1. Unit 6 Chapter 12 Chemical Quantities or
"Our Friend the Mole"

2. Stoichiometry
stoichiometry—using balanced chemical equations to obtain info

3. Particles of matter are too small and numerous to count
12.1 Counting Particles of Matter
Particles of matter are too small and numerous to count
SI unit of chemical quantity =
the mole (abbreviated mol)
The mole is a counting unit

4. How do you measure how much?
You can measure mass = grams
or volume = liters
or you can count pieces = MOLES

5. Moles 1 mole = 6.02 x 1023 representative particles
Treat it like a very large dozen
6.02 x is called Avogadro's number

6. Representative particles
= The smallest pieces of a substance
representative particles =
ATOMS
IONS
MOLECULES
FORMULA UNITS

7. Representative particles
For a molecular (covalent)
compounds it is a molecule
compound with all nonmetals
CO, BF3 , Cl2
1 mole = 6.02 x 1023 molecules

8. Representative particles
For an element it is an atom
Unless it is diatomic
one symbol, no charge: Br,Cs
1 mole = 6.02 x 1023 atoms

9. Representative particles
For an ionic compound it is
a formula unit
compound with metal and
nonmetal - KI, Na2SO4
1 mole = 6.02 x 1023 formula units

10. Representative particles
For ions it is 1 mol ions
one symbol with charge (monatomic) or
more than one symbol with charge (polyatomic: Na+ , N3- , (C2H3O2) –
1 mole ions = 6.02 x 1023 ions

11. Molar Mass = The mass of 1 mole of an element in grams.
We can make conversion factors from these.
Example - We can write this as g C = 1 mol
To change grams of a compound to moles of a compound.
Or moles to grams

12. Molar Masses the atomic masses on the periodic table
have a unit of amu (atomic mass unit)
GAM = gram atomic mass
= the atomic mass (listed on the periodic table) written in grams
1 atom Xe = u
GAM of Xe = g

13. molar mass—the mass, in g, of 1 mole of a substance Add up the gram atomic mass of all elements in compound to calculate molar mass of a substance

14. Find the molar mass of methane, CH4.
CH4 = 1(12.0) + 4(1.0) = = 16.0 g
Find the molar mass of calcium hydroxide, Ca(OH)2.
Ca(OH)2 = 1(40.1) + 2(16.0) + 2(1.0) = 74.1 g

15. 2. What is the molar mass of a molecule of H2?
1. What is the molar mass of 1 mole of phosphorus?
2. What is the molar mass of a molecule of H2?
3. What is the molar mass of a formula unit of MgCl2?
30.97 g
2.016 g
94 g

16. MOLAR VOLUME of any gas at STP: 22.4 L = 1 mol
Molar Volume: volume-to-mole and
mole-to-volume conversions
At STP, all gases occupy the same amount of space:
MOLAR VOLUME of any gas at STP:
22.4 L = 1 mol

17. dimensional analysis steps for success:
= using the units (dimensions) to solve problems
steps for success:
1) identify unknown (read carefully)
2) identify known (read carefully)
“Play checkers” with the units, moving them diagonally, canceling when appropriate. All units should cancel except those of the desired answer.
3) plan solution
4) calculate
5) check (sig.figs., units, and math)

18. CONVERSION FACTOR SUMMARY:
6.02 x 10^23 representative particles
1 MOLE
&

19. CONVERSION FACTOR SUMMARY:
MOLAR MASS (g) & MOLE
1 MOLE MOLAR MASS (g)
for a gas at STP:
22.4 L & 1 MOLE
1 MOLE L

20. Calculation question
How many molecules of CO2 are the in 4.56 moles of CO2 ?
4.56 moles CO2 X x 1023 molecules CO2
mole CO2
= x 1023 molecules CO2

21. For example How many moles is 5.69 g of NaOH?
5.69 g NaOH x 1 mole= mol NaOH
gNaOH
need to change grams to moles for NaOH
1mole Na = 22.99g 1 mol O = g 1 mole of H = 1.01 g
1 mole NaOH = g

22. Examples = 28.10 g C How much would 2.34 moles of carbon weigh?
2.34 mol C X g C
mol C
= g C

23. Calculation question
How many moles of salt is 5.87 x formula units?
5.87 X 1022 formula units NaCl X mole NaCl_
x 1023 f.u. NaCl
= moles NaCl

24. Examples 4.61 g Mg X _1 mol Mg 1 24.3 g Mg = 0.1897 mol Mg
How many moles of magnesium in 4.61 g of Mg?
4.61 g Mg X _1 mol Mg
g Mg
= mol Mg

25. example
What is the volume, in L, of mol of NO2 gas at STP?
0.495 mol NO2 x L NO2 = L NO2
mol NO2

26. How many moles are found in 84 L of neon gas at STP?
84 L Ne x mol Ne = mol Ne
L Ne

27. There are many types of mole problems:
1 step: r.p mol & mol r.p.
mass mol & mol mass
2 step: mass r.p. & r.p mass
mass volume & volume mass
r.p volume & volume r.p.

