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Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 3 (Chp 1,2,3):
Matter, Measurement,
& Stoichiometry
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall Inc.
Chemistry:
The study of matter and the changes it undergoes.
Ni
nickel
solid
metal
+
H2
HCl
hydrochloric acid
aqueous
solution
+
NiCl2
hydrogen nickel(II) chloride
gas
solid crystals
Quantitative
or
Qualitative
Matter
Atom: simplest particle with properties of element
Element: same type of atom (1 or more)
HH
H2
O O
O2
C
C
Na
C
C
C
Compound: different atoms bonded
molecule
H2O
CO2
NaCl
chromatography
distillation
(boiling)
Matter
separate
physically
Mixture
differences
or unevenly
mixed
filtering
Heterogeneous
Mixture
(suspensions/colloids)
Physical
changes
uniform
or evenly
mixed
Homogeneous
Mixture
(solutions)
cannot separate
physically
Pure
Substance
separate
chemically
cannot
separate
Chemical
changes
Compounds
Elements
salt, baking soda, oxygen, iron,
water, sugar
hydrogen, gold
NaCl NaHCO3
O2 Fe
H2O C12H22O11
H2 Au
Physical Separation:
Filtration: Separates heterogeneous
mixtures (solids from liquids).
Physical Separation:
Distillation: Separates solution by
boiling point differences.
Physical Separation:
Chromatography:
Separates solution by differences in solubility
(attractions).
Metric Prefixes
Prefix
Symbol
Multiplier Examples:
1,000,000,000 B
1,000,000 J
1,000 g
BASE UNIT:
1m 1L
1g
0.01 m
0.001 L
(light wavelength)
(atoms)
(nuclei)
0.000 001 g
0.000 000 001 m
Precision in Measurements
Measuring devices have different uses
and different degrees of precision.
(uncertainty)
% Error = |Experimental – Accepted| x100
Accepted
Significant Digits
• measured digits (using marks on instrument)
• last estimated digit (one digit past marks)
5.23 cm
• do not overstate the precision 5.230 cm
Significant Zeroes
0.0003700400 grams
0’s
1. All nonzero digits are significant.
2. Captive Zeroes between two
significant figures are significant.
3. Leading Zeroes at the beginning of
a number are never significant.
4. Trailing Zeroes:
Sig, if there’s a decimal point.
NOT, if there is no decimal point.
Scientific Notation
Power of 10 is the number of places
the decimal has been moved.
Examples:
42000 = 4.2 x 104
0.0508 = 5.08 x 10–2
positive power: move decimal right 
to obtain the original # in standard notation.
negative power: move decimal left 
to obtain the original # in standard notation.
Scientific Notation
1. Convert the numbers to scientific notation.
2.45 x 104
(i) 24500
9.85 x 10–4
(ii) 0.000985
(iii) 12002
1.2002 x 104
2. Convert to standard notation.
420,000
(i) 4.2 x 105
0.000215
(ii) 2.15 x 10-4
0.003
(iii) 3 x 10-3
Sigs Digs in Operations
+ or –
round answers to keep the
fewest decimal places
3.48 + 2.2 = 5.68
x or ÷
5.7
round answers to keep the
fewest significant digits
6.40 x 2.0 = 12.8
13
Sig Digs Practice
WS 1s
1. How many sig digs are in each number?
4
(i) 250.0
(ii) 4.7 x 10–5
2
(iii) 34000000
2
4
(iv) 0.03400
2. Round the answer to the correct sig digs.
(i) 34.5 x 23.46
809
(ii) 123/3
40
(iii) 23.888897 + 11.2
35.1
(iv) 2.50 x 2.0 – 3
2
WARM UP (for QUIZ!!!)
• Review WS 1s #1, 3, 10
• Complete WS 1a #1, 2, 8, 9, 10
Law of Definite Proportions
• 2 H’s & 1 O is ALWAYS water.
• Water is ALWAYS 2 H’s & 1 O.
• 2 H’s & 2 O’s is NOT water.
√ H 2O
X H2O2
 elemental formulas (composition)
of pure compounds cannot vary.
O
H
H
H
H
O
O
Law of Conservation of Mass
The total mass of substances present
at the end of a chemical process is the
same as the mass of substances
present before the process took place.
__H
2 2 + __O2
2 2O
__H
Balancing Equations!!!
Symbols of Elements
Mass Number
= p’s + n’s
12
C
6
Element Symbol
Atomic Number (Z)
= p’s
All atoms of the same element have the
same number of protons (same Z), but…
can have different mass numbers. HOW?
Isotopes
element: same or different
mass: same or different
why? same # of protons (& electrons),
but different # of neutrons
1
H
1
protium
2
H
1
deuterium
3
H
1
tritium
Average Atomic Mass
• average atomic mass: calculated as a
weighted average of isotopes by their
relative abundances.
