1. Magnet Chemistry Phillips 2016
The Mole
Magnet Chemistry Phillips 2016

2. Can you imagine a mole of donuts?
Measurements
Dozen = 12
Baker’s dozen = 13
Gross= 144
Mole (mol) = 6.02x1023
A mole of water molecules has a volume of only 18 mL!
Molecules are unbelievably small, so a lot of them doesn’t take up that much space.
The mole is useful in chemistry because it links the microscopic world of atoms, molecules, and ions, to the macroscopic world
Can you imagine a mole of donuts?

3. Avogadro’s Number
6.02x1023 is so important in chemistry it’s given its own name…Avogadro’s Number
In numerical form it looks like this:
602,000,000,000,000,000,000,000
A mole of paper would reach past Pluto
A mole of basketballs is the size of Earth
A mole of rice would cover the land masses of Earth to a depth of 75 meters!

4. Scientific Notation Practice
Using the mole requires that we are comfortable using scientific notation
Write the following in scientific notation
Write the following in standard notation
6000
0.0067
78,000,000
698,700
6.00x103
6.70x10-3
7.80x107
6.99x105
9.00x10-6
9.01x10-3
6.02x104
9.03x109
7.77x10-2
9.21x10-7
60,200
9,030,000,000
0.0777

5. The mole
The mole establishes a relationship between the atomic mass and the gram. Individual atoms and molecules are too small to be directly measured, so we refer to them using the unit amu. Using moles allows direct measurement.
The mass in grams of 1 mole of a substance is equal to its atomic mass.
6.02x1023 atoms Cu = g
6.02x1023 atoms H = g
6.02x1023 atoms Fe = g
Mole was based on Carbon. How many g of C are in 1 mol?

6. What’s in a Mole?
A mole of particles in an element is usually talking about atoms.
The number of molecules in a mole of any molecular compound is 6.02x1023
How many atoms are in a molecule of ammonia (NH3)?
4 atoms
1 N atom and 3 H atoms
Each molecule has 4 atoms
So 1 mole of ammonia gas contains 1 mole of NH3 molecules, but four times as many atoms, or 4 moles of atoms (1 mol N atoms and 3 mol H atoms)

7. Mole Conversions
Chemists measure amounts of substances by volume or mass.
Because the mole measures both a mass and a number of particles (and volume, but that will be covered later), it is the central unit in converting the amount of substance from one type of measurement to another.

8. Moles to Molecules
Molecules OR atoms
The number of particles in 1 mole of any substance is always the same- Avogadro’s number. or
How many atoms are in 3 mol of elemental Ne?
6.02x1023 particles
1 mol
1 mol
6.02x1023 particles
3 mol Ne 1
6.02x1023 atoms
1 mol
1.81x1024 atoms Ne

9. Unit Cancellation/ Dimensional Analysis Method (sometimes called ‘unit analysis’)
Write down units asked for in answer to the right
Write down the given value over 1 on the left
Apply one or more unit factors to cancel units
It’s as easy as 1-2-3!

10. Moles to Molecules (cont.)
PARTICLES
How many atoms of oxygen are in 6 mol of O2 molecules?
How many moles of I2are in 8.02x1020 molecules of I2?
Atoms Molecules?
Use the subscript
6.02x1023 particles
1 mol
Avogadro’s Number
1 mol
6.02x1023 particles
MOLES
6 mol O2 1
6.02x1023 molecules O2
1 mol O2
2 atoms O
1 molecule O2
7.22x1024 atoms O
1 mol I2
6.02x1023 molecules I2
8.02x1020 molec. I2 1
1.33x10-3 mol I2

11. Molar Mass
The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance.
The atomic mass of iron is amu.
Therefore, the molar mass of iron is g/mol.
Since oxygen occurs naturally as a diatomic, O2, the molar mass of oxygen gas is 2 times g or g/mol.

12. Calculating Molar Mass
The molar mass of a substance is the sum of the molar masses of each element and expressed as g/mol.
What is the molar mass of magnesium nitrate, Mg(NO3)2?
The sum of the atomic masses is:
( ) =
(62.01) = amu
The molar mass for Mg(NO3)2 is g/mol.

