1. Chapter 3.6-3.10 Student Notes
Stoichiometry

2. 3.5 Percent Composition of Compounds
Chapter 3
Table of Contents
3.1 Counting by Weighing
3.2 Atomic Masses
3.3 The Mole
3.4 Molar Mass
3.5 Percent Composition of Compounds
SEE PREVIOUS LESSONS ON OTHER SLIDE SHOW
3.6 Determining the Formula of a Compound
3.7 Chemical Equations
3.8 Balancing Chemical Equations
3.9 Stoichiometric Calculations: Amounts of Reactants and Products
3.10 The Concept of Limiting Reagent
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3. Pick up calculator, notes packet, and pre-lab packet.
Chapter 3
TUESDAY - B DAY - SEPT. 17, 2013
Table of Contents
Pick up calculator, notes packet, and pre-lab packet.
CW: Notes 3.1 to 3.5
CW: Finish Test part 2 ch.1-2 if not done yet
CW/HW: Read lab and complete Pre-lab questions. Note that the reading helps with the pre-lab questions due Thursday for lab. Try to be here by 7:45 for setup to give us time to discuss pre-lab and complete lab.
HW: Section pg. 117 #21, 26, 27, 29, 30, 31, 33, 35, 37, 39b,45b, 47b, 49b, 52, 57a-g, 61, 63 due Monday 9/23/13 to go over.

4. Determining the Formula of a Compound
Assignments Monday 9/23/13
Section 3.6
Determining the Formula of a Compound
Pick up handouts for use later & have out ch. 3 notes and HW ch. 3
CW: Notes finished together
CW: Empirical and Molecular Formula Race Game
CW/HW: Notes with computers.
HW: Formal Lab report due on Wednesday.
Be sure you completed homework previously assigned and turn in by Wednesday.
HW: Empirical Formula and Balancing Equations w/sheets due on Friday.
HW: Be reading chapter 3. The next test is over ch.3-4 and is tentatively scheduled for October 16th. Ch. 4 is mostly new material not covered in Gen. Chemistry.
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5. ASSIGNMENTS Ch. 3 Homework - Tuesday - 9/17/13
Section 3.5
Percent Composition of Compounds
ASSIGNMENTS Ch. 3 Homework - Tuesday - 9/17/13
HW: Section pg. 117 #21, 26, 27, 29, 30, 31, 33, 35, 37, 39b,45b, 47b, 49b, 52, 57a-g, 61, 63 due Monday 9/23/13 to go over.
You must show your work for most of these with calculations and units. No work for math problems will receive no points.
CW: Finish test part 2 if you haven’t.
HW: Read and complete Pre-lab Determination of the Molecular Weight of an Acid for lab on Thursday.
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6. Simplest whole-number ratio
Section 3.6
Determining the Formula of a Compound
Formulas
Empirical formula = CH
Simplest whole-number ratio
Molecular formula = (empirical formula)n [n = integer]
Molecular formula = C6H6 = (CH)6
Actual formula of the compound
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7. Formulas
Empirical formula: the lowest whole number ratio of atoms in a compound.
Molecular formula: the true number of atoms of each element in the formula of a compound.
molecular formula = (empirical formula)n [n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH

8. Formulas (continued)
Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3

9. Formulas (continued)
Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11

10. Chemical Formulas from Mass Percent Composition
We can “reverse” the process of finding percentage composition.
First we use the percentage or mass of each element to find moles of each element.
Then we can obtain the empirical formula by finding the smallest whole-number ratio of moles.
Find the whole-number ratio by dividing each number of moles by the smallest number of moles.

11. Empirical Formula Determination
Base calculation on 100 grams of compound. (If given masses, go to #2.)
Determine moles of each element in 100 grams of compound. (or sample given)
Divide each value of moles by the smallest of the values.
Multiply each number by an integer to obtain all whole numbers.

12. Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?
In 100 g sample, we would have
49.32 g Carbon
43.84 g Oxygen
6.85 g Hydrogen
Base calculation on 100 grams of compound.

