1. Review Unit 3 (Chp 1,2,3,4): Stoichiometry & Solutions
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Review Unit 3 (Chp 1,2,3,4): Stoichiometry & Solutions
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall Inc.

2. Uncertainty and Error % Error = |Accepted – Experimental| x100
Measuring devices have different uses and different degrees of precision.
% Error = |Accepted – Experimental| x100
Accepted

3. + or – x or ÷ Significant Figures 0.0003700400 3.48 + 2.2 = 5.68 5.7
Nonzero digits are significant.
Leading Zeroes are never significant.
Trailing Zeroes:
SIG, only if a decimal point.
+ or –
fewest decimal places
fewest significant figures =
5.68
5.7
Exact numbers are infinitely significant. How many boy students in here? Plus girls? How many sig figs?
x or ÷
6.40 x =
12.8 13

4. “Oxyanion” Names (elbO’s)C N O F Si P S Cl As Se Br Te I
perchlorate
ClO4–
chlorate
ClO3–
chlorite
ClO2–
hypochlorite
ClO–
nitrate
NO3–
nitrite
NO2– In
Out
Ion Name – 4
per-___-ate 3 4
___-ate
sulfate
SO42–
sulfite
SO32–
phosphate
PO43– 2 3
___-ite
hypo-___-ite 1

5. Naming Acids Ion Acid per-___-ate ___-ate ___-ite hypo-___-ite
add H+
Acid In
Out
Ion Name
Acid Name 4 –
per-___-ate 3
___-ate 2
___-ite 1
hypo-___-ite
per-___-ic acid
___-ic acid
___-ous acid
hypo-___-ous acid
nitrate
NO3–
nitrite
NO2–
Name Acids from these oxyanions:
perchlorate
ClO4–
chlorate
ClO3–
chlorite
ClO2–
hypochlorite
ClO–
sulfate
SO42–
sulfite
SO32–

6. (groups of 6.022x1023 particles)
Using Moles
Moles are the bridge from
the particle (micro) scale to
the real-world (macro) scale.
bridge
micro-
macro-
molar mass
Avogadro constant
Moles
(groups of 6.022x1023 particles)
Particles
(atoms)
(molecules)
(units)
Mass
(grams)
1 mol
# g
6.022x1023
1 mol
# g
1 mol
1 mol
6.022x1023

7. Stoichiometric Calculations
1) Convert grams to moles (if necessary)
using the molar mass (from PT)
2) Convert moles (given) to moles (wanted)
using the mol ratio (from coefficients)
3) Convert moles to grams
(OR Liters, atoms, …)
1 mol A .
grams A
_ mol B
mol A
grams B
1 mol B
grams A x x x =
1) molar mass
2) mole ratio
3) molar mass

8. Percent Composition and Empirical Formulas

9. Percent Composition by Mass
% element =
(# of atoms)(AW)
(FW)
x 100
So the percentage of carbon in ethane (C2H6) is…
%C =
(2)(12.01)
(30.07)
24.02
30.07 =
x 100
= 79.88% C

10. Types of Formulas CH3 C2H4O C2H6 C6H12O3 Empirical formulas:
the lowest ratio of atoms of each element in a compound.
Molecular formulas:
the total number of atoms of each element in a compound.
CH3
C2H4O
C2H6
C6H12O3
molecular mass = multiple of emp. form.
empirical mass

11. Calculating Empirical Formulas
from Mass % Composition
Steps (rhyme)
Percent to Mass
Mass to Mole
Divide by Small
Times ‘till Whole
assume 100 g
MM from PT
÷ moles by smallest to get mole ratio of atoms
CH4
x (if necessary) to get whole numbers of atoms
75 % C
75 g C
6.2 mol C
1 C
25 % H
25 g H
24.8 mol H
4 H

12. C2H5 C4H10 1) Percent to Mass 3) Divide by Small 2) Mass to Mole
4) Times ’till Whole
A hydrocarbon has is 17.34% H and 82.66% C by mass. Determine its empirical formula.
If MM is 58 g mol–1, what is the
Molecular Formula?
82.66 g C
17.34 g H
82.66 g C x = mol C
17.34 g H x = mol H
1 mol C
12.01 g C
1 mol H
1.008 g H
= 1  1 C
=  2.5 H
x 2
= 2 C
6.883 mol
6.883 mol
x 2
= 5 H
C2H5 58
29.06
molecular mass
empirical mass
= 2
2 (C2H5) =
C4H10

13. Step 1 is “combustion to mass”
Combustion Analysis
Hydrocarbons with C and H are analyzed through combustion with O2 in a chamber.
g C is from the g CO2 produced
g H is from the g H2O produced
g X is found by subtracting (g C + g H) from g sample
Step 1 is “combustion to mass”
Is this chemical or physical separation?

14. Combustion Analysis
A sample of a chlorohydrocarbon with a mass of g, containing C, H and Cl, was combusted in excess oxygen to yield g of CO2 and g of H2O.
Calculate the empirical formula of the compound.
If the compound has a MW of 193 g∙mol–1, what is the molecular formula?