28. Moles of Compounds
1 mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula for the compound.
Example:
1 mole of ammonia (NH3) has
1 mole of nitrogen atoms and
3 moles of hydrogen atoms.

29. Percent composition
composition of a compound is the percent by mass of each element in the compound.
Find the mass of each element
divide by the total mass of compound.

30. Determine the percent composition of calcium chloride (CaCl2).
Step 1: determine the molar masses of each element in compound and the compound
Ca = g
Cl2 = 2 x g = g
CaCl = g

31. ****make sure they add up to 100
Step 2: divide the molar mass of each element by the molar mass of the compound and times by 100
****make sure they add up to 100

32. Empirical Formula
The lowest whole number ratio of elements in a compound.
CH2
H2O

33. You can determine the empirical formula from percent composition and mole ratios
*** percent means “parts per hundred”
assume you have 100 g of the compound
Step 1: change the percent sign to g (grams)
Calculate the empirical formula of a compound composed of % C, % H, and %N

34. Step 2:calculate the number of moles for each element by converting grams to moles using
1 mole = molar mass
38.67 g C x 1mol C = mole C g C
16.22 g H x 1mol H = 16.1 mole H g H
45.11 g N x 1mol N = mole N g N

35. So empirical formula is CH5N
Step 3: divide each number of moles by the smallest number of moles
3.220 mole C = 1 mol C
3.220
16.1 mole H = 5 mol H
3.220 mole N = 1 mol N
So empirical formula is CH5N

36. Determine the empirical formula of a compound with percent composition of 38.4 % Mn, 16.8 % C, 44.7 % O
38.4 g Mn X 1mol Mn = mole Mn g Mn
16.8 g C X 1mol C = mole C g C
45.7 g O X 1mol O = mole O g O

37. So formula is MnC2O4 0.699 mole Mn = 1 mol Mn 0.699
1.339 mole C = 2 mol C
2.798 mole O = 4 mol O
So formula is MnC2O4

38. For many compounds, the empirical formula is not the true formula A molecular formula tells the exact number of atoms of each element in a molecule

39. Example: acetic acid
molecular formula = C2H4O2
empirical formula = CH2O
The molecular formula for a compound is always a whole-number multiple of the empirical formula.

40. To determine molecular formula you need to know the molar mass of the compound
Divide the actual molar mass by the molar mass of the empirical formula.
Multiply the empirical formula by this number.
molar mass of compound
molar mass of empirical formula

41. Example: A compound has an empirical formula of ClCH2 and a molar mass of g/mol. What is its molecular formula?
molar mass of ClCH2 = g/mol
= 2.01
X (ClCH2)
Empirical formula = ClCH2
Molecular formula = Cl2C2H4

42. Example
A compound has an empirical formula of CH2O and a molar mass of g/mol. What is its molecular formula?

43. Example
Ibuprofen is % C, 8.80 % H, % O, and has a molar mass of about 207 g/mol. What is its molecular formula?

44. Unit 6 Chemical Quantities Continued
Chapter 12.2
“Using the Mole"

45. remember our friend the mole?
Recall that the mole should be treated like a dozen
1 dozen = 12 pieces &
1 mole = pieces (or 6.02 X 1023 r.p)

46. remember our friend the mole?
You still need to know how and when to use:
 Avogadro’s number (6.02 x 1023), representative particles
 Molar mass, grams in one mole,
 Molar volume of a gas at STP (22.4 L)

47. Conversion factor review
1 mole = 6.02 x 1023 r.p 1 mole = molar mass (g) 1 mole = 22.4 L

48. So what about this silly mole?
Our useful friend the mole allows us to do calculations called stoichiometry—using balanced chemical equations to obtain info ex: 2C + O2 2CO

49. Mole - Mole (MOL – MOL) Conversions
new conversion factor –
# of mol A = # of mol B
# = coefficients in balanced equation

50. I.Mole - Mole (MOL – MOL) Conversions
the most important, most basic stoich calculation
B. uses the coefficients of a balanced equation to compare the amounts of reactants and products
C. coefficients are mole ratios

