Unit II Review Sheet
The Mole: Chapter 7
1. Chapter 7, which was the first part of Unit 2, dealt with the following
conversions:
a.
b.
c.
d.
e.
f.
g.
Mass to Moles (divide by the Molar Mass)
Moles to Mass (multiply by the Molar Mass)
Moles to Volume of a Gas @ STP (multiply by 22.4 L/mole)
Volume to Moles of a gas @ STP (divide by 22.4 L/mole)
Moles to molecules (multiply by 6.02 x 1023 molecules/mole)
Molecules to Moles (divide by 6.02 x 1023 molecules/mole)
Molecules to atoms & ions (multiply the number of molecules by the number of atoms &
ions)
2. When converting from molecules to atoms, or determining the molar mass of the diatomic
molecules, be sure to remember that the “Br I N Cl H O F” elements have two atoms per
molecule, ONLY WHEN THEY ARE GASES, BY THEMSELVES. In the lab that we did in class, for
example, Oxygen was a single atom in the MgO molecule.
3. Determine the molar density of a gas at STP
4. Calculate the percent composition by mass of a substance from its:
a. Chemical formula (i.e. C6H12O6)
Mass of Carbon = 6 x 12 g/mol = 72 g
Mass of Hydrogen = 12 x 1 g/mol = 12 g
Mass of Oxygen = 6 x 32 g/mol = 192 g
% C = [72 g/(72+12+192) g] x 100% = 26.1%
% H = [12 g/276 g] x 100% = 4.35 %
% O = [192 g/276 g] x 100% = 69.6 %
b. Experimental mass data (i.e. 10 grams C, 15 grams H, 41 grams O)
% C = [10 g/66 g] x 100% = 15.15%
% H = [15 g/66 g] x 100% = 22.7 %
% O = [41 g/66 g] x 100% = 62.1 %
5. Derive the empirical formula of a compound from
a. Experimental data: given grams of each element, calculate % composition of each element,
assume 100 grams, divide by each element’s Molar Mass, and “whole” the ratio.
b. Percent composition data: given % composition of each element, assume 100 grams, divide by
each element’s Molar Mass, and “whole” the ratio.
6. Calculate/determine the Molecular Formula from:
a. Experimental data: same as 5a., then divide “empirical molar mass” by the given, actual
Molar Mass to find the factor.
b. Percent Composition data: same as 5b., then divide “empirical molar mass” by the given,
actual Molar Mass to find the factor.
GHS Honors Chemistry
The start of Chapter 8: Chemical Equations
The Anatom y of a Chem ical
Reac tion
1. Identify the reactants and products in a chemical reaction.
Ingredients are c alled
the “REACTANTS”
S”
2 H2 + O2
2. Balance chemical equations when given the formulas for all
reactants and products in a chemical reaction.
Results are c alled the
“PRODUCTS”
S”
2 H2O

Reac ts with
Coeffic ient:
how many of
that molec ule
Subsc ript:
Number of
Atoms present
“Yields”
Yields” or
Produc es
Chapter 9
1. Chapter 9 continues with the concepts (from Chapter 7) of conversion from moles, to mass, to
molecules, and to volume of a gas @ STP, but now used these conversions of chemistry to predict
the amounts needed in a reaction.
2. Ultimately, after balancing a reaction, we can compare the reactants & products in “mole-ratios”
only, with the exception of a direct volume comparison of gases at STP. That said, we can convert
any of the previous units to moles, compare moles of reactants and products, and then convert
back to volume, mass, molecules, etc.
3. Whereas Chapter 7 dealt with “vertical conversions” of a reactant or product, Chapter 9 deals
with the comparison & conversions using the horizontal molar ratios.
2 C2H6
+
7 O2

4 CO2
Moles
Moles

Moles
Moles
Grams
Grams
Grams
Grams
Molecules
Molecules
Molecules
Molecules
Atoms
Atoms
Atoms
Atoms
Volume @ STP
Volume @ STP
Volume @ STP
Volume @ STP

+
6 H2O
Limiting Reagent Stoichiometry
4. Identify the limiting reagent in a chemical reaction and use it to calculate stoichiometric quantities
and the amount of excess reagents.
A) You can determine the limiting reagent by
determining the number of moles of product formed
from each of the reactant amounts given. The
reactant that produces the smaller number of moles
of product is your limiting reagent.
B) To calculate the grams of product for each of the
reactants given, simply convert the smaller number of
moles of product to grams using the molar mass of
that product.
Mg (s)
(s) + 2HCl(g)  MgCl2(aq) + H2(g)
??
 How
2.87 L
1.5 L
many gram s of solid Mg?
2.87L HCl
1 mol HCl
22.4L HCl
1 mol Mg
2 mol HCl
24.3 g Mg
1 mol Mg
= 1.6 g Mg
 How
muc h exc ess reagent
rem ains?
Exc ess Mg = 10.1 g – 1.6 g = 8.5 g Mg
GHS Honors Chem
C) To determine the excess reactant, calculate the amount of reactant needed to react with the
limiting reagent, using the molar ratios. The limiting reagent is entirely used up, along with the
stoichiometric amount of the “other” reactant. What is left over is the excess reactant.
GHS Honors Chemistry
5. Calculate the theoretical yield, and
percent yield for a given chemical reaction
using experimental data. The theoretical
yield can be determined from the reaction
stoichiometry. Assuming that you are
given the “actual or experimental,” the %
Yield is determined as follows:
% Yield = [Actual/Theoretical] x 100%
A “Yielding ” Problem
6.78 g of c opper is produc ed when
3.92 g of Al are reac ted with exc ess
c opper (II) sulfate.
Al + CuSO4  Al2(SO4)3 + Cu
2Al + 3 CuSO4  Al2(SO4)3 + 3Cu
6.78 g Al
 What is the ac tual yield?
 What is the theoretic al yield? 13.83 g Al
 What is the perc ent yield? 49%

GHS Honors Chem
GHS Honors Chemistry