• lithium-6 (6.015 amu), which has a
relative abundance of 7.50%, and
• lithium-7 (7.016 amu), which has a
relative abundance of 92.5%.
(6.015)(0.0750) + (7.016)(0.925) = 6.94 amu
Avg. Mass = (Mass1)(%) + (Mass2)(%) …
Mass Spectrometry
atomized,
ionized
magnetic
field
element
sample
isotopes
separated
WS Atomic
by
Structure
difference
Cl (avg at. Mass) = in mass
(35)(~0.75) + (37)(~0.25) = ?
~75%
~25%
Molecular (Covalent) Compounds
Covalent compounds
contain nonmetals that
“share” electrons to
form molecules.
(molecular compounds)
Diatomic Molecules
“H-air-ogens”
7
These seven elements occur naturally
as molecules containing two atoms.
Binary Molecular Compounds
• list less electronegative atom
first. (left to right on PT)
• use prefix for the number of
atoms of each element.
• change ending to –ide.
CO2: carbon dioxide
CCl4: carbon tetrachloride
pentoxide
N2O5: dinitrogen
________________
CuSO4∙5H2O
(ionic & covalent)
copper(II) sulfate pentahydrate
Cations
metals
lose e’s
(+) charge
(metal) ion
Anions
Ions nonmetals
gain e’s
(–) charge
(nonmetal)ide
Ionic Bonds
Attraction between +/– ions formed by
metals & nonmetals transferring e–’s.
Formulas of Ionic Compounds
• Compounds are electrically neutral, so the
formulas can be determined by:
– Crisscross the charges as subscripts (then erase)
– If needed, reduce to lowest whole number ratio.
Pb4+ O2–
Pb2O4
PbO2
Naming Ionic Compounds
1) Cation: Write metal name (ammonium NH4+)
For transition metals with multiple charges, write
charge as Roman numeral in parentheses.
Iron(II) chloride, FeCl2
Iron(III) chloride, FeCl3
2) Anion: Write nonmetal name with –ide
OR the polyatomic anion name. (–ate, –ite)
Iron(II) sulfide, FeS
Magnesium sulfate, MgSO4
Common
Polyatomic
Ions
* these 12 will be on
Quiz 1
- all 20 Polyatomic Ions
will be on Quiz 2
WS 2d
Name
Symbol
Charge
*ammonium
NH4+
1+
*acetate
C2H3O2–
1–
–
(ethanoate)
(CH3COO )
*hydroxide
OH–
1–
*perchlorate
ClO4–
1–
*chlorate
ClO3–
1–
chlorite
ClO2–
1–
hypochlorite
ClO–
1–
bromate
BrO3–
1–
iodate
IO3–
1–
*nitrate
NO3–
1–
nitrite
NO2–
1–
cyanide
CN–
1–
*permanganate
MnO4–
1–
*bicarbonate
1–
HCO3–
(hydrogen carbonate)
*carbonate
CO32–
2–
*sulfate
SO42–
2–
sulfite
SO32–
2–
*chromate
CrO42–
2–
dichromate
Cr2O72–
2–
*phosphate
PO43–
3–
“Oxyanion” Names (elbO’s)
C
Si
N O
P
S
As Se
Te
perchlorate
chlorate
chlorite
hypochlorite
ClO4–
ClO3–
ClO2–
ClO–
nitrate
nitrite
NO3–
NO2–
In Out
4 –
sulfate
sulfite
SO42–
SO32–
phosphate
PO43–
3
2
1
4
3
–
F
Cl
Br
I
Ion Name
per-___-ate
___-ate
___-ite
hypo-___-ite
Naming Acids
Ion
add H+
Acid
In Out
Ion Name
Acid Name
4
–
per-___-ate per-___-ic acid
3
4
___-ate
___-ic acid
2
3
___-ite
___-ous acid
1
–
hypo-___-ite hypo-___-ous acid
perchlorate
Name Acids
chlorate
from these
chlorite
oxyanions: hypochlorite
WS 2e
–
ClO4
ClO3–
ClO2–
ClO–
nitrate NO3–
nitrite NO2–
sulfate SO42–
sulfite SO32–
Anatomy of a Chemical Equation
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Anatomy of a Chemical Equation
CH4(g) + 2 O2(g)
Reactants appear on the
left side of the equation.
CO2(g) + 2 H2O(g)
Anatomy of a Chemical Equation
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Products appear on the
right side of the equation.
Anatomy of a Chemical Equation
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
States (s, l, g, aq) written in
parentheses next to each compound
Anatomy of a Chemical Equation
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Subscripts show how many
atoms of each element
Anatomy of a Chemical Equation
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Coefficients show the amount of each particle
and are inserted to balance the equation.
