1. Chemistry, The Central Science, 10th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten
Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice-Hall

2. Law of Conservation of Mass
“We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.”
--Antoine Lavoisier, 1789

3. Chemical Equations
Concise representations of chemical reactions

4. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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

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

7. 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.

8. Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation.

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

10. Subscripts and Coefficients Give Different Information
Subscripts tell the number of atoms of each element in a molecule
Coefficients tell the number of molecules

11. Balancing Equations Reactants Products # of atoms on left side
Use a table like the one below
Adjust coefficients ONLY (no changing subscripts!)
Use trial and error, updating the table as you go until the # of atoms of each element is equal on the right and left
Reactants 
Products
Element or Polyatomic Ion
# of atoms on left side
# of atoms on right side

12. Combination Reactions
Two or more substances react to form one product
Examples:
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
2 Mg (s) + O2 (g)  2 MgO (s)

13. 2 Mg (s) + O2 (g)  2 MgO (s)

14. Decomposition Reactions
One substance breaks down into two or more substances
Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)

15. Combustion Reactions Rapid reactions that produce a flame
Most often involve hydrocarbons reacting with oxygen in the air
Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)

16. Formula Weight (FW)
Sum of the atomic weights for the atoms in a chemical formula (the same # as molar mass, diff units)
So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
These are generally reported for ionic compounds

17. Molecular Weight (MW)
Sum of the atomic weights of the atoms in a molecule (the same # as molar mass, diff units)
For the molecule ethane, C2H6, the molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu

18. Molar Mass
By definition, these are the mass of 1 mol of a substance (i.e., g/mol)
The molar mass of an element is the mass number for the element that we find on the periodic table
The formula weight (in amu’s) will be the same number as the molar mass (in g/mol)

19. Percent Composition
One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:
% element =
(number of atoms)(atomic weight)
(FW of the compound)
x 100

20. Percent Composition So the percentage of carbon in ethane, C2H6 is…
(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu =
x 100
= 80.0%

21. Avogadro’s Number
6.02 x 1023
1 mole of 12C has a mass of 12 g

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

23. Mole Relationships
One mole of atoms, ions, or molecules contains Avogadro’s number of those particles
One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound

24. Calculating Empirical Formulas
One can calculate the empirical formula (the smallest whole number ratio of elements in a molecule) from the percent composition

25. 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.

26. 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

27. Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles:
These are the subscripts for the empirical formula:C7H7NO2
C: =  7
H: =  7
N: = 1.000
O: =  2
5.105 mol
mol
5.09 mol
1.458 mol

28. Combustion Analysis
Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this
C is determined from the mass of CO2 produced
H is determined from the mass of H2O produced
O is determined by difference after the C and H have been determined

29. Combustion Analysis Example
Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of g of isopropyl alcohol produces g of CO2 and g of H2O. Determine the empirical formula of isopropyl alcohol.
Now do a standard empirical formula problem. ANS: C3H8O

30. Molecular Formula Example
If you know the empirical formula and the molecular mass you can get the molecular formula by calculating the ratio of the empirical formula molar mass to the actual molar mass.
Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its molar mass is 62.1 g/mol.
What is the empirical formula of ethylene glycol? CH3O,
What is its molecular formula? Molar mass of CH3O is 31.04g/mol g/mol divided by 31.04g/mol is 2, so multiply the empirical formula by 2. The molecular formula is 2*(CH3O) = C2H6O2

31. Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles (NOT the ratio of masses!!) of reactants and products

32. Stoichiometric Calculations
From the mass of Substance A you can use the 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)

33. Stoichiometric Calculations
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…

34. 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)
sugar would be the limiting reactant

35. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount
In other words, it’s the reactant you’ll run out of first (in this case, the H2
The non-limiting reagent is the excess reagent
in this case, the O2

36. Solving Limiting Reagent Problems
Recognize it!
Convert each reactant to moles.
Use the mole ratio to figure out which reactant will run out first (limiting) and which will have some left over (excess)
Based on the limiting reagent calculate the amount of product formed.

37. Limiting reagent example (3.19)
Suppose a solution containing 3.50 g of Na3PO4 is mixed with a solution containing 6.40 g of Ba(NO3)2. How many grams of Ba3(PO4)2 can be formed?
Using the molar mass of each substance, we can calculate the number of moles of each reactant:
From the balanced equation, we have the following stoichiometric relations:

38. Use the mole ratio to convert one reactant to the other to determine how much is needed.
mol Na3PO4*(3Ba(NO3)2/ 2Na3PO4) = mol 3Ba(NO3)2
So mol of barium nitrate is needed to react with all of the sodium phosphate. But you only have mol 3Ba(NO3)2 present, so you will run out of barium nitrate before you use up all of the sodium phosphate. So 3Ba(NO3)2 is limiting and Na3PO4 is in excess.
Now calculate the mass of Ba3(PO4)2 produced based on the number of moles of 3Ba(NO3)2

39. Theoretical Yield
The theoretical yield is the amount of product that can be made in theory
In other words it’s the amount of product possible as calculated through the stoichiometry problem
This is different from the actual yield, the amount one actually produces and measures in the “real world”

40. Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100
A comparison of the amount actually obtained to the amount it was possible to make
Actual Yield
Theoretical Yield
Percent Yield = x 100

41. Percent Yield Example
Suppose a solution containing 3.50 g of Na3PO4 is mixed with a solution containing 6.40 g of Ba(NO3)2. How many grams of Ba3(PO4)2 can be formed? 4.92g from previous example
If the above reaction is carried out in the lab and 4.30g Ba3(PO4)2 is recovered, what is the % yield?
% Yield = (Actual/theoretical) *100%
= (4.30g/4.92g)*100% = 87.4%

42. Clean Up Directions Unplug hot plate, leave at bench
Discard filters, weighing dishes in trash
Discard test tubes in broken glass box
Clean glassware (use dish soap if needed)
Put beakers in cabinet #2, funnels in #4
Unscrew rings and clamps, put in drawer #1
Put ringstands underneath cabinet #1
Return EMPTY box to cart
Wipe down EMPTY lab bench
Put goggles away.