13. Empirical Formula Determination (part 2)
2. Determine moles of each element in 100 grams of compound.
Carbon: g / g/mol = mol
Oxygen: g / g/mol = mol
Hydrogen: 6.85 g / 1.01 g/mol = 6.78 mol

14. Empirical Formula Determination (part 3)
3. Divide each value of moles by the smallest of the values.
Carbon: mol /2.74 mol = 1.50
Oxygen: 2.74 mol /2.74 mol = 1.00
Hydrogen: 6.78 mol/2.74 mol = 2.47

15. Empirical Formula Determination (part 4)
4. Multiply each number by an integer to obtain all whole numbers.
Carbon: 1.50 x 2 = 3.0
Oxygen: 1.0 x 2 = 2.0
Hydrogen: 2.47 = 2.5 x 2 = 5
Empirical formula:
C3H5O2

16. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = g

17. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
2. Divide the molecular mass by the mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = g

18. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
3. Multiply the empirical formula by this number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = g
(C3H5O2) x 2 =
C6H10O4

19. Relating Molecular Formulas to Empirical Formulas
A molecular formula is a simple integer multiple of the empirical formula.
That is, an empirical formula of CH2 means that the molecular formula is CH2, or C2H4, or C3H6, or C4H8, etc.
So: we find the molecular formula by:
= integer (nearly)
molecular formula mass
empirical formula mass
We then multiply each subscript in the empirical formula by the integer.

20. Analyzing for Carbon and Hydrogen
Section 3.6
Determining the Formula of a Compound
Analyzing for Carbon and Hydrogen
Device used to determine the mass percent of each element in a compound.
Combustion Analysis Method for determining molecular mass
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21. Empirical & Molecular Formula Partner Team Race Activity
Section 3.6
Determining the Formula of a Compound
Empirical & Molecular Formula Partner Team Race Activity
Working with a partner, you will solve one problem at a time. Both of you will solve the problem on your paper and then decide if you agree. Consider the correct number of significant figures as well and race the answer to me when you think you have a consensus of the final answer. If you are correct, I will say “yes” and you will proceed to the next problem. If I say “no”, you will need to try again. If I say “no” Sig. Fig., your answer is right to the wrong significant figures. Use the AP periodic table values which are mostly to the hundredths for atomic mass.
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22. Determining the Formula of a Compound
Section 3.6
Determining the Formula of a Compound
Assignments Monday
CW: Notes finished together
CW: Empirical and Molecular Formula Race Game
CW/HW: Notes with computers.
HW: Formal Lab report due on Wednesday.
Be sure you completed homework previously assigned and turn in by Wednesday.
HW: Empirical Formula and Balancing Equations w/sheets due on Friday.
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23. Combustion Analysis Method for determining molecular mass
Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A g sample of menthol is combusted, producing g of CO2 and g of H2O. What is the empirical formula for menthol?

24. Combustion Analysis Method for determining molecular mass (cont)

25. Combustion Analysis Method for determining molecular mass (cont.)

26. More complicated example:
You do not have to record but do look over it.

27. Nasty example - you don’t have to record but look over.

28. Writing Chemical Equations
Notes 3.7
Writing Chemical Equations
A chemical equation is a shorthand description of a chemical reaction, using symbols and formulas to represent the elements and compounds involved.

29. C2H5OH + 3O2 → 2CO2 + 3H2O A representation of a chemical reaction:
Section 3.7
Chemical Equations
A representation of a chemical reaction:
C2H5OH + 3O2 → 2CO H2O
reactants products
Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow.
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30. The equation is balanced.
Section 3.7
Chemical Equations
C2H5OH + 3O2 → 2CO H2O
The equation is balanced.
All atoms present in the reactants are accounted for in the products.
1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.
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31. Section 3.7
Chemical Equations
The balanced equation represents an overall ratio of reactants and products, not what actually “happens” during a reaction.
Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed.
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32. Writing Chemical Equations
Sometimes additional information about the reaction is conveyed in the equation.

