1. Moles, Formulas, Reactions & Stoichiometry
Lecture 3 Topics Brown, chapter 3
1. Moles
Molecular mass (aka molecular weight)
2. Molar converstions
Percent composition
Empirical formulas
3. Stoichiometry: balancing chemical equations
4. Patterns of Chemical Reactivity
Combination
Decomposition
Combustion
Exchange
5. Stoichiometry & Conversions
Limiting reactants
Theoretical & percent yield

2. Stoichiometric conversions
Balanced equations can predict yields.
One reactant often limits the…
…theoretical yield of product.
The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws.
It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made.
This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory.
While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

3. Stoichiometry is not about mass!
How do we ‘interpret’ the meaning of chemical equations?
2Na(s) + 2H2O(l)  H2(g) + 2NaOH(aq)
There are several correct interpretations:
2 atoms Na + 2 molecules water become 1 molecule H2 gas + 2 molecules NaOH
2 dozen atoms + 2 dozen molecules become 1 dozen molecules + 2 dozen molecules
2 cases atoms + 2 cases molecules become 1 case molecules + 2 case molecules
2 MOLES atoms + 2 MOLES molecules become 1 MOLE molec. + 2 MOLES molec.
And an INCORRECT interpretation:
Stoichiometry counts moles & molecules. Not mass!
2 g Na + 2 g water become 1 g H2 gas + 2 g NaOH
How do you convert to grams?
2 moles Na g = 46 g
1 mole
1 mole H g = 2.0 g
1 mole
46 g + 36 g --> 2 g + 80 g
2 moles H2O g = 36 g
1 mole
2 moles NaOH g = 80.0 g
1 mole
p.98-99

4. Stoichiometry “mall map”
atoms X
molecules A
subsc
Av#
For the reaction:
A  B
Where A contains element X, and B contains element Y.
mass A
moles A MW
Stoic
mass B
moles B MW
Av#
atoms Y
subsc
molecules B p

5. Stoichiometric conversions
Combust C2H6 to produce water. How much ethane do you need to burn to produce 2.34 moles of water?
Steps:
A balanced chemical equation
Go to moles.
Use coefficients to change one reactant or product to another.
2C2H O2  4CO2 + 6H2O moles
2.34 mol H2O 2 mol C2H6 = 0.78 mol C2H6
6 mol H2O
How many grams of CO2 are produced by burning g of ethane?
Same balanced chemical equation! (see above)
155.0 g C2H mol C2H mol CO g CO2 = g CO2
30 g mol C2H mol CO2
p.100-2

6. Limiting reactants: used up or left over?
You want to make grilled cheese sandwiches and find 10 slices of bread & 4 slices of cheese. How many sandwiches can you make?
Each sandwich requires 2 slices of bread & 1 slice of Swiss: 2:1 ratio. (B2C)
So for 10 slices of bread you’d need 10/2 = 5 slices of Swiss… …but there are only 4 slices.
So cheese is limiting (sandwiches produced) & bread is in excess.
Since the product still includes O2, O2 was in excess & H2 was limiting
O2 + 2H2  2H2O p

7. Limiting reactants strategy
Keys to solving limiting reactant problems:
1. Start with a balanced chemical equation.
2. Determine which reactant is limiting (in terms of moles). 3. Base moles of product yield on moles of limiting reactant.
4. Run a race in which the two reactants compete to make product. The reactant that makes less product is limiting.
How much H2 gas can be made from 30.0 g of Na and 35.0 g of H2O?
2Na + 2H2O --> H NaOH 30.o g g
30.0 g mol Na 1 mol H g H2 = 1.31 g H2
22.99 g mol Na 1 mol
 limiting
Na H2O H2
Moles
Change
Left
Grams
30.5 g mol H2O 1 mol H g H2 = 1.71 g H2
18.02 g mol Na mol p

8. Examples: limiting reactants
How much water can be made from 150 g of H2 and 1500 g of O2?
2H2 + O2 --> 2H2O 150 g g
150 g mol H2 2 mol H2O g H2O = 1.3 x 103 g H2O made
2.02 g mol H mol
So H2 is limiting & dictates yield
1500 g mol O2 2 mol H2O g H2O = 1.7 x 103 g H2O made
32.00 g mol O mol
How much excess reactant is leftover?
1.3 x 103 g H2O 1 mole H2O 1 mole O g = 1200 g O2 used
18.02 g mole H2O 1 mole
Leftover O2 = 1500 g given – 1200 g used = 300 g leftover p

9. Percent yield
Percent yield compares actual yield to theoretical yield.
It’s a measure of how good the chemist’s yield was.
We’ve just learned to calculate theoretical yield in limiting reactions.
% yield = (actual/theoretical)(100)
How much Fe can be made from kg of rust and g CO? Rust is Fe2O3, MW g/mole
Fe2O CO --> 2Fe + 3CO2
103 g g ?
103 g Fe2O mol Fe2O mol CO g = g CO required so CO limiting
g mol Fe2O mol CO
445.2 g CO 1 mol CO mol Fe g = g Fe produced
28.00 g mol CO mol Fe
If the actual yield was g of Fe, what’s the percent yield?
g (100) = 84.45%
592.0 g p

10. Putting it all together!
How many grams of ethane does it take to make 1.00 kg of water if the process is 85% efficient? (C2H6; MW = g/mol)
You need to make 1.00 kg of water, but the rxn is only 85% efficient, so:
1.oo kg X kg --> kg must be made = 1.18x103 g H2O
85% %
2C2H O2 --> 4CO H2O
? g g
1.18E3 g H2O mol H2O 2 mol C2H g = 723 g of ethane (C2H6)
18.01 g mol H2O mol C2H6