1. 1411 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

2. Stoichiometry (stoy-key-OM-uh-tree)
stoichiometry - the area of study that examines the quantities of substances consumed and produced in chemical reactions
Built on an understanding of atomic masses, chemical formulas, and the law of conservation of mass

3. “produces” or “yields”
Chemical Equations
“produces” or “yields”
“reacts with”
“and”
Chemical equations are concise representations of chemical reactions.

4. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Reactants appear on the left side of the equation.
Products appear on the right side of the equation.

5. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
The states of the reactants and products are written in parentheses to the right of each compound.
(g) = gas (l) = liquid (s) = solid (aq) = aqueuos (water) solution
Sometimes, reaction conditions such as heat, symbolized Δ, are written above or below the arrow.

6. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.
balanced equation – must have equal # atoms on each side of the arrow
(The coefficient 1 is left off; no coefficient = understood 1)

7. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in one molecule.

8. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in one molecule.
Coefficients tell the number of molecules.
To find the # of each type of atom, multiply the coefficient by each subscript.

9. Examples
How many of each type of atom are there in the following formulas?
3 Fe2O3
4 Na2CO3
Ba(ClO3)2
3 (NH4)3PO4

10. Balancing Chemical Equations
Steps to follow
Determine the formulas of reactants and products
Write an unbalanced equation
Insert the coefficients that will provide equal numbers of atoms on each side of the arrow

11. Guidelines: Balancing Chemical Equations
Guidelines for balancing equations:
NEVER change or add subscripts (changes identity of chemical)
Adding a coefficient only changes the AMOUNT of a substance, rather than its identity
2 H2O = 2 molecules of water
3 H2O = 3 molecules of water
Balance only one element at a time.
Balance elements other than H and O first
Use the smallest whole-number coefficients
Ex: Zn HCl  ZnCl H2 good
2 Zn HCl  2 ZnCl H2 not good
Balance as a group those polyatomic ions that appear unchanged on both sides of arrow
CaO + HNO3  Ca(NO3)2 + H2O (unbalanced)

12. Balancing Chemical Equations: Question…
Which of the following would be the correct way to balance the oxygens in a chemical equation?
adding a subscript 2 to the end of CO to give CO2 or
adding a coefficient in front of the formula to give 2 CO

13. Balancing Chemical Equations
Fe3O4(s) H2(g) Fe(s) H2O(l) (unbalanced)
Work on one element at a time.
Don’t change the subscripts – only add coefficients.
F: Fe3O4(s) + H2(g) Fe(s) H2O(l)
O: Fe3O4(s) + H2(g) Fe(s) H2O(l)
H: Fe3O4(s) + 4H2(g) Fe(s) H2O(l)
Check that all atoms are balanced, and they are!

14. Examples Balance the following equations:
AgNO CaCl2  Ca(NO3) AgCl
P O2  P4O10
C9H O2  CO H2O

15. Reaction Types: Combination Reactions
Flares and some fireworks:
In this type of reaction two or more substances (often 2 elements) react to form one product
A + B  C
If reactants are metal and nonmetal, product is ionic solid.
Product is determined by predicting charges of ions.
Examples:
2 Al (s) + 3 Br2 (g)  2 AlBr3 (s)
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)

16. Reaction Types: Decomposition Reactions
In a decomposition one substance breaks down into two or more substances.
C  A + B
Often involves heat.
Ex: Many metal carbonates are heated to give metal oxides and CO2
Examples:
CaCO3 (s)  CaO (s) + CO2 (g) (CaO = lime)
2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g) (air bags)

17. Reaction Types: Combustion Reactions
These are generally rapid reactions that produce a flame.
Most often involve hydrocarbons reacting with O2(g) in the air
hydrocarbon: compound with either
C and H only
C, H, and O
If these completely combust, products are always CO2 and H2O
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C6H12O6 (g) + 6 O2 (g)  6 CO2 (g) + 6 H2O (g)

18. Examples: Reaction Types
Name the type of reaction, predict the product(s), and balance the equation:
CH3OH + O2
Na + S
BaCO3
Al(s) + Cl2(g)
When C8H8 burns in air Δ