33. Symbols Used in Chemical Equations
Section 1 Describing Chemical Reactions
Chapter 8
Symbols Used in Chemical Equations
Chapter menu
Resources
Copyright © by Holt, Rinehart and Winston. All rights reserved.

34. Writing and Balancing the Equation for a Chemical Reaction
Section 3.8
Balancing Chemical Equations
Writing and Balancing the Equation for a Chemical Reaction
Determine what reaction is occurring. What are the reactants, the products, and the physical states involved?
Write the unbalanced equation that summarizes the reaction described in step 1.
Balance the equation by inspection, starting with the most complicated molecule(s). The same number of each type of atom needs to appear on both reactant and product sides.
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35. Balancing Chemical Equations
must log-in to textbook to view video
Click here for visualization.
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36. Guidelines for Balancing Chemical Equations
If an element is present in just one compound on each side of the equation, try balancing that element first.
Balance any reactants or products that exist as the free element last.
In some reactions, certain groupings of atoms (such as polyatomic ions) remain unchanged. In such cases, treat these groupings as a unit.
At times, an equation can be balanced by first using a fractional coefficient(s). The fraction is then cleared by multiplying each coefficient by a common factor.

37. Section 3.8
Balancing Chemical Equations
Exercise
Which of the following correctly balances the chemical equation given below? There may be more than one correct balanced equation. If a balanced equation is incorrect, explain what is incorrect about it.
CaO + C → CaC2 + CO2
I. CaO2 + 3C → CaC2 + CO2
II. 2CaO + 5C → 2CaC2 + CO2
III. CaO + (2.5)C → CaC2 + (0.5)CO2
IV. 4CaO + 10C → 4CaC2 + 2CO2
II, III, and IV are correct. I is not correct because you cannot use subscripts to balance a chemical equation.
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38. Balancing Chemical Equations Concept Check
Section 3.8
Balancing Chemical Equations
Concept Check
Which of the following are true concerning balanced chemical equations? There may be more than one true statement.
I. The number of molecules is conserved.
II. The coefficients tell you how much of each substance you have.
III. Atoms are neither created nor destroyed.
IV. The coefficients indicate the mass ratios of the substances used.
V. The sum of the coefficients on the reactant side equals the sum of the coefficients on the product side.
Only III is correct.
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39. Subscripts must not be changed to balance an equation.
Section 3.8
Balancing Chemical Equations
Notice
The number of atoms of each type of element must be the same on both sides of a balanced equation.
Subscripts must not be changed to balance an equation.
A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction.
Coefficients can be fractions, although they are usually given as lowest integer multiples.
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40. Table 3.2 Information Conveyed by the Balanced Equation for the Combustion of Methane
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41. Stoichiometric Calculations
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Stoichiometric Calculations
Chemical equations can be used to relate the masses of reacting chemicals.
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42. Stoichiometric Equivalence and Reaction Stoichiometry
A stoichiometric factor or mole ratio is a conversion factor obtained from the stoichiometric coefficients in a chemical equation.
In the equation: CO(g) + 2 H2(g) → CH3OH(l)
1 mol CO is chemically equivalent to 2 mol H2
1 mol CO is chemically equivalent to 1 mol CH3OH
2 mol H2 is chemically equivalent to 1 mol CH3OH
1 mol CO
–––––––––
2 mol H2
1 mol CO
–––––––––––––
1 mol CH3OH
2 mol H2
–––––––––––––
1 mol CH3OH

43. Concept of Stoichiometric Equivalence
One car may be equivalent
to either 25 feet or 10 feet,
depending on the method of parking.
One mole of CO may be equivalent to one mole of CH3OH, or to one mole of CO2, or to two moles of CH3OH, depending on the reaction(s).

