1. UNIT 6
THE MOLE
Mole Video

2. Atoms are really, really small……
We can not work with individual atoms or AMU’s in the lab.
Atomic Mass Unit
WHY?
Because we can’t see things that small!

3. We as scientists work with portions of matter large enough for us to SEE and MASS on a balance using units of……
gramS

4. This presents a problem…..
So how would we keep track of that many atoms?
A pile of atoms large enough for us to see contains billions of atoms.
Copper (II) Sulfate

5. Chemists came up with a new unit.
The MOLE

6. Unit 6: The Mole
In Chemistry, the mole is a unit of measurement for an amount of a chemical substance
It’s a counting number
1 mol = 6.02  1023 = Avogadro’s Number
(mole is abbreviated mol…so much quicker to write)

7. The Mole 1 dozen eggs = 12 eggs 1 mole of eggs = 6.02 x 1023 eggs
1 ream of paper = 500 pieces of paper
1 mole of paper = 6.02 x paper

8. I did not discover the number. It was just named after me.
Avogadro’s Number
6.02 x 1023
He studied gases and discovered that no matter what the gas, there were the same number of molecules present.
Named in honor of
Amadeo Avogadro.

9. Units for Avogadro’s Number
For instance:
1 mole of pennies =
6.02 x 1023 pennies.
1 mole = 6.02 x 1023 (many different units)
This amount is equivalent to 7 stacks of pennies from the Earth to the moon.

10. 4 Different Units/ Particles
1) Atoms  Elements 2) Ions  +/- charge 3) Formula Units  Ionic Compounds/Acids 4) Molecules  Covalent Compounds/Diatomics

11. Examples 1 mole Mg = 6.02 x 1023 atoms (elements)
1 mole SO4-2 = x ions (has a +/-charge)
1 mole NaCl = 6.02 x formula units (ionic compounds)
1 mole Cl2 = 6.02 x molecules (covalent compounds)
(Diatomics = H2 , O2 , N2 , F2 , Cl2 , Br2 , I2)

12. Molar Mass The mass of 1 mole of an element or compound
The units are g/mol
To calculate the molar mass of a substance you find the sum of the masses for that substance
The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance.
The atomic mass of Fe is amu, so the molar mass of Fe is g/mol (1 mol = g)
Since oxygen occurs naturally as a diatomic, O2, the molar mass of oxygen gas is 2 times g or g/mol.

13. Calculating Molar Mass
Calculate the molar mass of NaCl. 6 Na
22.99 17 Cl
35.45
22.99 g g = g/mol
1 mole of NaCl (6.02 x 1023 formula units) has a mass of g

14. Calculating Molar Mass
Calculate the molar mass of K2O. 19 K
39.10 6 O
16.00
2(39.10) g g = g/mol
1 mole of K2O (6.02 x 1023 formula units) has a mass of g

15. Calculating Molar Mass
Calculate the molar mass of (NH4)2SO4. 7 N
14.01 1 H
1.02 16 S
32.07 6 O
16.00
2(14.01) g + 8(1.01) g + (32.07) + 4(16.00) g =
g/mol
1 mole of (NH4)2SO4 (6.02 x 1023 formula units) has a mass of g

16. Welcome to Mole Island 1 mol = molar mass 1 mole = 22.4 L @ STP
6.02 x 1023 particles

18. Mole – Particle Conversion
How many atoms are in moles of Iron (Fe)?
5. Use conversion factor that will allow you to “cancel” out moles and give you atoms.
Cancel out units and calculate using
significant digits.
4. Start with your “given” from the original problem.
Ask yourself the question – Can I convert directly from moles to atoms?
YES – there are
6.02 x 1023 atoms = 1 mole
3. Form conversion factors.
1 m0le
6.02 x 1023 atoms
6.02 x 1023 atoms
1 mole
6.02 x 1023 Fe atoms
10.0 moles Fe 1 x =
6.02 x 1024 Fe atoms
1 mole Fe

19. Mole – Particle Conversions
How many moles of potassium are in 1.25 × 1021 atoms K?
Step 1: we want moles K
Step 2: we have 1.25 × 1021 atoms K
Step 3: 1 mole K = 6.02 × 1023 atoms K
= 2.08 × 10-3 mol K
1.25 × 1021 atoms K ×
1 mol K
6.02 × 1023 atoms K