19. Formula Weight (FW)
Formula weight - sum of the atomic weights for the atoms in a chemical formula
generally reported for ionic compounds
The FW of an ionic compound is the mass of one formula unit (the empirical formula of an ionic compound)
Ex: FW of one formula unit of calcium chloride, CaCl2
Ca: 1( amu)
+ Cl: 2( amu)
amu

20. Molecular Weight (MW)
A molecular weight is the sum of the atomic weights of the atoms in a molecule
another name for formula weight, but it is usually only called MW for molecules
For the molecule ethane, C2H6, the molecular weight would be
C: 2( amu)
+ H: 6( amu)
amu

21. Example
Determine the formula weight (FW) or molecular weight (MW) for the following:
a) C5H6O
b) Al2(CO3)3

22. Percent Composition
To find the percentage of the mass of a compound that comes from each of the elements in the compound:
% element =
(number of atoms)(atomic weight)
(FW of the compound)
x 100

23. Percent Composition So the percentage of carbon in ethane (C2H6) is…
(2)( amu)
( amu)
amu
amu =
x 100
= %

24. Example: Percent Composition
What is the percent composition of hydrogen in NH4H2PO4?
5.26%

25. Mole
Even tiny samples in the lab contain MANY molecules, atoms, or ions
Ex: 1 teaspoon of water = 2 x 1023 water molecules!
So chemists devised a counting unit to relate numbers of molecules, atoms, etc to lab-sized amounts…. THE MOLE
Examples of other counting units (dozen = 12 objects, gross = 144 objects)
mole (abbr. mol) – the amount of matter that contains as many objects as there are atoms in 12 g of 12C
number of atoms = 6.02 x 1023 = “Avogadro’s number” = NA
objects = atom, molecules, ions, or whatever object you’re counting… people, apples, etc

26. 6.02 x 1023 particles = 1 mol particles
Avogadro’s Number
One mole of particles contains Avogadro’s number of those particles
“particles” are whatever you’re looking for (or are given) the number of.
usually atoms, ions, or molecules
6.02 x 1023 particles = 1 mol particles

27. Avogadro’s Number: A Conversion Factor!
Avogadro’s number allows us to interconvert:
Examples:
1 mol Ag atoms = 6.02 x 1023 Ag atoms or
1 mol H2O molecules = 6.02 x 1023 H2O molecules
1 mol NO3- ions = 6.02 x 1023 NO3- ions
6.02 x 1023 particles = 1 mol particles
6.02 x 1023 particles
1 mole of those same particles
1 mole of those same particles
6.02 x 1023 particles
number of particles
moles of those same particles
6.02 x 1023 Ag atoms
1 mole Ag atoms
1 mole Ag atoms
6.02 x 1023 Ag atoms
(the two conversion factors)
6.02 x 1023 H2O molecules
1 mole H2O molecules
1 mole H2O molecules
6.02 x 1023 H2O molecules
(the two conversion factors)
6.02 x 1023 NO3- ions
1 mole NO3- ions
1 mole NO3- ions
6.02 x 1023 NO3- ions
(the two conversion factors)

28. Avogadro’s Number: Examples
How many Cu atoms are in 0.50 mole of Cu?
How many moles of CO2 are in 2.50 x 1024 molecules CO2?
3.0 x 1023 Cu atoms
4.15 moles of CO2

29. Subscripts: Molecular Level and Molar Level
The subscripts in a formula show 2 things:
the number of each type of atom in one molecule (molecular level)
the moles of each element in 1 mole of compound (molar level)
Glucose
C6H12O6
In 1 molecule of C6H12O6 : 6 atoms C 12 atoms H 6 atoms O
In 1 mole of C6H12O6 : 6 moles C 12 moles H 6 moles O

