1. Chemical Quantities
Moles, % Compositions,

2. 3 ways to measure matter Number present By Mass/Weight By Volume
Count how many are there
Example: number of songs on your smartphone
By Mass/Weight
How many kg/lb present?
Example: buying produce at a supermarket
By Volume
How many gallons/liters present?
Example: buying gas

3. Using Apples as an Example
Remember Conversion Factors
A ratio equal to 1
1 dozen apples
12 apples
2.0 kg
0.2 bushel of apples
Fruit stand
Often by number present
1 dozen= 12 apples
Supermarket
Often by mass/weight
1 dozen= 2.0 kg
Orchard
Often by volume
1 dozen= 0.2 bushels

4. What is the mass of 90. average sized apples if 1 dozen of the apples has a mass of 2.0kg?
Identify what you want to change
Number of apples  dozens  mass
Identify your equalities (conversion factors)
1 dozen = 12 apples and kg = 1 dozen
Solve the problem using dimensional analysis
90. apples
1 dozen
2.0 kg
15 kg
12 apples
1 dozen

5. What is a mole?
A way to determine how many particles are present in chemistry because they are way too small to count
1 mol = x 1023 representative particles
Avogadro’s Number- Italian scientist Amedeo Avogadro who helped clarify the difference between atoms and molecules
Species present in a substance- this may vary depending on the substance (examples on next slide)

6. Examples of Representative Particles
Substance
Representative Particle
Chemical Formula
Representative Particles in 1.00 mol
Copper
atom Cu
6.022 X 1023
Water
molecule
H2O
Calcium Ion
ion
Ca2+
Calcium Fluoride
formula unit
CaF2

7. Conversion factor possibilities for a mol 1 mol = 6.022 x 1023 rp
rp= representative particle

8. Moles to Particles or Particles to Moles (1 Step Problem)
Number of RP (atoms, molecules, f.u.
Avogadro’s
number

9. Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, and tools. How many moles of Mg is 1.25 x 1023 atoms of Mg?
1.25 x 1023 atoms Mg

10. Atoms are your representative particles in this case
Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, and tools. How many moles of Mg is 1.25 x 1023 atoms of Mg?
1 mol Mg
1.25 x 1023 atoms Mg
6.022 x 1023 atoms Mg
Atoms are your representative particles in this case

11. Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, and tools. How many moles of Mg is 1.25 x 1023 atoms of Mg?
1 mol Mg
1.25 x 1023 atoms Mg
0.208 mol Mg
6.022 x 1023 atoms Mg

12. How many moles of Si are in 2.8 x 1024 atoms of silicon?
Answer 4.65 mol Si

13. Propane gas (C3H8) is often used for cooking and heating
Propane gas (C3H8) is often used for cooking and heating. How many carbon atoms are in 1 molecule of propane? How many hydrogen atoms are in 1 molecule of propane?
Answer: Carbon x 1024
Answer: Hydrogen x 1025

14. Propane gas (C3H8) is often used for cooking and heating
Propane gas (C3H8) is often used for cooking and heating. How many moles of carbon are in 1 mole of propane? How many moles of hydrogen are in 1 mole of propane?
Answer: Carbon x 1024
Answer: Hydrogen x 1025

15. Molar Mass
Atomic mass = mass of a single atom expressed in amu (atomic mass unit); relative values based on the isotopes; (decimals) on the periodic table
Molar mass is the mass of 1 mole of an element/compound/molecule; equal to the atomic mass
unit for molar mass is: g/mol (grams per mole)
Example: Carbon’s atomic mass is amu and its molar mass is g/mol

16. 1 MOLE

17. How to find the mass of a mole of a compound
You must know the chemical formula
Then add the atomic masses of the atoms that make up that compound
Example: Sulfur Trioxide
formula- SO3
1 sulfur atom and 3 oxygen atoms
Sulfur molar mass = g/mol
Oxygen molar mass = g/mol
= g/mol
Substituted g/mol for amu

18. The decomposition of hydrogen peroxide (H2O2) provides sufficient energy to launch a rocket. What is the molar mass of hydrogen peroxide?
Formula = H2O2
2 hydrogens and 2 oxygens

