f moles product
Relate moles NH4NO3 to moles product expected.
1 mol NH4NO3 f 2 mol H2O
Express this relation as the STOICHIOMETRIC FACTOR

8. = 11.3 mol H2O produced 454 g of NH4NO3 f N2O + 2 H2O
STEP 3 Convert moles reactant (5.67 mol) to amount (mol) of product
= mol H2O produced

9. ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!
454 g of NH4NO3 f N2O + 2 H2O
STEP 4 Convert amount of product (11.4 mol) to mass of product
Called the THEORETICAL YIELD
ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

10. GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS
Mass
product
Mass
reactant
Stoichiometric
factor
Moles
reactant
Moles
product

11. 454 g of NH4NO3 f N2O + 2 H2O
STEP 5 What mass of N2O is formed?
Total mass of reactants =
total mass of products
454 g NH4NO3 = ___ g N2O g H2O
mass of N2O = 250. g

12. 454 g of NH4NO3 f N2O + 2 H2O Amounts Table (from page 159)
Compound NH4NO3 N2O H2O
Initial (g)
Initial (mol)
Change (mol)
Final (mol)
Final (g)

13. 454 g of NH4NO3 f N2O + 2 H2O Compound NH4NO3 N2O H2O
Amounts Table (from page 159)
Compound NH4NO3 N2O H2O
Initial (g) 454 g 0 0
Initial (mol) 5.67 mol 0 0
Change (mol) (5.67)
Final (mol)
Final (g)
Note that matter is conserved!

14. STEP 6 Calculate the percent yield
454 g of NH4NO3 f N2O + 2 H2O
STEP 6 Calculate the percent yield
If you isolated only 131 g of N2O, what is the percent yield?
This compares the theoretical (250. g) and actual (131 g) yields.

15. 454 g of NH4NO3 f N2O + 2 H2O
STEP 6 Calculate the percent yield

16. PROBLEM: Using 5.00 g of H2O2, what mass of O2 and of H2O can be obtained?
2 H2O2(liq) f 2 H2O(g) + O2(g)
Reaction is catalyzed by MnO2
Step 1: amount (mol) of H2O2
Step 2: use STOICHIOMETRIC FACTOR to calculate amount (mol) of O2
Step 3: mass of O2

17. Reactions Involving a LIMITING REACTANT
In a given reaction, there is not enough of one reagent to use up the other reagent completely.
The reagent in short supply LIMITS the quantity of product that can be formed.

18. LIMITING REACTANTS 2 NO(g) + O2(g) f 2 NO2(g)
Products
2 NO(g) + O2(g) f 2 NO2(g)
Limiting reactant = ___________
Excess reactant = ____________

19. LIMITING REACTANTS
PLAY MOVIE
See demonstration of limiting reactants in Chemistry Now

20. LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq)
Zn + 2 HCl f ZnCl2 + H2
PLAY MOVIE 1 2 3
Rxn 1: Balloon inflates fully, some Zn left
* More than enough Zn to use up the mol HCl
Rxn 2: Balloon inflates fully, no Zn left
* Right amount of each (HCl and Zn)
Rxn 3: Balloon does not inflate fully, no Zn left.
* Not enough Zn to use up mol HCl

21. LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq)
Zn + 2 HCl f ZnCl2 + H2
0.10 mol HCl [1 mol Zn/2 mol HCl]
= mol Zn
Rxn 1 Rxn 2 Rxn 3
mass Zn (g)
mol Zn
mol HCl
mol HCl/mol Zn 0.93/ / /1
Lim Reactant LR = HCl no LR LR = Zn

22. Reaction to be Studied
2 Al(s) Cl2(g) f Al2Cl6(s)

23. PROBLEM: Mix 5. 40 g of Al with 8. 10 g of Cl2
PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
Mass
product
Mass
reactant
Stoichiometric
factor
Moles
reactant
Moles
product

24. Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

25. Reactants must be in the mole ratio
Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.
2 Al(s) Cl2(g) f Al2Cl6(s)
Reactants must be in the mole ratio