30. Subscripts: Another Conversion Factor!
The subscripts allow us to interconvert:
In 1 mole of C6H12O6 : 6 moles C 12 moles H 6 moles O
We can write 3 conversion factors from the subscripts of C6H12O6 :
Or we can flip them upside-down:
moles of
atoms or ions
moles of the compound the atoms or ions are contained in
6 mol C
1 mol C6H12O6
12 mol H
1 mol C6H12O6
6 mol O
1 mol C6H12O6
1 mol C6H12O6
6 mol C
1 mol C6H12O6
12 mol H
1 mol C6H12O6
6 mol O

31. Subscripts: Example
How many moles of O are in mole of aspirin, C9H8O4?
0.600 mole of O

32. Example: Avogadro’s Number + Subscripts
Calculate the number of H atoms in 2.00 mol H2O.
Our given is in units of moles
We want the number of particles (H atoms in this case)
The only way to relate number of particles to moles of those same particles is by using Avogadro’s number.
Remember: x 1023 particles = 1 mole of those same particles
So for this problem, we have: x 1023 H atoms = 1 mol H atoms
We can also relate the number of moles of H2O to number of moles of H atoms (by using the subscripts)
- So, 1 mol H2O = 2 mol H atoms
2 mol H atoms
1 mol H2O
6.02 x 1023 H atoms
1 mol H atoms
2.00 mol H2O x x
= x 1024 H atoms

33. Examples: Avogadro’s Number + Subscripts
Calculate the number of O atoms in 2.10 mol Al2(CO3)3.
Calculate the number of PO43- ions in 3.0 mol Mg3(PO4)2.
How many moles of Mg3(PO4)2 are in x 1020 formula units of Mg3(PO4)2?
1.14 x 1025 O atoms
3.6 x 1024 PO43- ions
2.409 x 10-4 formula units

34. Molar Mass
A mole is always the same number (6.02 x 1023), like a dozen is always 12, but 1-mole samples of different substances have different masses.
The molar mass is the mass of 1 mole of an element or compound (unit g/mol)
Equals
the atomic weight (for an element) that we find on the periodic table
molar mass of Li is g/mol
Or we can write, g Li = 1 mole of Li
the formula weight (for a compound)
molar mass of H2O is g/mol
Or we can write, g H2O = 1 mole of H2O
The formula weight (in amu) is same number as the molar mass (in g/mol), but different units

35. Molar Mass: Another Conversion Factor!
Molar mass allows us to interconvert:
6.941 g Li = 1 mole of Li
g H2O = 1 mole of H2O
mass of a substance
moles of that same substance
6.941 g Li
1 mol Li
1 mol Li
6.941 g Li
g H2O
1 mol H2O
1 mol H2O
g H2O

36. Molar Mass: Example How many moles of H2SO4 are in 0.2380 g of H2SO4?
What is the mass in grams of 2.13 x 10-2 moles of CO2?
2.427 x 10-3 mol H2SO4
0.937 g CO2

37. Using Moles
Moles provide a bridge from the molecular scale to the real-world scale

38. Avogadro’s Number + Subscripts + Molar Mass
Examples:
Avogadro’s Number + Subscripts + Molar Mass
How many H atoms are in 3.01 x 10-2 mg of H2O?
What is the mass in grams of Na+ ions in a sample of Na2O containing 1.02 x1020 Na2O formula units?
2.01 x 1018 H atoms
7.79 x 10-3 g Na+

39. Calculating Empirical Formulas
(Review): empirical formula tells relative number of atoms of each element
One can calculate the empirical formula from the percent composition.
The ratio of the number of moles of each element in a compound gives the subscripts in a compound’s empirical formula.
“Percent to mass, mass to mol, divide by small, multiply ‘til whole!”

40. 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.

41. Calculating Empirical Formulas
Assume g of para-aminobenzoic acid, find mass of each element:
C: % = g C
H: % = 5.14 g H
N: % = g N
O: % = g O

42. Calculating Empirical Formulas
From mass, calculate the number of moles of each element:
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

43. 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
Empirical formula:
C7H7NO2

44. Molecular Formula from Empirical Formula
If we are given molar mass of the compound, we can determine the molecular formula
The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in the empirical formula:
whole number multiple =
molar mass of actual compound
molar mass of empirical formula
Example:
The empirical formula of a compound is found to be NaS2O3. The molar mass of the compound is g/mol. Determine the molecular formula.
molar mass of empirical formula = g/mol
270.4 g/mol
135.1 g/mol
The molecular formula is Na2S4O6.
Multiply each subscript in empirical formula by 2 to get subscripts of molecular formula
= 2.001

45. Example: Empirical and Molecular Formulas
Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass. Its molar mass is 206 g/mol. Determine its empirical and molecular formulas.