51. D. the way to go from substance A to substance B
E. mol – mol is the only time the mole number in the conversion is not automatically 1. (Avogadro’s #, molar mass, and 22.4 L are all = to 1 mol)
MOL – MOL : # mol A # mol B
# mol B # mol A
# = coefficients

52. ***always start with a balanced chemical equation
MOL – MOL Conversions
# mol A # mol B
# mol B # mol A
ex: How many moles of carbon monoxide are produced when mol of oxygen reacts with carbon?
2C + O CO
***always start with a balanced chemical equation

53. How many moles of carbon monoxide are produced when 0
How many moles of carbon monoxide are produced when mol of oxygen reacts with carbon? C + O CO
Step 1; determine given and unknown given = mol O2 unknown = ? mol CO

54. How many moles of carbon monoxide are produced when 0
How many moles of carbon monoxide are produced when mol of oxygen reacts with carbon? C + O CO
Step 2; setup conversion using mole ratios (coefficients) in balanced equation
0.750 mol O2 X 2 mol CO = 1.5 mol CO
mol O2

55. Step 1; determine given and unknown given = 0.661 mol Al2O3
Find the number of moles of the reactants, given mol of product is formed Al + 3O Al2O3
Step 1; determine given and unknown
given = mol Al2O3
unknown = ? mol Al
= ? mol O2

56. 4Al + 3O Al2O3
Step 2; setup conversion using mole ratios (coefficients) in balanced equation
0.661 mol Al2O X 4 mol Al = mol Al
mol Al2O3
0.661 mol Al2O3 X 3 mol O2 = mol O2
mol Al2O3

57. II. MASS – MASS Conversions
– Using molar mass in stoich problems to predict masses of reactants and/or products
a balanced chemical equation can be used to compare masses of reactants and products
B. mass – mass cannot change which substance you are dealing with; only mol – mol can do that

58. PT = periodic table, molar mass # = coefficients, chemical equation
MASS – MASS Conversions
GIVEN g A X 1 mol A X # mol B X PT g B
PT g A # mol A mol B
PT = periodic table, molar mass
# = coefficients, chemical equation

59. How many grams of hydrochloric acid are made from the reaction of 0
How many grams of hydrochloric acid are made from the reaction of g of hydrogen gas with excess chlorine gas? H2 + Cl2 2HCl Known = 0.500g H2 Unknown = ? g HCl

60. H2 + Cl2 2HCl GIVEN g A X 1 mol A X # mol B X PT g B
0.500 g H2 X 1 mol H2 X 2 mol HCl X 36.5 gHCl g H2 1 mol H2 1 mol HCl = 18 g HCl
H2 + Cl HCl
GIVEN g A X 1 mol A X # mol B X PT g B
PT g A # mol A mol B

61. Calculate the numbers of grams of oxygen formed when 25
Calculate the numbers of grams of oxygen formed when 25.0 g of sodium nitrate decomposes into sodium nitrite and oxygen.
2NaNO NaNO2 + O2
25.0 g NaNO3 X 1 mol NaNO3 X 1 mol O2 X 32.0 g O2
g NaNO3 2 mol NaNO3 1 mol O2
= 20.3 g O2

62. “mole – mass” (mass – mole) calculations “MASS – MOLE”:
GIVEN g A X 1 mol A X # mol B
PT g A # mol A
“MOLE – MASS”:
GIVEN mol A X # mol B X PT g B
# mol A mol B
PT = periodic table, molar mass
# = coefficients

63. How many g of water are produced from the complete combustion of 0
How many g of water are produced from the complete combustion of mol of C2H2?
2C2H2 + 5O CO2 + 2H2O
mol C2H2 X 2 mol H2O X g H2O
mol C2H2 1 mol H2O
= 12.3 g H2O

64. Using the equation 2C2H2 + 5O2 4CO2 + 2H2O how many moles of O2 would be needed to produce g of CO2? g CO2 X 1 mol CO2 X 5 mol O g CO2 4 mol CO2 = 1.59 mol O2

65. “mass – volume” (volume – mass) calculations – “MASS – VOLUME”: STP) GIVEN g A X 1 mol A X # mol B X 22.4 L B 1 PT g A # mol A 1 mol B “VOLUME – MASS”: STP) GIVEN L A X 1 mol A X # mol B X PT g B L A # mol A 1 mol B PT = periodic table, molar mass # = coefficients

66. How many L of hydrogen are produced from the decomposition of 3
How many L of hydrogen are produced from the decomposition of 3.50 g of water at STP?
2H2O H2 + O2
3.50 g H2O X 1 mol H2O X 2 mol H2 X 22.4 L H2
g H2O 2 mol H2O 1 mol H2
= 4.36 L H2

67. Using the equation 2C2H2 + 5O2 4CO2 + 2H2O
how many liters of water vapor are produced when 5.02 g of C2H2 undergoes complete combustion?
5.02 g C2H2 x 1 mol C2H2 x 2 mol H2O x 22.4 L H2O
g C2H mol C2H mol H2O
= 4.32 L H2O

68. “volume – volume” calculations
GIVEN L A x 1 mol A x # mol B x 22.4 L B
L A # mol A mol B
# = coefficients (SHORT CUT: compare coefficients!)