19. The decomposition of hydrogen peroxide (H2O2) provides sufficient energy to launch a rocket. What is the molar mass of hydrogen peroxide?
Formula = H2O2
2 hydrogens and 2 oxygens
Hydrogen = 1.01 g/mol
Oxygen = g/mol

20. The decomposition of hydrogen peroxide (H2O2) provides sufficient energy to launch a rocket. What is the molar mass of hydrogen peroxide?
Formula = H2O2
2 hydrogens and 2 oxygens
Hydrogen = 1.01 g/mol
Oxygen = g/mol
= g/mol

21. Find the molar mass of calcium nitrate.
Answer- 164 g/mol

22. Mole to Mass Relationship
Molar mass of a substance is the mass in grams per mole of that substance
Use the molar mass of an element or compound to convert between the mass and the moles
The units for molar mass are already set up as a conversion factor
Conversion factors: g
1 mol
1 mol g
Molar mass = OR

23. Moles to Mass or Mass to Moles (1 Step Problem)
Molar mass
Mass of substance
of substance

24. Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide?

25. Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide?
Solve for molar mass of Al2O3 :
2 Al = g/mol g/mol = g/mol
3 O = g/mol g/mol g/mol = g/mol
Total= g/mol g/mol = g/mol

26. Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide?
Solve for molar mass of Al2O3 :
2 Al = g/mol g/mol = g/mol
3 O = g/mol g/mol g/mol = g/mol
Total= g/mol g/mol = g/mol
Use dimensional analysis:
9.45 mol

27. Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide?
Solve for molar mass of Al2O3 :
2 Al = g/mol g/mol = g/mol
3 O = g/mol g/mol g/mol = g/mol
Total= g/mol g/mol = g/mol
Use dimensional analysis: g
9.45 mol
1 mol

28. Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide?
Solve for molar mass of Al2O3 :
2 Al = g/mol g/mol = g/mol
3 O = g/mol g/mol g/mol = g/mol
Total= g/mol g/mol = g/mol
Use dimensional analysis: g
g = g
9.45 mol
1 mol

29. Calculate the mass, in grams, of 2.50 mol of Iron (II) hydroxide.
Answer- 225 g
Formula then molar mass then dimensional analysis

30. When iron is exposed to the air, it corrodes to form red-brown rust
When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide?

31. Figure out formula for Iron (III) oxide
When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide?
Figure out formula for Iron (III) oxide
Fe O2-
Fe2O3

32. When iron is exposed to the air, it corrodes to form red-brown rust
When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide?
Figure out formula for Iron (III) oxide
Fe O2-
Fe2O3
Calculate molar mass
2(55.85 g/mol) + 3(16.00 g/mol) = g/mol

33. When iron is exposed to the air, it corrodes to form red-brown rust
When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide?
Figure out formula for Iron (III) oxide
Fe O2-
Fe2O3
Calculate molar mass
2(55.85 g/mol) + 3(16.00 g/mol) = g/mol
Use dimensional analysis
92.2 g

34. When iron is exposed to the air, it corrodes to form red-brown rust
When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide?
Figure out formula for Iron (III) oxide
Fe O2-
Fe2O3
Calculate molar mass
2(55.85 g/mol) + 3(16.00 g/mol) = g/mol
Use dimensional analysis
92.2 g
1 mol g

35. When iron is exposed to the air, it corrodes to form red-brown rust
When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide?
Figure out formula for Iron (III) oxide
Fe O2-
Fe2O3
Calculate molar mass
2(55.85 g/mol) + 3(16.00 g/mol) = g/mol
Use dimensional analysis
92.2 g
1 mol
0.577 mol Fe2O3 g

36. Calculate the number of moles in 75.0g of dinitrogen trioxide.
Answer mol N2O3
Formula to molar mass to mol

37. Number of RP (atoms, molecules, f.u.
Mass to Particles or Particles to Mass
Mass to atoms or molecules
A 2-step problem (2 steps)
Number of RP (atoms, molecules, f.u.
Molar mass
Moles of Substance
Avogadro’s
Mass of substance
number
of substance
**Hint – If you do not see moles in the problem,
it is a 2 step problem** (2 steps)

38. Summary of Mole Conversions
Use Avogadro’s Number
Use Periodic Table
MOLE
Representative
Particle
Mass
Calculate Molar Mass
g/mol
6.022 x 1023
In grams
Atom, Molecule, Formula Unit A B C
IF YOU DON’T SEE “MOLES” IN THE PROBLEM- IT IS A 2 STEP PROBLEM!

