1. Stoichiometry
Stoichiometry: study of the quantitative relations between amounts of reactants and products.
Goals:
Perform stoichiometry calculations.
Understand the meaning of limiting reactant.
Calculate theoretical and percent yields of a chemical reaction.
Define and use molarity in solution stoichiometry.
Use stoichiometry to analyze a mixture of compounds or to determine the formula of a compound.

2. What is STOICHIOMETRY?
The study of the quantitative aspects of chemical reactions.
It rests on the principle of the conservation of matter.
2 Al(s) Br2(liq) > Al2Br6(s)
You must always begin with a balanced equation before carrying out a stoichiometry calculation.

3. Write a Chemical Equation
Phosphine, PH3 (g), combusts in oxygen gas to form gaseous water and solid tetraphosphorus decoxide.
______
+ _____
_____+ _______
Always check (and REcheck) the balancing

4. Information from a Balanced Equation
Equation: 2 H2 (g) + O2 (g) H2O (l)
Molecules: 2 molecules H2 + 1 molecule O2 2 molecules H2O
Mass (amu): amu H amu O amu H2O
Amount (mol): 2 mol H mol O mol H2O
Mass (g): g H g O g H2O

5. General Plan for Stoichiometry Calculations
Students should become familiar with stoichiometry calculations.

6. Mole relationships in Chemical Equations
Stoichiometric factor – relates the amounts of any two substances involved in a chemical reaction, on a mole basis.
C3H O CO H2O
propane

7. If 454 g of NH4NO3 decomposes, how much N2O and H2O are formed
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 ---> N2O H2O
STEP 2 Convert mass of reactant to moles of reactant. (454 g) --> moles
STEP 3 Convert moles reactant (5.68 mol) --> moles product. A) Relate moles using coefficients, write a STOICHIOMETRIC FACTOR.

8. NH4NO3 ---> N2O + 2 H2O
STEP 3 Convert moles reactant (5.68 mol) --> moles product. B) Multiply moles of reactant by the stoichiometric factor.
STEP 4 Convert moles product --> mass product = THEORETICAL YIELD

9. NH4NO3 ---> N2O + 2 H2O Total mass of reactants =
STEP 5 How much N2O is formed?
Total mass of reactants =
total mass of products

10. Students should become familiar with % yield calculations.
NH4NO3 ---> 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 and actual (131 g) yields.
Actual yield
% yield =
* 100
Theoretical yield
Students should become familiar with % yield calculations.

11. What is 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.

12. Which is the Limiting Reactant?
2 NO (g) + O2 (g) > 2 NO2 (g)
Reactants
Products
Limiting reactant is
Excess reactant is

13. Limiting Reactant Zn + 2 HCl ---> ZnCl2 + H2 1 2 3
React solid Zn with mol HCl (aq)
Zn + 2 HCl ---> ZnCl2 + H2
(See CD Screen 4.8) 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

14. Zn + 2 HCl ---> ZnCl2 + H2 1 2 3 0.10 mol of HCl, need ? mol Zn.
0.10 mol HCl 1 mol Zn
2 mol HCl
= mol Zn 1 2 3
Rxn 1 Rxn 2 Rxn 3
mass Zn (g)
mol Zn
mol HCl
mol HCl/mol Zn 0.93/1 2.00/1 5.00/1
Lim Reactant LR = HCl no LR LR = Zn

15. Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
2 Al Cl2 ---> Al2Cl6
Mass
product
Mass
reactant
Stoichiometric
factor
Moles
reactant
Moles
product

16. Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
2 Al Cl2 ---> Al2Cl6
STEP 1 FIND THE LIMITING REAGENT. Compare actual mole ratio of reactants to theoretical mole ratio.
Reactants must be in the mole ratio:
3 mol Cl2
If ratio > 3/2, not enough Al = LR
If ratio < 3/2, not enough Cl2 = LR
2 mol Al
Students should become familiar with calculations using the concept of limiting reagent.

17. Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
Calculate moles of each reactant.
We have 5.40 g of Al and 8.10 g of Cl2
Find the mole ratio of reactants:

18. Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
2 Al Cl2 ---> Al2Cl6
Limiting reactant = Cl2
BASE ALL CALCULATIONS on LR: Cl2
mass
Cl2
mass
Al2Cl6
moles
Cl2
moles
Al2Cl6

19. Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?
2 Al Cl2 ---> Al2Cl6
STEP 2 CALCULATE THE MASS OF THE PRODUCT.
Calculate moles of Al2Cl6 expected based on LR.

20. Mix 5.40 g of Al with 8.10 g of Cl2. How much of each reactant will remain when reaction is complete?
2 Al Cl2 ---> Al2Cl6
Cl2 was the limiting reactant. Therefore, Al was present in excess. But how much? First find how much Al was required (used). Then find how much Al is in excess.

21. Chemical Analysis Na2SO4(aq) + BaCl2(aq) --> NaCl(aq) + BaSO4(s) 2
An impure sample of the mineral thenardite contains Na2SO4. A mass of mineral sample weights g. The Na2SO4 in the sample is converted to insoluble BaSO4 by adding BaCl2. The recovered mass of BaSO4 is g. What is the mass percent of Na2SO4 in the mineral?
Na2SO4(aq) + BaCl2(aq) > NaCl(aq) + BaSO4(s) ? 2
0.177g

22. Chemical Analysis

23. General Plan for Stoichiometry Calculations

24. Chemical Analysis
Na2SO4(aq) + BaCl2(aq) > 2 NaCl(aq) + BaSO4(s)

25. Combustion Analysis of Hydrocarbons
Active Figure 4.9

26. Procedure for Calculating Empirical Formula
Grams
of each
element
Use
Molar
mass
Moles
of each
element
Calculate
mole ratio
Empirical
Formula
The central part of the calculation is determining the number of moles of each element in the compound.
Remember: in the mole ratio, divide by smaller number, then multiply ‘til whole.

27. CxHy + some oxygen ---> 0.379 g CO2 + 0.1035 g H2O
What is the empirical formula of a hydrocarbon, CxHy, if g burn and produce g of CO2 and g of H2O.
CxHy + some oxygen ---> g CO g H2O
First, recognize that all C in CO2 and all H in H2O is from CxHy.
1. Calculate amount (in moles) of C in CO2
2. Calculate amount (in moles) of H in H2O
3. Find ratio of mol H/mol C to find values of x and y in CxHy.

28. What is the empirical formula of a hydrocarbon, CxHy, if 0
What is the empirical formula of a hydrocarbon, CxHy, if g burn and produce g of CO2 and g of H2O.
1. Calculate amount (moles) of C in CO2
2. Calculate amount (moles) of H in H2O
3. Ratio of mol H/mol C to find values of x and y in CxHy.

29. Practice Calculate amount of Ti:
4.67 Titanium (IV) oxide, TiO2, is heated in hydrogen gas to give water and a new titanium oxide, TixOy. If g of TiO2 produces g of TixOy, what is the formula of the new oxide?
Calculate amount of Ti:

30. Practice… TiO2 + H2 H2O + TixOy 2. Calculate amount of O:
3. Calculate the molar ratio of O to T

31. How are Reactions in Solution Quantified?
In solution we need to define the -
SOLVENT
the component whose physical state is preserved when solution forms
SOLUTE
the other solution component

32. Students should become familiar with calculations using MOLARITY.
What is Molarity?
The amount of solute in a solution is given by its concentration.
Moles of solute
Molarity =
Liters of solution
Students should become familiar with calculations using MOLARITY.

33. Calculate molarity of a solution of 5
Calculate molarity of a solution of 5.00 g of NiCl2•6 H2O dissolved in enough water to make 250 mL of solution.
M =
mol L
STEP 1 Calculate the number of moles of solute
STEP 2 Calculate the molarity of the solution
M =
mol L

34. How many IONS are in the Solution?
NiCl2(aq) --> Ni2+(aq) + 2 Cl-(aq)

35. What mass of oxalic acid, H2C2O4, is required to make 250 mL of a 0
What mass of oxalic acid, H2C2O4, is required to make 250 mL of a M solution?
moles
Volume (L)
M =
moles = M•V
STEP 1 Calculate moles of acid required.
STEP 2 Calculate mass of acid required.

