Download Stoichiometry Notes

Name:
Stoichiometry Notes (Chapter 9)
I. Problems Involving Compounds
a. Compounds are measured in molecules (or formula units) and single elements are measured in
atoms.
1 mole of CO2 = 6.022 X 10 23 molecules
1 mole of Ag = 6.022 X 10 23 atoms
A mole is 6.022 X 10 23 of any substance!!!
Practice Problems:
1. How many atoms of carbon are in 25.0 grams of carbon?
2. How many molecules of calcium nitrate are in 0.200 moles of calcium nitrate?
b. Recall that the molar mass of a compound is the mass, in grams, of one mole of that compound.
determine the molar mass of a compound, add together the molar mass of all of the
elements in the compound.
Practice problems:
3. What is the molar mass of nitrogen dioxide?
To
c. Problems involving molar mass of compounds.
4. What is the mass in grams of 1.25 moles of sulfur trioxide?
5. When bees sting they release the compound isopentyl acetate, C7H14O2. How many
molecules of isopentyl acetate are there in 2.00 grams of C7H14O2?
II. Stoichiometry - shows the relationship between reactants and products in a chemical
reaction. Now we will use our knowledge of chemical reactions and equations to solve
problems.
a. One new conversion we need to know is the mole ratio – it relates the amounts of
moles of any two substances involved in a chemical reaction.
2Al2O3 
4Al +
3O2
2 mol Al2O3
2 mol Al2O3
4 mol Al
3 mol O2
4 mol Al
3 mol O2
Writing the inverse of any of the above would make a total of 6 mole ratios.
Practice Problem:
7. For the following chemical equation write all possible mole ratios.
2HgO  2Hg + O2
b. Three types of problems involving stoichiometric calculations (mole ratios) are:
1. Mole to mole problems (1 step problems)
2. Mole to mass problems (2 step problems)
3. Mass to mass problems (3 step problems)
To solve these problems first write a balanced equation for the reaction. Second
set-up a dimensional analysis problem, starting with the known quantity written
as a fraction. Use correct mole ratios to solve the problem.
Practice Problems:
8. According to the following equation how many moles of lithium hydroxide are
required to react with 20.0 moles of CO2?
CO2
+ 2LiOH  Li2CO3 + H2O
9. How many moles of ammonia, NH3 are produced when 6.00 moles of hydrogen gas
react with excess nitrogen gas?
3H2 + N2  2NH3
10. What mass, in grams, of glucose is produced when 3.00 moles of water react
with carbon dioxide?
6CO2 + 6H2O  C6H12O6 + 6O2
11. What mass in grams of magnesium oxide is produced when 2.00 moles of
magnesium react with oxygen?
12. The reaction below is run using 824 grams of NH3 and excess oxygen, how many
moles of NO are formed? How many moles of H2O are formed?
4NH3 + 5O2  4NO + 6H2O
13. If mercury (II) oxide decomposes, how many grams of mercury (II) oxide are
needed to produce 125 grams of oxygen?
14. How many grams of SnF2 are produced from the reaction of 30.0 grams of HF
with Sn?
Sn + 2HF  SnF2 + H2
15. What mass of aluminum is produced by the decomposition of 5000. grams of
Al2O3?
16. How many molecules of hydrogen chloride gas will be produced when 33.5 grams
of chlorine combine with excess hydrogen?
III. Percent Yield
a. Theoretical yield – maximum amount of product that can be produced from a given
amount of reactant. This value can be calculated from working mass to mass problems.
Example: In Practice Problem 15 you determined that 2646 grams of Al could be
produced from 5000. grams of Al2O3. This will occur if the reaction occurs without ANY
type of error!!
b. Actual yield – actual amount of a product obtained from a reaction.
experimentally obtained. It cannot be more than the theoretical yield.
It must be
Example: Let’s say we carry out the reaction in Practice Problem 15 in a lab
and we only obtain 2100. grams of Al from 5000. grams of Al2O3.
c. Percent yield = Actual yield
X 100
Theoretical yield
Percent yield = 2100. grams X 100 = 79.36%
2646 grams
17. When 36.8 grams of C6H6 react with an excess of Cl2, the actual yield of C6H5Cl
is 38.8 grams. What is the percent yield of C6H5Cl?