39. Calculate the number of formula units in 78.4g of sodium chloride.
This is a two step problem. First go from grams to moles, then moles to formula units.

40. Calculate the number of formula units in 78.4g of sodium chloride.
This is a two step problem. First go from grams to moles, then moles to formula units.
Sodium Chloride’s chemical formula is NaCl.
Its molar mass is: g/mol g/mol= g/mol

41. Calculate the number of formula units in 78.4g of sodium chloride.
This is a two step problem. First go from grams to moles, then moles to formula units.
Sodium Chloride’s chemical formula is NaCl.
Its molar mass is: g/mol g/mol= g/mol
1 mol NaCl
78.4 g NaCl
58.44 g NaCl

42. Calculate the number of formula units in 78.4g of sodium chloride.
This is a two step problem. First go from grams to moles, then moles to formula units.
Sodium Chloride’s chemical formula is NaCl.
Its molar mass is: g/mol g/mol= g/mol
1 mol NaCl
78.4 g NaCl
6.022 x 1023 fu
58.44 g NaCl
1 mol NaCl

43. Calculate the number of formula units in 78.4g of sodium chloride.
This is a two step problem. First go from grams to moles, then moles to formula units.
Sodium Chloride’s chemical formula is NaCl.
Its molar mass is: g/mol g/mol= g/mol
1 mol NaCl
78.4 g NaCl
6.022 x 1023 fu
8.34 x 1023 fu
58.44 g NaCl
1 mol NaCl

44. How many grams are in 94.8 molecules of SO2?
Answer:
64.06 is SO2 molar mass

45. Mole to Volume Relationship
Avogadro’s Law: equal volumes of gases at the same temperature and pressure contain equal numbers of particles
Volume of a gas can change with a change in temperature and pressure so it is usually measured at STP (standard temperature and pressure)
0oC (273 K) and 1 atm.
At STP: 1 mol= x 1023 particles = 22.4 L
22.4L is called the molar volume of a gas
Molar volume is used to convert between number of moles of a gas and the volume of the gas at STP

46. Conversion Factors for mol and volume ratio 1 mol = 22.4 L

47. Sulfur dioxide (SO2) is a gas produced by burning coal
Sulfur dioxide (SO2) is a gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters, of 0.60 mol of SO2 gas at STP.
Known from problem
0.60 mol

48. Conversion factor from gas at STP
Sulfur dioxide (SO2) is a gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters, of 0.60 mol of SO2 gas at STP.
0.60 mol
22.4 L
1 mol
Conversion factor from gas at STP

49. Sulfur dioxide (SO2) is a gas produced by burning coal
Sulfur dioxide (SO2) is a gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters, of 0.60 mol of SO2 gas at STP.
0.60 mol
22.4 L
13 L SO2
1 mol

50. What is the volume of 0.960 mol CH4 at STP?
At STP, how many moles are in 1.00 x 103 L of C2H6?
Answers L CH4
mol C2H6

51. The density of a gaseous compound containing carbon and oxygen is found to be g/L at STP. What is the molar mass?

52. The density of a gaseous compound containing carbon and oxygen is found to be g/L at STP. What is the molar mass?
I know:
Density = g/L
1 mol of gas at STP = 22.4L

53. The density of a gaseous compound containing carbon and oxygen is found to be g/L at STP. What is the molar mass?
I know:
Density = g/L
1 mol of gas at STP = 22.4L
1.964 g
1 L

54. The density of a gaseous compound containing carbon and oxygen is found to be g/L at STP. What is the molar mass?
I know:
Density = g/L
1 mol of gas at STP = 22.4L
22.4 L
1.964 g
1 L
1 mol

55. The density of a gaseous compound containing carbon and oxygen is found to be g/L at STP. What is the molar mass?
I know:
Density = g/L
1 mol of gas at STP = 22.4L
22.4 L
1.964 g
43.99 g/mol
1 L
1 mol

56. What is the density of krypton gas at STP?
Answer g/L
Look up Krypton’s molar mass (83.8 g/mol) and use 22.4 L

57. Summary of mole conversions
L, mL, etc….
Summary of mole conversions
atoms,
molecules,
formula units
g, kg, etc..

58. Percent Composition % by mass of element = mass of element x 100
Percent composition is the percent by mass of each element in a compound
Using the mass of the element you solve using the following formula
% by mass of element = mass of element x 100
mass of compound
Where you may have seen something like this before?

