CHEMISTRY CP
Name: _______KEY______________________
Period: _____
UNIT 5: TEST REVIEW SHEET
MOLE CONVERSIONS, EMPIRICAL/MOLECULAR FORMULA, STOICHIOMETRY
MOLES AND MOLE CONVERSIONS
1. Things to Know:
a. What is a mole?
♦ Avogadro’s number of particles; 1Mole = 6.02 × 1023 particles = gram formula
mass of an element/compound
b. What is Avogadro’s number?
♦ 6.02 × 1023
c. Be able to determine the empirical formula when given the molecular formula. What is
the empirical formula of C6H6 and N2O4?
♦ Molecular Formula: C6H6
Molecular Formula: N2O4
Empirical Formula: CH
Empirical Formula: NO2
d. The relationship between the weight of one atom of an element and the weight of
6.02×1023 atoms of the element. If a single molecule of H2O has a mass of 18 a.m.u.,
what is the mass of 6.02×1023 molecules?
♦ They have the same numerical value, but one atom will have a mass in a.m.u.
While 6.02 × 1023 atoms (1 Mole) will have a mass in grams
1 molecule/atom of H2O = 18.0 amu
6.02 × 1023 atoms = 1 Mole = 18.0 g
e. The relationship between empirical formula mass and molecular mass.
♦ The molecular mass is a whole number multiple of the empirical formula mass.
The empirical formula is always in lowest or simplest terms
2. Calculate the formula mass in a.m.u. of the following:
a. CaCO3
♦ Ca: 40.1 × 1 = 40.1 amu CaCO3 = 40.1 amu + 12.0 amu + 48.0 amu = 100.1 amu
C: 12.0 × 1 = 12.0 amu
O: 16.0 × 3 = 48.0 amu
b. KNO3
♦ K: 39.1 × 1 = 39.1 amu
KNO3 = 39.1 amu + 14.0 amu + 48.0 amu = 101.1 amu
N: 14.0 × 1 = 14.0 amu
O: 16.0 × 3 = 48.0 amu
3. Calculate the gram formula mass (formula mss in grams) of the following:
a. Na2CO3
♦ Na: 23.0 × 2 = 46.0 g
Na2CO3 = 46.0 g + 12.0 g + 48.0 g = 106 g/M
C: 12.0 × 1 = 12.0 g
O: 16.0 × 3 = 48.0 g
b. AlPO4
♦ Al: 27.0 × 1 = 27.0 g
AlPO4 = 27.0 g + 31.0 g + 64.0 g = 122.0 g/M
P: 31.0 × 1 = 31.0 g
O: 16.0 × 4 = 64.0 g
4. Make the following conversions.
a. What is the number of moles of CaS in 144 grams of the compound?
144 grams
♦ Ca: 40.1 g
1 Mole
CaS
S: + 32.1 g
= 2.0 M
72.2 grams CaS
CaS = 72.2 g/M
b. What is the mass in grams of 18.06×1023 molecules of CO2?
♦ C: 12.0 g × 1 = 12.0 g
O: 16.0 g × 2 = 32.0 g
18.06×1023mol
1 Mole
CO2 = 44.0 g/M
6.02×1023 mol
44.0g/M
1 Mole
= 132g
c. How many molecules of CO2 are in .25 moles?
0.25 M CO2
6.02×1023 mol
1M
= 1.51×1023 molecules
d. How many ions of Cl- are in 1 mole of CaCl2?
♦ 1 Mole CaCl2 (Ratio of Ca to Cl- is 1 : 2)
1 Mole CaCl2 × 2 Cl- = 2 Moles of Cl2 Moles Cl-
6.02×1023 mol
1M
= 12.04×1023
molecules
EMPIRICAL/MOLECULAR FORMULA/PERCENT COMPOSITION
5. Find the empirical formula of a compound which is 49.5% Fe, and 50.5% F.
♦ Step 1: Convert to grams
Fe: 49.5 grams
F: 50.5 grams
Step 2: Convert grams to Moles
Fe: 49.5 grams = 0.887 M
F: 50.5 grams = 2.66 M
55.8 grams
19.0 grams
Step 3: Divide by Smallest Moles
Fe: 0.887M = 1
F: 2.66M = 3
EF: FeF3
0.887M
0.887M
6. The empirical formula of a compound is CH and its molecular mass is 78 a.m.u. What is the
molecular formula?
♦ C: 12.0. amu
n = Molar Mass____ = 78 amu__ = 6
MF: C6H6
H: + 1.0 amu
EF Molecular Mass 13.0 amu
CH = 13.0 amu
7. A sample of a compound is composed of 63.0 grams of Rb and 5.9 grams of Oxygen. What is
the empirical formula of the compound?