44. Outline of Simple Reaction Stoichiometry
Note: preliminary and/or follow-up calculations may be needed.

45. Calculating Masses of Reactants and Products in Reactions
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant or product to moles of that substance.
3. Use the balanced equation to set up the appropriate mole ratios.
4. Use the appropriate mole ratios to calculate the number of moles of desired reactant or product.
5. Convert from moles back to grams if required by the problem.
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46. Outline of Stoichiometry Involving Mass
… we’ve added a conversion from mass at the beginning …
To our simple stoichiometry scheme …
Substances A and B may be two reactants, two products, or reactant and product.
… and a conversion to mass at the end.
Think: If we are given moles of substance A initially, do we need to convert A to grams?

47. Calculating Masses of Reactants and Products in Reactions
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
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48. HW: Balancing Equations practice quick w/sheet due FRI
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
WEDNESDAY - SEPT 25, 2013
CW: Notes 3.9
Turn in Formal Lab Report from last Thursday’s titration lab AND HW 3.1 to go over any questions
CW: Limiting Reactants problem w/sheet handout - HW if not finished in class due FRI
HW: Balancing Equations practice quick w/sheet due FRI
HW: Complete the empirical vs. molecular formula and % composition w/sheet from Monday - due Friday
CW/HW: pre-lab moles vs. molar mass graphical with limiting reactants - using graphing to determine unknown.
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49. Stoichiometry
“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”
Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

50. Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2 → 2CO H2O
reactants
products
When the equation is balanced it has quantitative significance:
1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

51. Calculating Masses of Reactants and Products
Balance the equation.
Convert mass or volume to moles, if necessary.
Set up mole ratios.
Use mole ratios to calculate moles of desired substituent.
Convert moles to mass or volume, if necessary.

52. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.
1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2
Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?

53. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?
4 Al O2 2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3
g Al2O3 =
? g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
6.50 x 2 x ÷ ÷ 4 =
12.3 g Al2O3

54. Working a Stoichiometry Problem
Question- How many grams of water can be produced from 10. grams of H2 with excess O2?
You try. =

55. Working a Stoichiometry Problem
Question- How many grams of water can be produced from 10. grams of H2 with excess O2?
Answer: 2H2 + O2 --> 2H2O
2 mol H2O
10 g H2
1 mol H2
18.02 g H2O =
89.2=89 g
1 mol H2O
2.02 g H2
2 mol H2

56. Limiting Reactant
The limiting reactant is the reactant
that is consumed first, limiting the amounts of products formed.
The reactant left over is in EXCESS. (remember nuts and bolts mini-lab)

57. Limiting Reactants
Many reactions are carried out with a limited amount of one reactant and a plentiful amount of the other(s).
The reactant that is completely consumed in the reaction limits the amounts of products and is called the limiting reactant, or limiting reagent.
The limiting reactant is not necessarily the one present in smallest amount.

58. Limiting Reactant Analogy
… how many packaged meals can we make?
If we have 10 sandwiches, 18 cookies, and 12 oranges …

59. Limiting Reactant Analogy
… how many packaged meals can we make?
There would only be 9 because you need 20 cookies to use all of the sandwiches 1:2 ratio and you only need 10 oranges with 10 sandwiches so oranges and sandwiches are in excess.
If we have 10 sandwiches, 18 cookies, and 12 oranges …

60. Molecular View of the Limiting Reactant Concept
Why is ethylene left over, when we started with more bromine than ethylene? (Hint: count the molecules.)
What mass of ethylene is left over after reaction is complete? (Hint: it’s an easy calculation; why?)
When 28 g (1.0 mol) ethylene reacts with …
… 128 g (0.80 mol) bromine, we get …
… 150 g of 1,2-dibromoethane, and leftover ethylene!

61. Recognizing and Solving Limiting Reactant Problems
We recognize limiting reactant problems by the fact that amounts of two (or more) reactants are given.
One way to solve them is to perform a normal stoichiometric calculation of the amount of product obtained, starting with each reactant.
The reactant that produces the smallest amount of product is the limiting reactant.