46. Example: Empirical and Molecular Formulas
Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass. Its molar mass is 206 g/mol. Determine its empirical and molecular formulas.
Empirical formula = C13H18O2
Molecular formula = C13H18O2

47. Combustion Analysis
“Empirical” means based on observation and experiment
So chemists have techniques to determine empirical formula from experiments
Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this.
mass of C - determined from the mass of CO2 produced
mass of H - determined from the mass of H2O produced
mass of O - determined by difference after the masses of C and H have been determined
mass of O = mass of sample – (mass of C + mass of H)
CxHyOz + O2  CO2 + H2O (not balanced)

48. Combustion Analysis Problem
Isopropyl alcohol is composed of C, H, and O. Combustion of g isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol.
Plan: To determine the empirical formula of isopropyl alcohol, CxHyOz, we need to determine number of moles of C, H, and O.
CxHyOz + O2  CO2 + H2O (not balanced)
We will first use mole concept to find mass of C and mass of H.
mass of C present in CO2 = mass of C present in isopropyl alcohol
mass of H present in H2O = mass of H present in isopropyl alcohol
mass of O = mass of isopropyl alcohol – (mass of C + mass of H)
Finally calculate moles of C, H, and O from masses of C, H, and O to give subscripts for empirical formula

49. Combustion Analysis Problem
Isopropyl alcohol is composed of C, H, and O. Combustion of g isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol.
We will first use mole concept to find mass of C and mass of H.
mass of C: g CO2 x
mass of C present in CO2 = g C = mass of C present in isopropyl alcohol
mass of H: g H2O x
mass of H present in H2O = g H = mass of H present in isopropyl alcohol
mass of O = mass of isopropyl alcohol – (mass of C + mass of H)
mass of O = g – (0.153 g g) = g O
1 mol CO2
44.0 g CO2
1 mol C
1 mol CO2
12.0 g C
1 mol C x x
= g C
1 mol H2O
18.0 g H2O
2 mol H
1 mol H2O
1.01 g H
1 mol H x x
= g H

50. Combustion Analysis Problem
Isopropyl alcohol is composed of C, H, and O. Combustion of g isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol.
Finally calculate moles of C, H, and O from masses of C, H, and O to give subscripts for empirical formula
moles C = g C x = mol C
moles H = g H x = mol H
moles O = g O x = mol O
CxHyOz  C3H8O
1 mol C
12.0 g C
= 2.98 mol C  3
0.0043
1 mol H
1.01 g H
= 7.98 mol H  8
0.0043
1 mol O
16.0 g O
= 1 mol O
0.0043

51. Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products.
This mole-to-mole ratio will serve as a conversion factor in calculations.
Ex: For above chemical equation, some mole-to-mole ratios are:
2 mol H mol O mol H2O
2 moles H2
2 moles H2O
2 moles H2O
2 moles H2
2 moles H2
1 mole O2
1 mole O2
2 moles H2O

52. Stoichiometric Calculations Coefficients: Another Conversion Factor!
Just like subscripts, coefficients in a chemical equation also apply to both the molecular and molar levels
The coefficients in the balanced equation give the ratio of moles of reactants and products.
This mole-to-mole ratio allows us to interconvert:
Ex: For above chemical equation, some conversion factors are:
2 mol H mol O mol H2O
moles of one substance in a chemical equation
moles of another substance in the chemical equation
2 moles H2
2 moles H2O
2 moles H2O
2 moles H2
2 moles H2
1 mole O2
1 mole O2
2 moles H2O