C6H6 + C12  C6H5Cl + HCl
Actual Yield=
Theoretical Yield=
18. If 75 grams of CO react to produce 68.4 grams of CH3OH, what is the percent
yield of CH3OH?
CO + 3H2  CH3OH
Stoichiometry Notes (Chapter 9) PART 2
IV. Limiting Reactants- the reactant that limits the amounts of the other reactants that can
combine and the amount of the product that can form in a chemical reaction. The substance
that is not used up completely in a reaction is called the excess reactant.
Think about making a hamburger: let’s say you need 2 buns, 1 piece of meat, and 1 piece of
lettuce to make one hamburger. If you have 16 buns, 16 pieces of meat, and 20 pieces of
lettuce, how many hamburgers can we make? What limits how many we can make? What is
in excess?
You should be able to recognize a limiting reactant problem because they have 2
amounts given – one for each reactant! To solve these problems first write the
balanced equation for the reaction. Second determine which of the reactants the
limiting reactant is. (To determine which the limiting reactant is use both amounts of
reactants given and setup two dimensional analysis problems to solve for the product.
The smaller answer will come from the limiting reactant and will be the correct
answer.)
Practice Problems:
19. Silicon dioxide (quartz) is usually quite uncreative but reacts readily with hydrogen
fluoride according to the following equation.
SiO2 (s) + 4HF (g)  SiF4 (g) + 2H2O (l)
If 2 moles of HF are exposed to 4.5 moles of SiO2, which is the limiting reactant?
20. If 20.5 grams of chlorine is reacted with 20.5 grams of sodium, which reactant is in
excess?
21. Using what you learned from problem #20, how much sodium chloride would be
produced if 20.5 grams of chlorine is reacted with 20.5 grams of sodium?
22. Magnesium sulfate reacts with potassium. How much magnesium would be
produced if 35 grams of magnesium sulfate is exposed to 45 grams of potassium?
V. Calculating Empirical Formulas from Chemical (Combustion) Analysis
When a compound containing C and H is subject to combustion with oxygen in a special
combustion apparatus all the C is converted to CO2 and the H is converted to H2O. The
amount of C produced can be determined by measuring the amount of CO2 produced; the
amount of H produced can be determined by the amount of H2O produced.
Example: Consider the combustion of isopropyl alcohol. The sample is known to contain only
C, H and O. Combustion of 0.255 grams of it produces 0.561 grams of CO2 and 0.306 grams of
H2O. Find the empirical formula.
CxHyOz
0.255g
Carbon:
+
O2 
H2O
0.306g
12.01g C
=
?g C
44.01g CO2
0.561g CO2
Hydrogen: 2.02g H
=
18.02g H2O
Oxygen:
CO2 +
0.561g
0.255g
CxHyOz
=
?=0.154g C
?g H
0.306g H2O
0.154g C
C
+
?=0.034g H
0.034g H
H
+
?
O
?=0.067g O
To find empirical formula remember the rhyme: % to mass, mass to mole, divide
by small, multiply till whole.
Carbon: ?mol C  0.154g C
1
X
1mol C = 0.0128mol/0.0042mol = 3.05
12.01g C
Hydrogen: ?mol H  0.034g H
1
Oxygen: ?mol O  0.067g O
1
X
1mol H = 0.0340mol/0.0042mol = 8.095
1.01g H
X
1mol O = 0.0042mol/0.0042mol = 1
16.00g O
Within experimental error, the most likely empirical formula is: C3H8O1
Practice Problem:
23. An elemental analysis showed that an unknown sample contained only C, H, and O.
A 0.1014g sample of the compound was burned in a C-H combustion train to produce 0.1486g of
CO2 and 0.0609g of H2O. Determine the empirical formula for this compound.
24. A 0.1153 gram sample of a pure hydrocarbon was burned in a C-H combustion train
to produce 0.3986 grams of CO2 and 0.0578 grams of H2O. Determine the empirical formula for
this hydrocarbon.