15. 1 mol CO2 44.01 g CO2 1 mol C 1 mol CO2 12.01 g C 1 mol C
6.274 g CO2 x x x
? g C
= g C
Step 1: “combustion to mass”
1 mol H2O
18.02 g H2O
2 mol H
1 mol H2O
1.008 g H
1 mol H x x
3.212 g H2O x
? g H
= g H
4.599 g sample – (1.712 g C) – ( g H) =
? g Cl
= g Cl

16. C2H5Cl C6H15Cl3 1 mol C 12.01 g C 0.1425 mol C 1.712 g C x = = 2 C
1 mol H
1.008 g H
mol H
g H x = =
5 H
mol
1 mol Cl
35.45 g Cl
mol Cl
2.528 g Cl x = =
1 Cl
mol
C2H5Cl
If the compound has a MW of 193 g∙mol–1, what is the molecular formula? MW EW
193
64.51
= 3
C6H15Cl3

17. Limiting Reactant 2 H2 + O2 2 H2O H2 O2 Before After
Initial: ? mol ? mol ? mol
Change:
End: 10 7
–10 –5
+10
2 mol
In other words, it’s the reactant you’ll run out of first
Which is the limiting here (H or O)? How do you know?
0 mol
10 mol
Does limiting mean smallest amount of reactant?
No!
O2 is in smallest amount, but…
H2 is in smallest “stoichiometric” amount

18. Limiting Reactant 2 Al + 3 CuCl2  2 AlCl3 + 3 Cu
convert reactant A to reactant B to compare
If available < needed (limiting)
If available > needed (excess)
Solid aluminum metal is reacted with aqueous copper(II) chloride in solution
2 Al CuCl2  2 AlCl Cu
54.0 g Al mol CuCl2
(Which is limiting?)
1 mol Al
26.98 g Al
3 mol CuCl2
2 mol Al
54.0 g Al x x =
3.00 mol CuCl2
(4.50 mol CuCl2) available > needed (3.00 mol CuCl2)
CuCl2 is excess
Al is limiting 18

19. Theoretical Yield
theoretical yield: the maximum amount of product that can be formed
calculated by stoichiometry
limited by LR (use LR only to calculate)
different from actual yield (or experimental), amount recovered in the experiment
limiting
1 mol Al
26.98 g Al
3 mol Cu
2 mol Al
63.55 g Cu
1 mol Cu
191 g
54.0 g Al x x x = Cu
produced

20. (calculate using the LR only)
Percent Yield
A comparison of the amount actually obtained to the amount it was possible to make
%Yield = x 100
Actual
Theoretical
(calculate using the LR only)
NOT % Error:
% Error = |Accepted – Experimental| x100
Accepted

21. Percent Yield 4 Al + 3 O2 2 Al2O3 WS MC #6-11 FR #2b
When 23.4 g of Al are allowed to burn in excess oxygen, 39.3 g of aluminum oxide are formed. What is the percentage yield?
1mol Al
26.98 g Al
2 mol Al2O3
4 mol Al
g Al2O3
1 mol Al2O3
23.4 g Al x x x
= 44.2 g Al2O3
39.3 g
44.2 g
%Yield = x 100 =
88.9 %

22. Strong Acids: Only 6 strong acids: Nitric (HNO3) Sulfuric (H2SO4)
Hydrochloric (HCl)
Hydrobromic (HBr)
Hydroiodic (HI)
Perchloric (HClO4)
Strong Acids:
HI + H2O  H3O+ + I–
proton (H+) donors

23. ase Strong Bases: The strong bases are soluble hydroxides (OH–) of…
Group 1 (Li,Na,K)
CBS (Ca, Ba, Sr)
Mg(OH)2 not as soluble
Strong Bases:
OH– + H3O+  H2O + H2O
proton (H+) acceptors
ase

24. Salts: Ionic Solids: (metal-nonmetal)
dissociate (dissolve) by separation into ions
Electrolytes:
ions in solution that conduct electricity

25. Electrolytes: Strong, Weak, or Non?
(ions conduct electricity)
Compound
metal-nonmetal
nonmetals (Covalent)
Ionic
Molecular
Acid
(H____)
Not Acid
STRONG
KBr
CaI2
FeCl3
NaOH
Ca(OH)2
(strong bases)
STRONG
(6)
WEAK
(& NH3)
NON
C11H22O11
C2H5OH
H2O
HCl, HBr, HI
HNO3
H2SO4
HClO4
CH3COOH
HNO2 HF

26. Solubility Rules ALWAYS Soluble ions: *
Li+, Na+, K+, ... Group I (alkali metals)
NH4+ ammonium
NO3– nitrate *
Common Precipitates form with: examples
Ag+, Pb2+, Hg2+ (AP/H) AgCl, PbI2
OH– (hydroxide) Cu(OH)2
CO32– (carbonate) CaCO3
SO42– (sulfate) BaSO4

27. Preparing a Solution 1-mass solute 2-add solvent, swirl to dissolve
3-add solvent to mark 27

28. Dilution M1V1 = M2V2 1-calc M1V1=M2V2 2-pipet V1 from concentrated
3-fill to mark w DI 28

29. Dilution
M1V1 = M2V2
What volume of a stock solution of concentration 8.0 M CuSO4 is needed to prepare 2.0 L of 3.0 M CuSO4 ? 29

30. Solution Stoichiometry
Rxn: A(aq) + 2 B(aq)  C(aq) + 2 D(aq)
molar mass A
molarity A (M)
g A
mol A
L of A
g A
1 mol A
mol A
1 L
mol-to-mol ratio
g B
1 mol B
mol B
1 L
molar mass B
molarity B (M)
g B
mol B
L of B 30

31. Transition Metal Ion ColorsWS
MC #1-5, 12-15
FR #3
Color from e– movement in unfilled d orbitals.
Filled or Empty d sublevel has no e– movement so ion is colorless.