26. Deciding on the Limiting Reactant
2 Al(s) Cl2(g) f Al2Cl6(s) If
There is not enough Al to use up all the Cl2
Limiting reagent = Al

27. Deciding on the Limiting Reactant
2 Al(s) Cl2(g) f Al2Cl6(s) If
There is not enough Cl2 to use up all the Al
Limiting reagent = Cl2

28. Step 2 of LR problem: Calculate moles of each reactant
We have 5.40 g of Al and 8.10 g of Cl2

29. Find mole ratio of reactants
2 Al(s) Cl2(g) f Al2Cl6(s)
Ratio should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio.
Limiting reactant is Cl2

30. Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
2 Al(s) Cl2(g) f Al2Cl6(s)
Limiting reactant = Cl2
Base all calcs. on Cl2
mass
Cl2
mass
Al2Cl6
moles
Cl2
moles
Al2Cl6

31. CALCULATIONS: calculate mass of Al2Cl6 expected.
Step 1: Calculate moles of Al2Cl6 expected based on LR.
Step 2: Calculate mass of Al2Cl6 expected based on LR.

32. How much of which reactant will remain when reaction is complete?
Cl2 was the limiting reactant. Therefore, Al was present in excess. But how much?
First find how much Al was required.
Then find how much Al is in excess.

33. Calculating Excess Al 0.114 mol = LR
2 Al + 3 Cl2
products
0.114 mol = LR
0.200 mol
Excess Al = Al available - Al required
= mol mol
= mol Al in excess

34. Chemical Analysis
See Active Figure 4.6

35. Chemical Analysis
An impure sample of the mineral thenardite contains Na2SO4.
Mass of mineral sample = g
The Na2SO4 in the sample is converted to insoluble BaSO4.
The mass of BaSO4 is g
What is the mass percent of Na2SO4 in the mineral?

36. Chemical Analysis Na2SO4(aq) + BaCl2(aq) f 2 NaCl(aq) + BaSO4(s)
0.177 g BaSO4 (1 mol/233.4 g) = 7.58 x 10-4 mol BaSO4
7.58 x 10-4 mol BaSO4 (1 mol Na2SO4/1 mol BaSO4) = 7.58 x 10-4 mol Na2SO4
7.58 x 10-4 mol Na2SO4 (142.0 g/1 mol) = g Na2SO4
(0.108 g Na2SO4/0.123 g sample)100% = 87.6% Na2SO4

37. Determining the Formula of a Hydrocarbon by Combustion
See Active Figure 4.7
PLAY MOVIE

38. Using Stoichiometry to Determine a Formula
Burn g of a hydrocarbon, CxHy, and produce g of CO2 and g of H2O.
CxHy + some oxygen f g CO g H2O
What is the empirical formula of CxHy?

39. Using Stoichiometry to Determine a Formula
CxHy + some oxygen f g CO g H2O
First, recognize that all C in CO2 and all H in H2O is from CxHy.
0.379 g CO2
+O2
Puddle of CxHy
0.115 g
1 CO2 molecule forms for each C atom in CxHy
g H2O
1 H2O molecule forms for each 2 H atoms in CxHy

40. Using Stoichiometry to Determine a Formula
CxHy + some oxygen f g CO g H2O
First, recognize that all C in CO2 and all H in H2O is from CxHy.
1. Calculate amount of C in CO2
8.61 x 10-3 mol CO2 f x 10-3 mol C
2. Calculate amount of H in H2O
5.744 x 10-3 mol H2O f x 10-2 mol H

41. Using Stoichiometry to Determine a Formula
CxHy + some oxygen f g CO g H2O
Now find ratio of mol H/mol C to find values of x and y in CxHy.
1.149 x mol H/ 8.61 x 10-3 mol C
= mol H / 1.00 mol C
Multiply by 3 to get whole number coefficients.
Therefore, we have 4 mol H / 3 mol C
Empirical formula = C3H4

