2. The Mole (mol)
The mole is a very large number used for counting particles.
6.022 x 1023 of anything is a mole
Defined as the number of atoms in exactly 12 grams of carbon-12.
6.022 x 1023 is called Avogardro’s Number

3. Representative Particles
ATOMS are the smallest particle of an ELEMENT
MOLECULES are the smallest particle of MOLECULAR COMPOUNDS (Covalent)  
FORMULA UNITS are the smallest particle of IONIC COMPOUNDS
1 mol = x 1023 atoms, molecules, formula units

4. Mole Conversions Example 1:
Convert 1.35 moles of carbon disulfide(CS2) into molecules?

5. Mole Conversions Example 2:
How many moles are in 3.59 x 1021 formula units of sodium nitrate, NaNO3

6. Molar Mass (molecular weight)
Mass of 1 mole of a substance.
Represented by M & has the units of grams/mol
To determine the molar mass of an element, look at the element’s atomic mass from the periodic table.
To determine the molar mass of a compound, add up the atomic masses of the elements that make it up.

7. Find the Molar Mass of A single carbon atom. 1 mole of carbon atoms
1 mole of chlorine
12.01 amu
12.01 grams
35.45 x 2 = grams

8. Examples Example 6: What is the molar mass of sulfur dioxide?
Example 7: Calculate the molar mass of Al2O3
2 (26.98) + 3(16.00) = g/mol

9. Mole Conversions 1 mol = molar mass (g) Example 8:
Example 8:
How many moles of calcium carbonate are in a stick of chalk with a mass of 14.8 g.

10. Mole Conversions 1 mol = molar mass (g) Example 9:
Example 9:
How many atoms are in g of potassium ?

11. Mole Conversions 1 mol = molar mass (g) Example 10:
Example 10:
How many hydrogen atoms are in 72.5 g of C3H8O ?

12. Gases & The Mole Many of the chemicals we deal with are gases.
They are difficult to weigh.
Need to know how many moles of gas we have.
Two things effect the volume of a gas
Temperature and pressure
Compare at the same temp. and pressure.

13. Standard Temperature and Pressure
0 ºC and 1 atm pressure
abbreviated STP
At STP 1 mole of gas occupies 22.4 L
Called the molar volume
Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

14. Mole Conversions 1 mol = 22.4 L Example 11
Example 11
What is the volume of 4.59 mole of CO2 gas at STP?

15. Percent Composition Percent of each element a compound is composed of.
Find the mass of each element, divide by the total mass, multiply by a 100.
Easiest if you use a mole of the compound.

16. Example 12: Calculate the mass percent of carbon in C2H6O
So the percentage of carbon in ethane is…
%C =
(2)(12.01 amu)
(46.08 amu)
24.02 amu
46.08 amu =
x 100
= 52.1 %
© 2009, Prentice-Hall, Inc.

17. Calculate the % composition of water in the hydrate, (CaSO4 · 2 H2O )
Example 13:
Calculate the % composition of water in the hydrate, (CaSO4 · 2 H2O )
Ca 1 x g = g
Cl x g = g g/ g
H2O 2 x 18.02g = g
24.50% H2O
Copyright © Cengage Learning. All rights reserved

18. Emprical Formula
From percent composition, you can determine the empirical formula.
Empirical Formula the lowest ratio of atoms in a molecule.
Based on mole ratios.

19. Calculating Empirical Formulas
If you are given a % composition, assume you have 100 grams and switch the % unit to g. If you are given the actual number of grams of each element, use those instead
Convert the grams of each element to moles of each element.
Divide each mole amount by the smallest mole amount.
If your answers are within .1 of whole number, round up or down and these numbers become your subscripts.
If your answers are not within .1 of a whole number, you must find the smallest number that you can multiply by all the mol amounts you have just calculated to get them to be all whole numbers or within .1 of a whole number.
© 2009, Prentice-Hall, Inc.

20. Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.
© 2009, Prentice-Hall, Inc.

21. Calculating Empirical Formulas
Assuming g of para-aminobenzoic acid,
C: g x = mol C
H: g x = 5.09 mol H
N: g x = mol N
O: g x = mol O
1 mol
12.01 g
14.01 g
1.01 g
16.00 g
© 2009, Prentice-Hall, Inc.

22. Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles:
C: =  7
H: =  7
N: = 1.000
O: =  2
5.105 mol
mol
5.09 mol
1.458 mol
© 2009, Prentice-Hall, Inc.

23. Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
© 2009, Prentice-Hall, Inc.

24. Calculating Empirical Formulas
Example 14
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.
© 2009, Prentice-Hall, Inc.

25. Calculating Empirical Formulas
Example 15
A compound has the following percent composition: 31.9%K, 29.0 % Cl and 39.1%O. What is the empirical formula?
© 2009, Prentice-Hall, Inc.