53. Stoichiometric Calculations
Examples:
How many moles of O2 react with 3.02 moles of H2?
How many moles of H2O are produced from 2.4 moles of O2?
1 mol O2
2 mol H2
3.02 mol H2 x
= 1.51 mol O2
2 mol H2O
1 mol O2
2.4 mol O2 x
= 4.8 mol O2

54. Stoichiometric Calculations: Summary
This chapter’s 4 new conversion factors, always relating moles:
Avogadro’s number
Subscripts
Molar Mass
Coefficients
number of particles
moles of those same particles
moles of
atoms or ions
moles of the compound the atoms or ions are contained in
mass of a substance
moles of that same substance
moles of one substance in a chemical equation
moles of another substance in the chemical equation

55. Stoichiometric Calculations
Starting with the mass of substance A you can use the mole-to-mole ratio of the coefficients of A and B to calculate the mass of substance B formed (if it’s a product) or used (if it’s a reactant).

56. Stoichiometric Calculations
Calculate the mass in grams of water produced when starting when 1.00 g of C6H12O6 combusts.
C6H12O6 + 6 O2  6 CO2 + 6 H2O =
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams.

57. Limiting Reactants: How Many Cookies Can I Make?
You can make cookies until you run out of one of the ingredients.
Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).

58. Limiting Reactants: How Many Cookies Can I Make?
In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.

59. Limiting Reactant Bread and cheese sandwich example 2 Bd + Ch  Bd2Ch
- We have 10 slices of bread and 7 slices of cheese
The limiting reactant is the reactant present in the smallest stoichiometric amount; the reactant that produces the least amount (fewest moles) of product!!!
In other words, it’s the reactant you’ll run out of first
Here, it’s the bread because
10 slices of bread makes 5 sandwiches
7 slices of cheese makes 7 sandwiches
The bread limits the number of sandwiches!
There will be cheese left over, so cheese would be the excess reactant - the one that produces more product (more moles) of product!!!

60. Limiting Reactant 2 H2 + O2  2 H2O
Suppose we have 10 moles H2 and 7 moles O2.
To determine which reactants are the limiting reactant and excess reactant, we calculate moles of product from the mole-to-mole ratio for both reactants and see which is less:
We can determine how much of the excess reactant was actually used up and how much we have left over:
2 mol H2O
2 mol H2
10 mol H2 x
= 10 mol H2O
limiting reactant = H2
2 mol H2O
1 mol O2
7 mol O2 x
= 14 mol H2O
excess reactant = O2
1 mol O2
2 mol H2
10 mol H2 x
= 5 mol O2 actually reacted
We will have 2 mol O2 leftover: 7 mol O2 (started with) – 5 mol O2 (actually reacted)

61. Theoretical Yield
The theoretical yield is the maximum amount of product that can be made.
In other words it’s the amount of product possible as calculated through the stoichiometry problem.
This is different from the actual yield, which is the amount one actually produces and measures.

62. Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100
One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield).
Actual Yield
Theoretical Yield
Percent Yield = x 100

63. Limiting Reactant: Example
Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made commercially by a controlled reaction between cyclohexane (C6H12) and O2:
2 C6H12(l) O2(g)  2 H2C6H8O4(l) + 2 H2O(g)
The reaction is carried out with 25.0 g of cyclohexane and 35.0 g of O2. After the reaction, 33.5 g of adipic acid are formed.
What is the limiting reactant?
How much of the excess reactant remains after the reaction?
What is the theoretical yield of adipic acid?
What is the actual yield of adipic acid?
What is the percent yield of adipic acid?

64. Limiting Reactant: Example
Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made commercially by a controlled reaction between cyclohexane (C6H12) and O2:
2 C6H12(l) O2(g)  2 H2C6H8O4(l) + 2 H2O(g)
The reaction is carried out with 25.0 g of cyclohexane and 35.0 g of O2. After the reaction, 33.5 g of adipic acid are formed.
What is the limiting reactant?
How much of the excess reactant remains after the reaction?
What is the theoretical yield of adipic acid?
What is the actual yield of adipic acid?
What is the percent yield of adipic acid?
cyclohexane
10.5 g
43.4 g
33.5 g
77.2 %