20. Mole – Particle Conversion
How many molecules are in 2.50 moles of C12H22O11?
6.02  1023
molecules
1 mol
2.50 mol
= 1.51  1024
molecules
C12H22O11

21. Mole – Particle Conversions
6.02 x “particles”
1 mole OR
1 mole
6.02 x 1023 “particles”
Particle unit = atom, ion, formula unit, or molecule

23. Mole – Mass Conversions
How many moles are in 26 g of carbon?
26 g C
1 mol C
12.01 g C
= 2.2 mol C

24. Mole – Mass Conversions
How many grams are in 3.00 moles of MgCl2?

25. Mole-Volume Conversion
A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present?
We want moles, we have volume.
Use molar volume of a gas: 1 mol = 22.4 L
4.50 L CH4 ×
= mol CH4
1 mol CH4
22.4 L CH4

26. Mole Unit Factors We now have three interpretations for the mole:
1 mol = 6.02 × 1023 particles
1 mol = molar mass
1 mol = 22.4 L at STP for a gas
This gives us 3 unit factors to use to convert between moles, particles, mass, and volume.

27. QOD #5 What is the mass of 1.33 moles of titanium, Ti?
We want grams, we have 1.33 moles of titanium.
Use the molar mass of Ti: 1 mol Ti = g Ti
= 63.7 g Ti
1.33 mole Ti ×
47.87 g Ti
1 mole Ti

29. 2-Step Conversion Examples
Find the mass of 2.1  1024 molecules of NaHCO3.
2.1  1024
molecules
1 mol
6.02  1023
molecules
84.01 g
1 mol
= 290 g NaHCO3

30. 2 – Step Conversions
How many O2 molecules are present in g of oxygen gas?
We want molecules O2, we have grams O2.
Use Avogadro’s number and the molar mass of O2
0.470 g O2 ×
1 mol O2
32.00 g O2
6.02×1023 molecules O2
1 mole O2 ×
8.84 × 1021 molecules O2

31. QOD #6 What is the molar mass of aluminum phosphate?
Identify the representative particle of the following:
S NO3
CaCl2 CO32-
3. What volume would be in moles of carbon dioxide?

32. Mass-Volume Calculation
What is the mass of 3.36 L of ozone gas, O3, at STP?
We want mass O3, we have 3.36 L O3.
Convert volume to moles then moles to mass:
3.36 L O3 × ×
22.4 L O3
1 mol O3
48.00 g O3
= 7.20 g O3

33. Molecule-Volume Calculation
How many molecules of hydrogen gas, H2, occupy L at STP?
We want molecules H2, we have L H2.
Convert volume to moles and then moles to molecules:
0.500 L H2 ×
1 mol H2
22.4 L H2
6.02×1023 molecules H2
1 mole H2 ×
= 1.34 × 1022 molecules H2

34. Law of Definite Composition
The law of definite composition states that “Compounds always contain the same elements in a constant proportion by mass”.
Sodium chloride is always 39.3% sodium and 60.7% chlorine by mass, no matter what its source.
Water is always 11.2% hydrogen and 88.8% oxygen by mass.

35. Law of Definite Composition
Figure: 04-08
Title:
Law of Definite Composition
Caption:
A drop of water, a glass of water, and a lake of water all contain hydrogen and oxygen in the same percent by mass, that is, 11.2% hydrogen and 88.8% oxygen.
Notes:
The Law of Definite Composition states that no matter its source, a compound contains the same elements in the same percents by mass.
A drop of water, a glass of water, and a lake of water all contain
hydrogen and oxygen in the same percent by mass.

36. Percent Composition
The percent composition of a compound lists the mass percent of each element.
For example, the percent composition of water, H2O is:
11% hydrogen and 89% oxygen
All water contains 11% hydrogen and 89% oxygen by mass.

37. Percent Composition
Find the molar mass (total mass) of the compound/sample
Divide the mass of each element by the molar mass (total mass)
Multiply by 100 to make it a %

38. Calculating Percent Composition
There are a few steps to calculating the percent composition of a compound. Lets practice using H2O.
Assume you have 1 mole of the compound.
One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen.
2(1.01 g H) + 1(16.00 g O) = molar mass H2O
2.02 g H g O = g H2O

39. Calculating Percent Composition
Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water:
2.02 g H
18.02 g H2O
× 100% = 11.2% H
16.00 g O
18.02 g H2O
× 100% = 88.79% O

40. Percent Composition Problem
TNT (trinitrotoluene) is a white crystalline substance that explodes at 240°C. Calculate the percent composition of TNT, C7H5(NO2)3.
7(12.01 g C) + 5(1.01 g H) + 3 (14.01 g N g O) = g C7H5(NO2)3
84.07 g C g H g N g O = g C7H5(NO2)3.