26. Extra Practice
A substance has the following composition by mass: % Na ; % B ; % H. What is the empirical formula of the substance?
© 2009, Prentice-Hall, Inc.

27. Empirical To Molecular Formulas
Empirical is lowest ratio.
Molecular is actual molecule.
Need Molar mass.
Ratio of empirical to molar mass will tell you the molecular formula.
Must be a whole number because...

28. Example 16
A compound is made of only sulfur and nitrogen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its formula?

29. Exercise
The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol.
What is the empirical formula?
C3H5O2
What is the molecular formula?
C6H10O4
Assume 100.0g g of carbon is mol C (49.3/12.01) g of hydrogen is mol H (6.9/1.008) g of oxygen is mol O (43.8/16.00). The ratio of C:H:O is 1.5:2.5:1 (C: 4.105/ = 1.5; H: 6.845/ = 2.5). Multiplying each by 2 to get a whole number becomes a ratio of 3:5:2. The empirical formula is therefore C3H5O2. The molar mass of the empirical formula is g/mol, which goes into the molar mass of the molecular formula 2 times (146/73.07). The molecular formula is therefore C6H10O4.
Copyright © Cengage Learning. All rights reserved

30. Sample is burned completely to form CO2 and H2O
Pure O2 in
CO2 is absorbed
Sample is burned completely to form CO2 and H2O
H2O is absorbed
Combustion Analysis

31. How To:
Use the mass of CO2 to covert to moles of CO2 to moles of C to grams of C.
Use the mass of H2O and convert to moles of H2O to mole H to g H
Subtract the combined mass of C and H from the mass of the sample to get the mass of oxygen. Convert it to moles.
Simplify the mole ratios of C, H and O to get the empirical formula

32. Example 17
A gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O g of CO2 and g of H2O are produced. What is the empirical formula?

33. Chemical Equations Are sentences.
Describe what happens in a chemical reaction.
Reactants ® Products
Equations should be balanced.
Have the same number of each kind of atoms on both sides because ...

34. Balancing Chemical Equations
Must have the same number of each kind of atom on both sides of the equation must exist
__ CH4 + __ O  __ CO2 + __ H2O C H O

35. Balancing Chemical Equations
Add coefficients to make atoms of each element equal on both sides
2 CH4 + 4 O  2 CO H2O C H O

36. Example 18 __ Ca(OH) 2 + __ H3PO4  __ H2O + __ Ca3(PO4) 2
__ Cr + __ S8  __ Cr2S 3
16 Cr + 3 S8  8 Cr2S 3
__ KClO 3  ___ KCl + __ O 2
2 KClO 3  KCl O 2

37. Example 18 cont.
Translate this and then balance: Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas
Fe2S3 + HCl  FeCl3 + H2S
1 Fe2S3 + 6 HCl  2 FeCl H2S

38. Steps to solve problems
Stoichiometry
Given an amount of either starting material or product, determining the other quantities.
Steps to solve problems
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
4. Convert moles of sought quantity into desired units

39. Meaning of Coefficients
2H2 + O2 ® 2H2O
2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water.
2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form 2 x (6.02 x 1023) molecules of water.
2 moles of hydrogen and 1 mole of oxygen form 2 moles of water.

40. Conversion Factors to USE!
1 mol = x 1023 particles
molar mass: 1 mol = molar mass (g)
mole ratio: ? Mol A = ? Mol B
molar volume: 1 mol = 22.4

41. Mole Ratio:
conversion factor between 2 different amounts in a balance equation
2 Al2O3 ® 4Al + 3O2
every time we use 2 moles of Al2O3 we make 3 moles of O2
2 moles Al2O3
3 mole O2 or
3 mole O2
2 moles Al2O3

42. Mole to Mole Conversions
1 step process using a mole ratio to convert from moles of substance A to moles of substance B
How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?
2 Al2O3 ® 4Al + 3O2
3.34 moles Al2O3
3 mole O2 =
5.01 moles O2
2 moles Al2O3

43. 4 Types of Problems Example 19 Mole- Mole
How many moles of sulfur trioxide can be produced if 8.1 moles of oxygen react with sulfur S + 3 O2  2 SO3

44. 4 Types of Problems Mole-Mass:
2 step process using a mole ratio to convert from moles of substance A to moles of substance B and then the molar mass conversion factor to convert between moles of B to grams of B
No example

45. 4 Types of Problems
Mass-Mole: 2 step process using the molar mass conversion factor to convert from mass of A to moles of A and then the mole ratio to convert from moles A to moles of B
Example 20
One way of producing oxygen O2 involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of potassium chlorate is decomposed. How many moles of O2 are produced? KClO 3  2 KCl O 2