♦ Step 1: Convert grams to Moles
Rb: 63.0 grams = 0.737 M
O: 5.9 grams = 0.369 M
85.47 grams
16.0 grams
Step 3: Divide by Smallest Moles
Rb: 0.737 = 2
O: 0.369 = 1
EF: Rb2O
0.369
0.369
8. A compound has an empirical formula of C2H8N and a molar mass of 46 grams per mole.
What is the molecular formula
n = 1, MF = C2H8N
9. What is the percentage composition of CaCO3?
♦ CaCO3
Ca: 40.1 × 1 = 40.1 g
% Ca: (40.1 g / 100.1 g) × 100 = 40.0%
C: 12.0 × 1 = 12.0 g
% C: (12.0 g / 100.1 g) × 100 = 12.0%
O: 16.0 × 3 = 48.0 g_
% O: (48.0 g / 100.1 g) × 100 = 48.0%
CaCO3
= 100.1 g
10. What is the percentage by mass of oxygen in Ammonium nitrate, NH4NO3?
♦ NH4NO3
N: 14.0 × 2 = 28.0 g
% O: (48.0 g / 80 g) × 100 = 60.0%
H: 1.0 × 4 = 4.0 g
O: 16.0 × 3 = 48.0 g_
NH4NO3 = 80.0 g
MIXED STOICHIOMETRY PROBLEMS
11. Given the equation: Zn + 2 HCl  ZnCl2 + H2
a. How many moles of hydrogen will be formed when 4 moles of HCl are reacted?
♦ Mole Ratio: H2 = 1 = x Moles H2
x = 2 Moles H2
HCl
2
4 Moles HCl
b. How many moles of Zn will be reacted to form 3 moles of ZnCl2?
♦ Mole Ratio: Zn = 1 = x Moles Zn
x = 3 Moles Zn
ZnCl2 1
3 Moles ZnCl2
c. How many moles of HCl will be needed to react completely with 3 moles of Zn?
♦ Mole Ratio: Zn = 1 = 3 Moles Zn
x = 6 Moles HCl
HCl
2
x Moles HCl
12. According to the following reaction, calculate the number of grams of sodium hydroxide that
will be produced if 8.215 grams of sodium are reacted with excess water.
2 Na + 2 H2O  2 NaOH + H2
♦ Mass to Mass Problem with Balanced Equation
Step 1: Convert grams to Moles
8.215 g Na × 1 Mole Na = 0.357 Moles Na
23.0 g Na
Step 2: Mole Ratio
Na
= 2
= 0.357 Moles Na
NaOH
2
x Moles NaOH
=
x = 0.357 Moles NaOH
Step 3: Convert Moles to grams
0.357 Moles NaOH × 40 g NaOH
1 Mole
=
x = 14.28 grams NaOH
13. According to the following reaction, if 35 grams of C6H12O6 react with 40 grams of O2 how
many grams of CO2 will be produced? What is the limiting reactant?
C6H12O6 + 6 O2  6 CO2 + 6 H2O
♦ Step 1: Balance the equation
Step 2: Convert grams of reactants into Moles and compare
35 g C6H12O6 × 1 Mole C6H12O6
= 0.194 Moles C6H12O6 **Limiting Reactant**
180 g C6H12O6
40 g O2 × 1 Mole O2
32 g O2
= 1.25 Moles O2
Step 3: Use Limiting Reactant to solve for Moles of CO2
C6H12O6
= 1
= 0.194 Moles
=
CO2
6
x Moles
x = 1.164 Moles CO2
Step 4: Convert Moles to grams
1.164 Moles CO2
× 44 g CO2
1 Mole
x = 51.2 grams CO2
=
14. The amount of product in a given reaction turns out to be only 2.34 grams although the
theoretical amount was calculated to be 3.12 grams. What is the percentage yield?
♦ Actual yield
× 100
Theoretical yield
2.34 grams × 100 = 75%
3.12 grams
15. In the reaction of 10 grams of NaHCO3 with excess HCl, 4.5 grams of NaCl is actually
produced. What is the percentage yield from this reaction?
NaHCO3 + HCl  NaCl + H2O + CO2
♦ Step 1: Convert grams of reactant into Moles
10 g NaHCO3 × 1 Mole NaHCO3 = 0.119 Moles NaHCO3
84 g NaHCO3
Step 2: Set Up Mole Ratio and Solve
NaHCO3
= 1
= 0.119 Moles
NaCl
1
x Moles
=
x = 0.119 Moles NaCl
Step 3: Convert Moles to grams
0.119 Moles NaCl
× 58.5g NaCl
1 Mole
=
x = 6.96 grams NaCl
Step 4: Find Percent Yield
4.5 grams × 100 = 64.7%
6.96 grams
16. Excess copper reacts with 60.5 grams of silver nitrate to produce copper (II) nitrate and silver.
Determine the number of grams of Cu(NO3)2 produced. The reaction is:
Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag
♦ Mass to Mass Problem with Balanced Equation
Step 1: Convert grams to Moles
60.5 g AgNO3 × 1 Mole AgNO3 = 0.356 Moles AgNO3
169.9 g AgNO3
Step 2: Mole Ratio
AgNO3
= 2
Cu(NO3)2
1
= 0.356 Moles AgNO3
x Moles Cu(NO3)2
Step 3: Convert Moles to grams
0.179 Moles Cu(NO3)2 × 187.6 g Cu(NO3)2
1 Mole
=
=
0.179 Moles Cu(NO3)2
33.5 grams Cu(NO3)2