41. Percent Composition of TNT
84.07 g C
g TNT
× 100% = 37.0% C
1.01 g H
g TNT
× 100% = 2.2% H
42.03 g N
g TNT
× 100% = 18.5% N
96.00 g O
g TNT
× 100% = 42.3% O

42. Calculating Mass of an Element
Remember % Composition = 𝑃𝑎𝑟𝑡 𝑊ℎ𝑜𝑙𝑒 X 100 What is the mass of hydrogen in 378 g H2O? H = 1.01 𝑥 𝑔 = 2.02 𝑔 𝐻 𝑔 𝐻2𝑂 So, set-up a proportion: 2.02 𝑔 𝐻 𝑔 𝐻2𝑂 = ? 𝑔 𝐻 378 𝑔 𝐻2𝑂 2.02 𝑥 = 42.4 g H

43. Empirical Formulas
The empirical formula of a compound is the simplest whole number ratio of atoms of each element in a compound.
The molecular formula of benzene is C6H6
The empirical formula of benzene is CH.
The molecular formula of octane is C8H18
The empirical formula of octane is C4H9.

44. How to Calculate Empirical Formulas
1. Determine the mass (g) of each element present (this is usually given).
2. Calculate the number of moles of each element (convert grams  moles).
Divide each mole value by the smallest number of moles to obtain the simplest whole number ratio.
Use whole number to write empirical formula (whole # = subscript in formula).

45. Example Problem
A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine the empirical formula. 2.34g N x 1 mol = moles N = g g O x 1 mol = moles O = g Formula:

46. Empirical Formulas from Percent Composition
We can also use percent composition data to calculate empirical formulas.
Assume that you have 100 grams of sample.
Benzene is 92.2% carbon and 7.83% hydrogen, what is the empirical formula.
If we assume 100 grams of sample, we have g carbon and 7.83 g hydrogen.

47. Empirical Formulas from Percent Composition
Calculate the moles of each element:
1 mol C
12.01 g C
92.2 g C ×
= mol C
1 mol H
1.01 g H
7.83 g H ×
= mol H
The ratio of elements in benzene is C7.677H7.752 Divide by the smallest number to get the formula.
7.677 C
= C1.00H1.01 = CH
7.752 H

48. Another Example
A g sample of radium metal was heated to produce g of radium oxide. What is the empirical formula (Ra?O?)?
We have g Ra and = g O
Formula = RaO
1 mol Ra
g Ra
1.640 g Ra ×
= mol Ra = 1
0.007
1 mol O
16.00 g O
0.115 g O ×
= mol O = 1

49. Molecular Formula
The formula that states the actual number of each kind of atom found in one molecule of the compound.
Covalent compounds have molecular formulas

50. Molecular Formula 𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝐸.𝐹. 𝑀𝑎𝑠𝑠 = ?
You have to find the empirical formula before getting the molecular formula!!!
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 𝐸.𝐹. 𝑀𝑎𝑠𝑠 = ?
Multiply the answer by the subscripts in the empirical formula to get the molecular formula

51. Calculating Molecular Formula
Find empirical formula mass (same thing as calculating molar mass)
Find molar mass (g/mol) – this is usually given in the problem
Divide molar mass by empirical formula mass and get a whole number
Multiply each subscript in the empirical formula by the whole number. This is your molecular formula.

52. Example Problem
A compound has an empirical formula of NO2. The colorless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?

53. A compound has 30.43% nitrogen and
69.57% oxygen and has a molecular molar
mass of 92.0 g/mol. What is its molecular
formula?
A compound has 15.8% carbon and 84.2% sulfur. What is its molecular formula if its molecular molar mass is 228 g/mol?
A 0.238g sample of carbon combines with a 0.634g sample of oxygen. What is the molecular formula if the molecular molar mass is 44.0 g/mol?

54. Hydrate Videos
Hydrates