13. Molar Mass Practice Ne = 20.18 g/mol O2 = 16.00*2 = 32.00 g/mol
U = g/mol
NaOH = = g/mol
CO2 = *2 = g/mol
Al2(CO3)3 = 26.98*2 + (12.01*3) + (16.00*9) = g/mol

14. Moles to Grams Conversions
If you know the mass of a substance, you can calculate the number of moles.
You have 11.2 g of NaCl. How many moles is that?
1. determine the molar mass of NaCl (using the periodic table)
Na= 23.0 g/mol Cl= 35.5 g/mol
23.0 g/mol g/mol = 58.5 g/mol (get these #’s from the periodic table)

15. Moles to Grams Conversions
Molar mass of NaCl = 58.5 g/mol
Set up a conversion factor (a fraction whose value is equal to 1)
The units we need should be in the numerator and the units you already know in the denominator.

16. Moles to Grams Conversions
You can also determine the mass of a sample if you know the number of moles
2.50 mol of NaCl is how many grams?
Set up a conversion factor 

17. Moles to Grams Conversions
MASS
How many moles are in 14 g LiOH?
How many moles are in 15 g N2?
moles x g = g mol
Molar Mass
g x mol = moles g
MOLES
14 g LiOH 1
1 mol LiOH
23.95 g LiOH
0.58 mol LiOH
15 g N2 1
1 mol N2
28.02 g N2
0.54 mol N2

18. Moles to Grams Conversions
How many grams are in 4 mol H2O2?
How many grams are in 56 mol CaCO3?
4 mol H2O2 1
g H2O2
1 mol H2O2
g H2O2
56 mol CaCO3 1
g CaCO3
1 mol CaCO3
g CaCO3

19. Mole Calculations II = 87.8 g Pb
What is the mass of 2.55 × 1023 atoms of lead?
We want grams, we have atoms of lead.
Use Avogadro’s number and the molar mass of Pb
2.55 × 1023 atoms Pb ×
1 mol Pb
6.02×1023 atoms Pb
207.2 g Pb
1 mole Pb ×
= 87.8 g Pb

20. Mole Calculations II 8.84 × 1021 molecules O2
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

21. Multistep Conversions
You want to impress your date by boasting you know how many molecules of table sugar are in the cake you just made. You need 250 g of sugar (C12H22O11). How many sucrose molecules will be in the cake?
Plan: Convert the mass to moles using the molar mass and then convert to moles using Avogadro’s number.
MASS
PARTICLES
Avogadro’s Number
Use molar mass
MOLES
250 g C12H22O11 1
1 mol C12H22O11
342.3 g C12H22O11
6.02x1023 molecules
1 mol C12H22O11
4.4x1023 molec. C12H22O11

22. Multistep Conversions
If you burned 4.0x1024 molecules of methane (CH4) during a laboratory experiment, what mass of methane did you use?
Plan: Convert your given # of molecules to moles using Avogadro’s number, then convert the moles to grams using the molar mass of methane. ( *4)
4.0 x 1024 molec. CH4 1
1 mol CH4
6.02x1023 molec. CH4
g CH4
1 mol CH4
g CH4

23. Moles and Gases
At the same temperature and pressure, equal volumes of gases contain the same number of gas particles.
1 mole of any gas at 0oC and 1 atm (Standard Temperature and Pressure; STP) has a volume of L.
This volume, 22.4 L/mol, is called molar volume.

24. Gas Density- not on this test
The density of gases is much less than that of liquids.
We can calculate the density of any gas at STP easily.
The formula for gas density at STP is:
= density, g/L
molar mass in grams
molar volume in liters

25. Calculating Gas Density not on this test
What is the density of ammonia gas, NH3, at STP?
First we need the molar mass for ammonia;
(1.01) = g/mol
The molar volume NH3 at STP is 22.4 L/mol.
Density is mass/volume:
= g/L
17.04 g/mol
22.4 L/mol

26. Molar Mass of a Gas 1.96 g 1.00 L 22.4 L 1 mole × = 43.9 g/mol
We can also use molar volume to calculate the molar mass of an unknown gas.
1.96 g of an unknown gas occupies 1.00L at STP. What is the molar mass?
We want g/mol, we have g/L.
1.96 g
1.00 L
22.4 L
1 mole ×
= g/mol

27. Mole Unit Factors 1 mol = 6.02 × 1023 particles 1 mol = molar mass
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.

28. Mole-Volume Calculation
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

29. 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

30. 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

31. Moles and Gases: A Summary
MASS
PARTICLES
You try it: A student fills a 1.0-L flask with CO2 at STP. How many molecules of gas are in the flask?
Plan: 1) Convert from volume to moles using molar volume. 2) Then convert from moles to molecules using Avogadro’s number.
Molar
Mass
Avogadro’s Number
MOLES
Molar Volume (22.4 L/mol)
VOLUME of gases at STP

32. Moles and Gases Let’s see how you did…
A chemical reaction produces 0.82 mole of oxygen gas. What volume will that gas occupy at STP?
1.0-L CO2 1
1 mol CO2
22.4 L CO2
6.02 x 1023 molec CO2
1 mol CO2
2.7 x molec. CO2
0.82 mol O2 1
22.4 L O2
1 mol O2
18 L O2

33. 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.