36. 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.

37. Preparing Solutions

38. Notice that the amount of NaOH (moles of NaOH) present did not change.
You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
M = moles/L
Notice that the amount of NaOH (moles of NaOH) present did not change.

39. moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution
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?
Amount of NaOH in original sol. = M • V =
Amount of NaOH in final sol. must also =
Volume of final solution =
moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution

40. You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
Conclusion:
add _______of water to the 50.0 mL of 3.0 M NaOH to make _________of 0.50 M NaOH.
A shortcut:
Cinitial • Vinitial = Cfinal • Vfinal
Principle of dilution: addition of solvent does not change the amount of solute in a solution but does change its concentration.

41. Solution Stoichiometry
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?

42. General Plan for Solution Stoichiometry
M = moles / volume
Moles = M * volume

43. 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?
Mass
zinc
Moles
zinc
Stoichiometric
factor
Moles
HCl
Volume
HCl

44. 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
Step 2: Calculate amount of Zn
Step 3: Use the stoichiometric factor

45. 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 4: Calculate volume of HCl req’d

46. What is pH? It is a concentration scale.
pH: a way to express acidity – the concentration of ____in solution.
Low pH: high [H+]
High pH: low [H+]
Acidic solution pH < 7
Neutral pH = 7
Basic solution pH > 7

47. pH = log (1/ [H+]) = - log [H+] The pH Scale In a neutral solution,
[H+] = [OH-] = 1.00 x 10-7 M at 25 oC
pH = - log [H+] = -log (1.00 x 10-7) = - [0 + (-7)]
= 7
See CD Screen 5.17 for a tutorial
See book Appendix A.3 for more on logs

48. pH and [H+]
If the [H+] of soda is 1.6 x 10-3 M, the pH is ____?

49. pH and [H+]
If the pH of Coke is 3.12, what is the [H+]?

50. What is a Titration?
Titration – procedure in which two reactants in solution react in the precise proportions shown by the chemical equation for the reaction.
Buret – a calibrated instrument used in a titration. It is a graduated, long glass tube calibrated to deliver precise volumes of solution through a stopcock valve.
Equivalence point – the point in a titration at which one reactant has been exactly consumed by addition of the other reactant.

51. What is a Titration?
Phenolphthalein indicator

52. Acid-Base Titration
How do we measure the concentration of an acid in a solution?
A measured volume of a solution of an acid of unknown concentration is transferred to a flask.
A solution of a base of known concentration is added carefully from a buret until the reaction of the acid with the base is just complete.
Equivalence point of the titration – the point at which the acid is just neutralized. At that point, the number of moles of OH- added equals the number of moles of H+ that were in the sample of acid.
The equivalence point is determined with an indicator dye – a substance that changes color as the reaction is completed (litmus, phenolphthalein).

53. 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 (M) of the NaOH?
Step 1: Write a balanced chemical equation:
Step 2: Calculate amount (moles) of H2C2O4
Step 3: Calculate amount (moles) of NaOH req’d: use stoichiometric factor.

54. 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 (M) of the NaOH?
M = moles/L

55. Practice
What volume of M FeCl3 contains 12.5 g FeCl3?

56. __H2C4H4O5(aq) + __NaOH(aq) ---> __Na2C4H4O5(aq) + __H2O(l)
Apples contain malic acid, H2C4H4O g of apple requires mL of M NaOH for titration. What is weight % of malic acid?
__H2C4H4O5(aq) + __NaOH(aq) ---> __Na2C4H4O5(aq) + __H2O(l)

57. 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: Write a balanced chemical equation:
Step 2: Calculate amount (moles) of NaOH
Step 3: Calculate amount (moles) of malic acid titrated: use stoichiometric factor.

58. 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 4: Calculate the mass of H2C2O4 titrated
Step 5: Calculate the weight % H2C2O4

59. Remember Go over all the contents of your textbook.
Practice with examples and with problems at the end of the chapter.
Practice with OWL tutor.
Work on your assignments for Chapters 4, 5.
Practice with the quiz on CD of Chemistry Now.