62. Consider the following reaction: P4 (s) + 5 O2 (g) 2P2O5 (s)
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Exercise
Consider the following reaction:
P4 (s) + 5 O2 (g) P2O5 (s)
If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with?
(6.25 g P4)(1 mol P4 / g P4)(5 mol O2 / 1 mol P4)(32.00 g O2 / 1 mol O2) = g O2
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63. 8.07 g if you do everything and then round at the end
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Exercise
Consider the following reaction:
P4 (s) + 5 O2 (g) P2O5 (s)
If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with?
6.25 g / 124 g/mol = mol 1:5 ratio so
x 5 =0.252 mol x 32 g/mol = 8.06 g O2
8.07 g if you do everything and then round at the end
(6.25 g P4)(1 mol P4 / g P4)(5 mol O2 / 1 mol P4)(32.00 g O2 / 1 mol O2) = g O2
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64. Identify limiting reactants & excess reactant
Section 3.10
The Concept of Limiting Reagent
Concept Check
Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation:
2H2 + O2 → 2H2O
Identify limiting reactants & excess reactant
a) 2 moles of H2 and 2 moles of O2
2 moles of H2 and 3 moles of O2
2 moles of H2 and 1 mole of O2
d) 3 moles of H2 and 1 mole of O2
e) Each produce the same amount of product.
The correct answer is “e”. All of these reaction mixtures would produce two moles of water.
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65. Section 3.10
The Concept of Limiting Reagent
Notice
We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation.
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66. Section 3.10
The Concept of Limiting Reagent
Concept Check
You know that chemical A reacts with chemical B. You react 10.0 g of A with g of B.
What information do you need to know in order to determine the mass of product that will be produced?
See the “Let’s Think About It” slide next to get the students going.
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67. Where are we going? How do we get there?
Section 3.10
The Concept of Limiting Reagent
Let’s Think About It
Where are we going?
To determine the mass of product that will be produced when you react 10.0 g of A with 10.0 g of B.
How do we get there?
We need to know:
The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation.
The molar masses of A, B, and the product they form.
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68. 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation:
Section 3.10
The Concept of Limiting Reagent
Exercise
You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is
20.0 g/mol, and C is 25.0 g/mol? They react according to the equation:
A + 3B → 2C
8.33 g of C is produced.
Note: Use the box animations to assist in explaining how to solve the problem.
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69. Section 3.10
The Concept of Limiting Reagent
Percent Yield
An important indicator of the efficiency of a particular laboratory or industrial reaction.
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70. 85.0 g Consider the following reaction: P4(s) + 6F2(g) → 4PF3(g)
Section 3.10
The Concept of Limiting Reagent
Exercise
Consider the following reaction:
P4(s) + 6F2(g) → 4PF3(g)
What mass of P4 is needed to produce g of PF3 if the reaction has a 64.9% yield?
% yield = actual / theoretical
0.649 = 85.0 g / t t = = 131 g
131 g PF3 / 88 g/mol = 1.49 mol (1.5 mol)
ratio 4 mol PF3 to 1 mol P4 1.49/4=.3725 mol P4
.3725 mol x 124 g/mol = = 46.2 g P4
85.0 g ?
? g
64.9% = (85.0 g PF3 / x)(100); x = g PF3
( g PF3)(1 mol PF3 / g PF3)(1 mol P4 / 4 mol PF3)( g P4 / 1 mol P4) = 46.1 g of P4 needed
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71. Consider the following reaction: P4(s) + 6F2(g) → 4PF3(g)
Section 3.10
The Concept of Limiting Reagent
Exercise
Consider the following reaction:
P4(s) + 6F2(g) → 4PF3(g)
What mass of P4 is needed to produce g of PF3 if the reaction has a 64.9% yield?
46.1 g P4
64.9% = (85.0 g PF3 / x)(100); x = g PF3
( g PF3)(1 mol PF3 / g PF3)(1 mol P4 / 4 mol PF3)( g P4 / 1 mol P4) = 46.1 g of P4 needed
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72. Yields of Chemical Reactions
The theoretical yield of a chemical reaction is the calculated quantity of product in the reaction.
The actual yield is the amount you actually get when you carry out the reaction.
Actual yield will be less than the theoretical yield, for many reasons … can you name some?
actual yield
Percent yield = ––––––––––––– × 100
theoretical yield