34. 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.

35. 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.

36. 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

37. 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

38. Percent Composition Problem
TNT (trinitrotoluene) is a white crystalline substance that explodes at 240°C. Calculate the percent composition of TNT, C7H5(NO2)3.

39. Solution Percent Composition Problem
TNT (trinitrotoluene) is a white crystalline substance that explodes at 240°C. Calculate the percent composition of TNT, C7H5(NO2)3.
First, find molar mass of TNT:
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.

40. Percent Composition of TNT
84.07 g C
g TNT
× 100% = 37.01% C
1.01 g H
g TNT
× 100% = 2.22% H
42.03 g N
g TNT
× 100% = 18.50% N
96.00 g O
g TNT
× 100% = 42.26% O

41. Try one more: Percent Composition
What is the percent carbon in C5H8NO4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats?
a) %C
b) %C
c) %C

42. Chemical Formulas of Compounds
Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions).
NO atoms of O for every 1 atom of N
1 mole of NO2 : 2 moles of O atoms to every 1 mole of N atoms
If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

43. Types of Formulas Empirical Formula
The formula of a compound that expresses the smallest whole number ratio of the atoms present.
*Ionic formula is always empirical formula
Molecular Formula
The formula that states the actual number of each kind of atom found in one molecule of the compound.

44. To obtain an Empirical Formula
1. Determine the mass in grams of each element present, if necessary.
2. Calculate the number of moles of each element.
3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.
If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers
* Be careful! Do not round off numbers prematurely

45. Sol’n: require mole ratios, so convert grams to moles
A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance.
Sol’n: require mole ratios, so convert grams to moles
moles of N = 2.34g of N = moles of N
14.01 g/mole
moles of O = g = moles of O
16.00 g/mole
Formula:
(HONORS only)

46. Calculation of the Molecular Formula
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?

47. Empirical Formula from % Composition
A substance has the following composition by mass: % Na ; % B ; % H
What is the empirical formula of the substance?
Consider a sample size of 100 grams
This will contain grams of B and grams H
Determine the number of moles of each
Determine the simplest whole number ratio

48. More Empirical & Molecular Formulas
The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule.
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.

49. Another Exp: Empirical Formulas
We can calculate the empirical formula of a compound from its composition data.
We can determine the mole ratio of each element from the mass to determine the formula Ex: radium oxide, Ra?O?.
A g sample of radium metal was heated to produce g of radium oxide. What is the empirical formula?
We have g Ra and = g O.

50. Calculating Empirical Formulas
The molar mass of radium is g/mol and the molar mass of oxygen is g/mol.
1 mol Ra
g Ra
1.640 g Ra ×
= mol Ra
1 mol O
16.00 g O
0.115 g O ×
= mol O
We get Ra O Simplify the mole ratio by dividing by the smallest number.
We get Ra1.01O1.00 = RaO is the empirical formula.

51. Another Exp: 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.

52. Empirical Formulas from Percent Composition
Calculate the moles of each element:
1 mol C
12.01 g C
92.2 g C ×
= 7.68 mol C
1 mol H
1.01 g H
7.83 g H ×
= 7.75 mol H
The ratio of elements in benzene is C7.68H7.75. Divide by the smallest number to get the formula.
7.68 C
= C1.00H1.01 = CH
7.75 H

53. Another Exp: Molecular Formulas
The empirical formula for benzene is CH. This represents the ratio of C to H atoms of benzene.
The actual molecular formula is some multiple of the empirical formula, (CH)n.
Benzene has a molar mass of 78 g/mol. Find n to find the molecular formula. = CH
(CH)n
78 g/mol
13 g/mol
n = 6 and the molecular formula is C6H6.

54. Conclusions
Avogadro’s number is 6.02 × 1023 and is one mole of any substance.
The molar mass of a substance is the sum of the atomic masses of each element in the formula.
At STP, 1 mole of any gas occupies 22.4 L.

55. Conclusions Continued
We can use the following flow chart for mole calculations:

56. Conclusions Continued
The percent composition of a substance is the mass percent of each element in that substance.
The empirical formula of a substance is the simplest whole number ratio of the elements in the formula.
The molecular formula is a multiple of the empirical formula.