59. When a g sample of a compound containing only magnesium and oxygen is broken down, 5.40g of oxygen and 8.20 g of magnesium are obtained. What is the percent composition of this compound?

60. When a 13.60g sample of a compound containing only magnesium and oxygen is decomposed, 5.40g of oxygen is obtained. What is the percent composition of this compound?
Percent composition of oxygen:
5.40g x 100 = 39.7%
13.60g

61. When a 13.60g sample of a compound containing only magnesium and oxygen is decomposed, 5.40g of oxygen is obtained. What is the percent composition of this compound?
Percent composition of oxygen:
5.40g x 100 = 39.7%
13.60g
Percent composition of magnesium:
8.20g x 100 = 60.3%
13.60g

62. When a 13.60g sample of a compound containing only magnesium and oxygen is decomposed, 5.40g of oxygen is obtained. What is the percent composition of this compound?
Percent composition of oxygen:
5.40g x 100 = 39.7%
13.60g
Percent composition of magnesium:
8.20g x 100 = 60.3%
13.60g
The compound is 39.7% oxygen and 60.3% magnesium.

63. Percent Composition from a Chemical Formula
Percent by mass of an element
Molar mass of element =
x 100
Molar mass of compound

64. Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane.

65. Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane.
Calculate molar mass:
Carbon= 3 x g/mol = g/mol
Hydrogen= 8 x 1.01 g/mol = 8.08 g/mol
Total molar mass of propane = = g/mol

66. Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane.
Calculate molar mass:
Carbon= 3 x g/mol = g/mol
Hydrogen= 8 x 1.01 g/mol = 8.08 g/mol
Total molar mass of propane = = g/mol
Percent Carbon
(36.03 ÷ 44.11) x 100 = 81.8% Carbon

67. Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane.
Calculate molar mass:
Carbon= 3 x g/mol = g/mol
Hydrogen= 8 x 1.01 g/mol = 8.08 g/mol
Total molar mass of propane = = g/mol
Percent Carbon
(36.03 ÷ 44.11) x 100 = 81.68% Carbon
Percent Hydrogen
(8.08 ÷ 44.11) x 100 = 18.3% Hydrogen

68. Example: Hydrogen peroxide
Empirical Formulas
Empirical formulas are the lowest whole number ratio of the atoms or moles of elements in a compound
May or may not be the same as the molecular formula
Example: Hydrogen peroxide

69. Molecular Formula vs Empirical Formula
Either the same as the empirical formula or is a simple whole number multiple of the empirical formula
molecular formula empirical formula
H2O2 = HO
C6H12O6 =
N2O6 =
H2O =
C2H10O2 =
N3O2 =
P4O10 =

70. The percent composition of a compound can be used to calculate the empirical formula of that compound
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?

71. A compound is analyzed and found to contain 25. 9% nitrogen and 74
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?
What we know:
Nitrogen = 25.9%
Oxygen = 74.1%

72. A compound is analyzed and found to contain 25. 9% nitrogen and 74
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?
What we know:
Nitrogen = 25.9%
Oxygen = 74.1%
Step 1:
Assume 100 g and calculate moles of each, so simply change % to mass (g) for each
Nitrogen = 25.9g
Oxygen = 74.1g

73. A compound is analyzed and found to contain 25. 9% nitrogen and 74
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?
What we know:
Nitrogen = 25.9%
Oxygen = 74.1%
Step 1:
Assume 100 g and calculate moles of each, so simply change % to mass (g) for each
Nitrogen = 25.9g
Oxygen = 74.1g
Step 2: Convert mass to moles of each
Moles of Nitrogen
25.9 g
14.0 g
1 mol
1.85 mol N
Moles of Oxygen
74.1 g
16.0 g
1 mol
4.63 mol O