46. Example 20
A 25.5 g sample of potassium chlorate is decomposed. How many moles of O2 are produced?
2KClO 3  2 KCl O 2
1 mol KClO 3
32.00 g O 2
25.5 g KClO 3
1 mol O 2
g KClO 3
2 mol KClO 3
1 mol O 2
=.312 mol O2 =

47. 4 Types of Problems
Mass-Mass: 3 step process using the molar mass conversion factor to convert from mass of A to moles of A, then a mole ratio used to convert from moles A to moles of B, then a molar mass conversion factor to convert between moles of B to mass of B
Example 21
If 209 g of methanol are used up in the combustion, what mass of water is produced?
2CH3OH + 3O2  2CO2 + 4H2O

48. Periodic Table
Balanced Equation
Periodic Table
Mass g A
MolesA
MolesB
Mass g B
Decide where to start based on the units you are given and stop based on what unit you are asked for

49. Example 21
If 209 g of CH3OH are used up in the combustion, what mass of water is produced?
2CH3OH + 3O2 ® 2CO2 + 4H2O
1 mol CH3OH
18.02 g H2O
209 gCH3OH
4 mol H2O
32.05 gCH3OH
2 molCH3OH
1 mol H2O
=235 g H2O =

50. We can also change Liters of a gas to moles At STP
0ºC and 1 atmosphere pressure
At STP 22.4 L of a gas = 1 mole

51. Example 22
If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP?
2H2O ® 2H2 + O2
1 mol H2O
1 mol O2
22.4 L O2
6.45 g H2O
2 mol H2O
1 mol O2
18.02 g H2O
= L O2

52. Limiting Reagent
If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make?
The limiting reagent is the reactant you run out of first.
The excess reagent is the one you have left over.
The limiting reagent determines how much product you can make

53. Limiting Reagent Steps to determine the LR (limiting reagent)
Pick a product to convert to
Convert each reactant to the selected product in moles or grams
The reactant that gives the smallest answer will be the limiting reactant
Once you determine the LR, you should always start with the amount given of it to answer any other parts

54. How much excess reagent?
Use the limiting reagent to find out how much excess reagent you used
Subtract that from the amount of excess you started with

55. Cu is Limiting Reagent = 19.0 g Cu2S
Example 23: If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?
2Cu + S ® Cu2S
Cu is Limiting Reagent
1 mol Cu
1 mol Cu2S
g Cu2S
10.6 g Cu
63.55g Cu
2 mol Cu
1 mol Cu2S
= 13.3 g Cu2S
= 13.3 g Cu2S
1 mol S
1 mol Cu2S
g Cu2S
3.83 g S
32.06g S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S

56. Short Cut to Finding the Limiting reagent
1. Take each reactant amount and convert to moles. If already in
moles move to the next step.
2. Divide each mole amount by the coefficient of that substance from th e balanced chemical equation.
3. Whichever is the smallest number will identify the limiting reagent (LR)
4. Use the original amount of the reactant from the problem to begin your conversion process to determine either how much excess reagent you have left over or how much product you will form.

57. Example 24
10.0 g Al reacts with 35.0 g Cl2 gas to produce AlCl3. Which reactant is the limiting? Which is in excess? How much product is produced?
2Al + 3 Cl2 ® 2AlCl3
1 mol Al
= mol Al/2 = mol
10.0 g Al
26.98 g Al
1 mol Cl2
35.0 g Cl2
= mol Cl2/3 = mol
70.90 g Cl2
Cl2 is LR Al is ER

58. 10. 0 g Al reacts with 35. 0 g Cl2 gas to produce AlCl3
10.0 g Al reacts with 35.0 g Cl2 gas to produce AlCl3. Which reactant is the limiting? Which is in excess? How much product is produced?
2Al + 3 Cl2 ® 2AlCl3
1 mol Cl2
2 mol AlCl3
g AlCl3
35.0 g Cl2
3 mol Cl2
1 mol AlCl3
70.90 g Cl2
Smallest Amount 
= 43.9 g AlCl3

59. Yield How much you get from an chemical reaction

60. Type of Yield
The theoretical yield is the amount you would make if everything went perfect.
The actual yield is what you make in the lab.

61. % yield = what you got x 100% what you could have got
Percent Yield
% yield = Actual x 100% Theoretical
% yield = what you got x 100% what you could have got

62. Example 25
Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine g of aluminum bromide are produced. What are the three types of yield.

63. Density of a Gas D = M /V for a gas the units will be g / L
We can determine the density of any gas at STP if we know its formula.
To find the density we need the mass and the volume.
If you assume you have 1 mole than the mass is the molar mass (PT)
At STP the volume is 22.4 L.

64. Examples Find the density of CO2 at STP. D= 44.02 g/22.4 L = 1.97 g/L
You Try:
Find the density of CH4 at STP.