69. How many L of sulfur trioxide are produced from the reaction of 36
How many L of sulfur trioxide are produced from the reaction of 36.1 L of oxygen with sulfur dioxide at STP?
2SO2 + O SO3
36.1 L O2 x 1 mol O2 x 2 mol SO3 x 22.4 L SO3
L O2 1 mol O mol SO3
= 72.2 L SO3
SHORTCUT: coefficient of O2 = 1 coefficient of SO3 = 2 so 36.1 L x 2 = 72.2 L SO3

70. SHORTCUT: coefficients = 1 mol HCl to 1 mol CO2.
How many liters of CO2 are produced from L of HCl reacting with NaHCO3?
NaHCO3 + HCl NaCl + CO2 + H2O
0.252 L HCl x 1 mol HCl x 1 mol CO2 x 22.4 L CO2
L HCl mol HCl mol CO2
= L CO2
SHORTCUT: coefficients = 1 mol HCl to 1 mol CO2.
so L HCl = L CO2

71. mass –particle (particle – mass) calculations “MASS – PARTICLE”:
GIVEN g A x 1 mol A x # mol B x 6.02 x 1023 r.p. B
PT g A # mol A mol B
“PARTICLE – MASS”:
GIVEN r.p. A x 1 mol A x # mol B x PT g B
x 1023 r.p. A # mol A mol B
PT = periodic table, molar mass
# = coefficients

72. How many molecules of NH3 are produced from reacting 2
How many molecules of NH3 are produced from reacting 2.07 g of H2 with excess N2?
N2 + 3H NH3
2.07 g H2 x 1 mol H2 x 2 mol NH3 x 6.02 x 1023 NH3
g H mol H mol NH3
= 4.2 x 1023 molecules NH3

73. Limiting Reactants
Rarely are the reactants in a chemical reaction present in the exact mole ratios specified in the balanced equation.
Usually, one or more of the reactants are present in excess, and the reaction proceeds until all of one reactant is used up.

74. The reactant that is used up is called the limiting reactant.
The left-over reactants are called excess reactants.

75. Limiting Reactant To figure out the limiting reactant:
You must do the stoichiometric calculation with all reactants.
The one that produces the least amount of product is the limiting reactant.

76. Ex: 4Al + 3O2  2Al2O3
What is the limiting reactant when 35 g of aluminum reacts with 35 g of oxygen? How much aluminum oxide is formed in this reaction?
35 g Al X 1 mol Al X 2 mol Al2O3 =
g Al mol Al
35 g O2 X 1 mol O2 X 2 mol Al2O3 = 1.1 mol Al2O3
g O mol O2
0.65 mol Al2O3
Aluminum is limiting reactant
0.65 mol Al2O3 X g Al2O = g Al2O3
mol Al2O3

77. Percent Yield theoretical yield —amount of product predicted by the math (theory) actual yield —amount of product obtained in lab

78. percent yield —percentage of product recovered; comparison of actual and
theoretical yields
% YIELD = ACTUAL YIELD X 100
THEORETICAL YIELD

79. 35. 0 g of product should be recovered from an experiment
35.0 g of product should be recovered from an experiment. A student collects 22.9 g at the end of the lab. What is the percent yield?
22.9 g X 100 = 65.4%
35.0 g

80. = THEORETICAL YIELD = 3.19 g NaCl % YIELD = 2.89 g NaCl X 100 = 90.6%
What is the percent yield if 2.89 g of NaCl is produced when 1.99 g of HCl reacts with excess NaOH? Water is the other product.
HCl + NaOH NaCl + H2O
Actual yield = 2.89 g NaCl
Theoretical yield = ?
1.99 g HCl x 1 mol HCl x 1 mol NaCl x 58.5 g NaCl
g HCl 1 mol HCl mol NaCl
= THEORETICAL YIELD = g NaCl
% YIELD = 2.89 g NaCl X = 90.6%
3.19 g NaCl

81. The End