74. A compound is analyzed and found to contain 25. 9% nitrogen and 74
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?
What we know:
Nitrogen = 25.9%
Oxygen = 74.1%
Assume 100 g and calculate moles of each
Step 3: Divide each molar quantity by the smallest number of moles to get 1 mole for the element with the smallest number
Moles of Nitrogen
25.9 g
14.0 g
1 mol
1.85 mol N
1.85 mol = 1 mol N
1.85 mol
4.63 mol = 2.5 mol O
Moles of Oxygen
74.1 g
16.0 g
1 mol
4.63 mol O

75. A compound is analyzed and found to contain 25. 9% nitrogen and 74
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?
What we know:
Nitrogen = 25.9%
Oxygen = 74.1%
Assume 100 g and calculate moles of each
Divide each molar quantity by the smallest number of moles to get 1 mole for the element with the smallest number
Moles of Nitrogen
25.9 g
14.0 g
1 mol
1.85 mol N
1.85 mol ÷ 1.85 mol = 1 mol N
4.63 mol ÷ 1.85 mol = 2.5 mol O
Moles of Oxygen
Step 4: Multiply each by the lowest whole number that converts both to a whole number (if necessary)
74.1 g
16.0 g
1 mol
4.63 mol O
1 mol N x 2 = 2 mol N
2.50 mol O x 2 = 5 mol O

76. A compound is analyzed and found to contain 25. 9% nitrogen and 74
A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?
What we know:
Nitrogen = 25.9%
Oxygen = 74.1%
Assume 100 g and calculate moles of each
Divide each molar quantity by the smallest number of moles to get 1 mole for the element with the smallest number
Moles of Nitrogen
25.9 g
14.0 g
1 mol
1.85 mol N
1.85 mol ÷ 1.85 mol = 1 mol N
4.63 mol ÷ 1.85 mol = 2.50 mol O
Moles of Oxygen
Multiply each by the lowest whole number that converts both to a whole number
74.1 g
16.0 g
1 mol
4.63 mol O
1 mol N x 2 = 2 mol N
2.50 mol O x 2 = 5 mol O
Step 5: Write empirical formula: N2O5

77. Calculate the empirical formula for a compound that is 67. 6% Hg, 10
Calculate the empirical formula for a compound that is 67.6% Hg, 10.8% S, and 21.6% O.
Answer- HgSO4
Change % to grams
Calculate moles based on molar masses
Divide by lowest number of moles
Multiply to get each to a whole number

78. Calculate the molecular formula of a compound whose molar mass is 60
Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N.

79. Calculate the empirical formula mass
Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N.
Calculate the empirical formula mass
C= 12.0 g/mol
H= 1.0g/mol x 4= 4.0 g/mol
N= 14.0 g/mol
Total= 30.0 g/mol

80. Calculate the molecular formula of a compound whose molar mass is 60
Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N.
Calculate the empirical formula mass
Divide the molar mass of the compound by the empirical formula mass
60.0 g/mol
30.0 g/mol
C= 12.0 g/mol
H= 1.0g/mol x 4= 4.0 g/mol
N= 14.0 g/mol 2 =
Total= 30.0 g/mol

81. Calculate the molecular formula of a compound whose molar mass is 60
Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N.
Calculate the empirical formula mass
Divide the molar mass of the compound by the empirical formula mass
60.0 g/mol
30.0 g/mol
C= 12.0 g/mol
H= 1.0g/mol x 4= 4.0 g/mol
N= 14.0 g/mol 2 =
Multiply the empirical formula by this value to figure out the molecular formula
Total= 30.0 g/mol
CH4N x 2 =

82. Calculate the molecular formula of a compound whose molar mass is 60
Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N.
Calculate the empirical formula mass (efm)
Divide the molar mass of the compound by the efm
60.0 g/mol
30.0 g/mol
C= 12.0 g/mol
H= 1.0g/mol x 4= 4.0 g/mol
N= 14.0 g/mol 2 =
Multiply the empirical formula by this value to figure out the molecular formula
Total= 30.0 g/mol
CH4N x 2 =
C2H8N2

83. Find the molecular formula of ethylene glycol, which is used as antifreeze. The molar mass is 62.0 g/mol, and the empirical formula is CH3O.
Answer- C2H6O2