42. Quantitative Aspects of Reactions in Solution Sections 4.5-4.7
PLAY MOVIE

43. Terminology In solution we need to define the SOLVENT
the component whose physical state is preserved when solution forms
SOLUTE
the other solution component

44. Concentration of Solute
The amount of solute in a solution is given by its concentration.
Concentration (M) = [ …]

45. 1. 00 L of water was used to make 1. 00 L of solution
1.00 L of water was used to make 1.00 L of solution. Notice the water left over.
CCR, page 174

46. Preparing a Solution
See Active Figure 4.9

47. PROBLEM: Dissolve 5. 00 g of NiCl2•6 H2O in enough water to make 250
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250. mL of solution. Calculate molarity.
Step 1: Calculate moles of NiCl2• 6H2O
Step 2: Calculate molarity
[NiCl2·6 H2O] = M
PLAY MOVIE

48. The Nature of a CuCl2 Solution: Ion Concentrations
CuCl2(aq) f
Cu2+(aq) + 2 Cl-(aq)
If [CuCl2] = 0.30 M, then
[Cu2+] = 0.30 M
[Cl-] = 2 x 0.30 M

49. moles = M•V USING MOLARITY
What mass of oxalic acid, H2C2O4, is required to make 250. mL of a M solution?
Because
Conc (M) = moles/volume = mol/V
this means that
moles = M•V

50. USING MOLARITY moles = M•V
What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a M
solution?
moles = M•V
Step 1: Calculate amount (mol) of acid required.
( mol/L)(0.250 L) = mol
Step 2: Calculate mass of acid required.
( mol )(90.00 g/mol) = g

51. Preparing Solutions
Weigh out a solid solute and dissolve in a given quantity of solvent.
Dilute a concentrated solution to give one that is less concentrated.

52. Preparing a Solution by Dilution
See Figure 4.10

53. PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
Add water to the 3.0 M solution to lower its concentration to 0.50 M
Dilute the solution!

54. PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
But how much water
do we add?

55. The important point is that
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
How much water is added?
The important point is that
moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution

56. M • V = Amount of NaOH in original solution =
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution =
M • V =
(3.0 mol/L)(0.050 L) = mol NaOH
Amount of NaOH in final solution must also = 0.15 mol NaOH
Volume of final solution =
(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L
or mL

57. PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
Conclusion:
add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

58. Preparing Solutions by Dilution
A shortcut
Cinitial • Vinitial = Cfinal • Vfinal

59. pH, a Concentration Scale
pH: a way to express acidity -- the concentration of H3O+ in solution.
Low pH: high [H3O+]
High pH: low [H3O+]
Acidic solution pH < 7
Neutral pH = 7
Basic solution pH > 7

60. The pH Scale pH = log (1/ [H3O+]) = - log [H3O+]
In a neutral solution,
[H3O+] = [OH-] = 1.00 x 10-7 M at 25 oC
pH = - log [H3O+] = -log (1.00 x 10-7) = - [0 + (-7)] =
See book Appendix A.3 for more on logs
See GO CHEMISTRY module on pH

61. [H3O+] and pH If the [H3O+] of soda is 1.6 x 10-3 M, the pH is ____
Because pH = - log [H3O+]
then
pH= - log (1.6 x 10-3)
pH = -{log (1.6) + log (10-3)}
pH = -{ )
pH = 2.80

62. pH and [H3O+] If the pH of Coke is 3.12, it is __________.
Because pH = - log [H3O+] then
log [H3O+] = - pH
Take antilog and get
[H3O+] = 10-pH
[H3O+] = = x 10-4 M

63. SOLUTION STOICHIOMETRY Section 4.7
Zinc reacts with acids to produce H2 gas.
Have 10.0 g of Zn
What volume of 2.50 M HCl is needed to convert the Zn completely?

64. GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS
Mass
HCl
Mass
zinc
Moles
zinc
Stoichiometric
factor
Moles
HCl
Volume
HCl

65. Zinc reacts with acids to produce H2 gas. If you have 10
Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?
Step 1: Write the balanced equation
Zn(s) HCl(aq) f ZnCl2(aq) + H2(g)
Step 2: Calculate amount of Zn
Step 3: Use the stoichiometric factor

66. Zinc reacts with acids to produce H2 gas. If you have 10
Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?
Step 3: Use the stoichiometric factor
Step 4: Calculate volume of HCl req’d

67. ACID-BASE REACTIONS Titrations
H2C2O4(aq) NaOH(aq) f
acid base
Na2C2O4(aq) H2O(liq)
Carry out this reaction using a TITRATION.
Oxalic acid,
H2C2O4

68. Setup for titrating an acid with a base
See Active Figure 4.14

69. Titration 1. Add solution from the buret.
2. Reagent (base) reacts with compound (acid) in solution in the flask.
3. Indicator shows when exact stoichiometric reaction has occurred.
4. Net ionic equation
H3O+(aq) OH-(aq)
f 2 H2O(liq)
5. At equivalence point
moles H3O+ = moles OH-
PLAY MOVIE

70. LAB PROBLEM #1: Standardize a solution of NaOH — i. e
LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.
1.065 g of H2C2O4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

71. 1. 065 g of H2C2O4 (oxalic acid) requires 35
1.065 g of H2C2O4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?
Step 1: Calculate amount of H2C2O4
Step 2: Calculate amount of NaOH req’d

72. 1. 065 g of H2C2O4 (oxalic acid) requires 35
1.065 g of H2C2O4 (oxalic acid) requires mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?
Step 1: Calculate amount of H2C2O4
= mol acid
Step 2: Calculate amount of NaOH req’d
= mol NaOH
Step 3: Calculate concentration of NaOH
[NaOH] = M

73. LAB PROBLEM #2: Use standardized NaOH to determine the amount of an acid in an unknown.
Apples contain malic acid, C4H6O5.
C4H6O5(aq) NaOH(aq) f Na2C4H4O5(aq) H2O(liq)
76.80 g of apple requires mL of M NaOH for titration. What is weight % of malic acid?

74. 76. 80 g of apple requires 34. 56 mL of 0. 664 M NaOH for titration
76.80 g of apple requires mL of M NaOH for titration. What is weight % of malic acid?
Step 1: Calculate amount of NaOH used.
C • V = (0.664 M)( L)
= mol NaOH
Step 2: Calculate amount of acid titrated.
= mol acid

75. 76. 80 g of apple requires 34. 56 mL of 0. 664 M NaOH for titration
76.80 g of apple requires mL of M NaOH for titration. What is weight % of malic acid?
Step 1: Calculate amount of NaOH used.
= mol NaOH
Step 2: Calculate amount of acid titrated
= mol acid
Step 3: Calculate mass of acid titrated.

76. 76. 80 g of apple requires 34. 56 mL of 0. 663 M NaOH for titration
76.80 g of apple requires mL of M NaOH for titration. What is weight % of malic acid?
Step 1: Calculate amount of NaOH used.
= mol NaOH
Step 2: Calculate amount of acid titrated
= mol acid
Step 3: Calculate mass of acid titrated.
= g acid
Step 4: Calculate % malic acid.

77. Spectrophotometry

78. An Absorption Spectrophotometer
See Figure 4.16

79. Spectrophotometry
Amount of light absorbed by a sample depends on path length and solute concentration.
Same concs but different path lengths
Different concs of Cu2+
Concentration
Path length

80. Spectrophotometry
BEER-LAMBERT LAW relates amount of light absorbed and the path length and solute concentration.
A = absorbance l = path length
 = molar absorptivity c = concentration
There is a linear relation between A and c for a given path length and compound.
This means you can find unknown solution concentration if A is measured.

81. Spectrophotometry
To use the Beer-Lambert law you must first calibrate the instrument.
The calibration plot can be used to find the unknown conc of a solution from a measured A.