73. Example - % Yield 2CO(g) + O2(g) --> 2CO2(g)
Calculate the % yield if 69.1g of CO combines with excess O2 to form an experimental yield of 48.3L of

74. Example - % Yield 2CO(g) + O2(g) --> 2CO2(g)
Calculate the % yield if 69.1g of CO combines with excess O2 to form an experimental yield of 48.3L of
69.1 g / 28 g⋆mol-1 = 2.47 mol CO thus creates same amount of moles of CO2. For gases only, 22.4 L = 1 mol of any gas, so mol x 22.4 L / 1 mol = 55.3 L (theoretically from stoichiometry)
% yield = 48.3 / (x 100) = 87.3%

75. 2CO(g) + O2(g) --> 2CO2(g)
Calculate the % yield if 69.1g of CO combines with excess O2 to form an experimental yield of 48.3L of 4
2.47 mol 4 4 4 8
87.3% yield

76. Actual Yield of ZnS Is Less than the Theoretical Yield

77. Solutions and Solution Stoichiometry
Solute: the substance being dissolved.
Solvent: the substance doing the dissolving.
Concentration of a solution: the quantity of a solute in a given quantity of solution (or solvent).
A concentrated solution contains a relatively large amount of solute vs. the solvent (or solution).
A dilute solution contains a relatively small concentration of solute vs. the solvent (or solution).
“Concentrated” and “dilute” aren’t very quantitative …

78. Molar Concentration
Molarity (M), or molar concentration, is the amount of solute, in moles, per liter of solution:
moles of solute
Molarity = ––––––––––––––
liters of solution
A solution that is 0.35 M sucrose contains moles of sucrose in each liter of solution.
Keep in mind that molarity signifies moles of solute per liter of solution, not liters of solvent.

79. Add more water to reach the 1.000 liter mark.
Preparing M KMnO4
Weigh mol (1.580 g) KMnO4.
Add more water to reach the liter mark.
Dissolve in water. How much water? Doesn’t matter, as long as we don’t go over a liter.

80. Example 3.23 Example 3.24 Example 3.25
What is the molarity of a solution in which 333 g potassium hydrogen carbonate is dissolved in enough water to make 10.0 L of solution?
Example 3.24
We want to prepare a 6.68 molar solution of NaOH (6.68 M NaOH).
(a) How many moles of NaOH are required to prepare L of M NaOH?
(b) How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH?
Example 3.25
The label of a stock bottle of aqueous ammonia indicates that the solution is 28.0% NH3 by mass and has a density of g/mL. Calculate the molarity of the solution.

81. Dilution of Solutions
Dilution is the process of preparing a more dilute solution by adding solvent to a more concentrated one.
Addition of solvent does not change the amount of solute in a solution but does change the solution concentration.
It is very common to prepare a concentrated stock solution of a solute, then dilute it to other concentrations as needed.

82. Dilution Calculations …
… couldn’t be easier.
Moles of solute does not change on dilution.
Moles of solute = M × V
Therefore …
Mconc × Vconc = Mdil × Vdil

83. Visualizing the Dilution of a Solution
We start and end with the same amount of solute.
Addition of solvent has decreased the concentration.

84. Example 3.26
How many milliliters of a 2.00 M CuSO4 stock solution are needed to prepare L of M CuSO4?

85. Solutions in Chemical Reactions
Molarity provides an additional tool in stoichiometric calculations based on chemical equations.
Molarity provides factors for converting between moles of solute (either reactant or product) and liters of solution.

86. Adding to the previous stoichiometry scheme …
If substance A is a solution of known concentration …
… we can start with molarity of A times volume (liters) of the solution of A to get here.
If substance B is in solution, then …
… we can go from moles of substance B to either volume of B or molarity of B. How?

87. Limiting Reactants Click here to watch visualization.
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