Reaction
Types
Combination
A + B → AB
Demo: MgO
2 Mg(s)
+
2→1
O2(g)  2 MgO(s)
Decomposition
AB → A + B
1→2
(50 milliseconds!)
2 NaN3(s)  2 Na(s) + 3 N2(g)
Replacement Reactions
(or “Displacement”)
Single Replacement
AB + C → A + CB
(aq)
+
(s) 
(s)
+
video
(aq)
Double Replacement
AB + CD → AD + CB
Pb(NO3)2(aq) + KI(aq)  PbI2(s) + KNO3(aq)
Demo:
Combustion
CxHy + _O2  _CO2 + _H2O
• Often involve hydrocarbons
reacting with oxygen in the air
WS 4a
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
Formula
Weights
Formula Weight
(FW)
Molecular Weight
(MW)
• Sum of the
atomic weights
for the atoms in a
chemical formula
• Formula Weight of
calcium chloride,
CaCl2, is…
• Sum of the
atomic weights for
the atoms in a
molecule or
compound
• Molecular Weight of
ethane, C2H6, is…
Ca: 1(40.08 amu)
+ Cl: 2(35.45 amu)
110.98 amu
C: 2(12.01 amu)
+ H: 6(1.008 amu)
30.07 amu
Percent Composition
(# of atoms)(AW)
x 100
% element =
(FW)
One can find the percent by mass
of a compound of each element in
the compound by using this equation.
Percent Composition
So the percentage of carbon in
ethane (C2H6) is…
%C =
=
(2)(12.01)
(30.07)
24.02 x 100
30.07
= 79.88% C
Moles
Avogadro Constant
• One mole of particles
contains the Avogadro
constant of those particles
6.022 x 1023
Mole Relationships
• One mole of atoms, ions, or molecules contains the
Avogadro constant of those particles 6.022 x 1023
In 1 mol Na2CO3 , how many…
• Na atoms?
• C atoms?
• O atoms?
• How many donuts in 1 mol of donuts?
• How many boogers in 1 mol of boogers?
Which has more atoms, 1 mol CH3 or 1 mol NH3 ?
How about CH3CH2OH or H2SO4 ?
Molar Mass
• the mass of 1 mol of a substance (g/mol)
– molar mass (in g/mol) of an element is the
atomic mass (in amu) on the periodic table
– formula weight (amu) of a compound
same number as the
molar mass (g/mol) of 1 mole of particles
of that compound
Using Moles
Moles are the bridge from
the particle (micro) scale to
the real-world (macro) scale.
bridge
macro-
Mass
molar
mass
Moles
microAvogadro
constant
Particles
1 mol
#g
(groups of
(atoms)
6.022x1023
(molecules)
23
particles) 6.022x10
(units)
1 mol
#g
1 mol
1 mol
6.022x1023
(grams)
Using Moles
1.What is the mass of 1 mole of copper(II)
bromide, CuBr2?(63.55) + 2(79.90) = 223.35 g
= 6.022 x 1023 particles
2.How many moles are there in 112 g of
copper(II) bromide, CuBr2?
1 mol CuBr2
= 0.501 mol
112 g CuBr2 x
223.35 g CuBr2
CuBr2
3.How many particles present in each of the
questions #1 & #2 above?
6.022 x 1023 particles
= 3.02 x 1023
0.501 mol x
1 mol
particles
Stoichiometry:
calculations of quantities in chemical rxns
–how much reactant is consumed or
–how much product is formed
•Balanced chemical equations show the amount
of: atoms, molecules, moles, and mass
Most important are the ratios of reactants and
products in moles, or…
mol-to-mol ratios
Stoichiometric Calculations
Rxn:
gA
A(aq) + 2 B(aq)  C(aq) + 2 D(aq)
molar
mass A
g A
1 mol A
?
1 mol A
OR
?
gA
?
g B
1 mol B
molar
gB
mass B
mol A
Coefficients of
balanced equation
OR
mol-to-mol 2 mol B
1 mol A
ratio
mol B
1 mol A
2 mol B
Stoichiometric problems have 1-3 Steps: (usually)
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 (if necessary)
using the molar mass (from PT)
grams A x
1 mol A
x _ mol B x
grams A
mol A
.
grams B =
1 mol B
1) molar mass 2) mole ratio 3) molar mass
Stoichiometric Calculations
Example : g of A  g of B
HW p. 114 #58
Solid magnesium is added to an aqueous
solution of hydrochloric acid. What mass of H2
gas will be produced from completely reacting
18.0 g of HCl with magnesium metal?
Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)
mole ratio B/A
molar mass B
2.016 g H2
1 mol HCl x 1 mol H2
x
18.0 g HCl x
36.46 g HCl
2 mol HCl
1 mol H2
g of A
molar mass A
= ____
g Hg2 H2
0.498
Finding
Empirical
Formulas
Types of Formulas
• Empirical formulas:
the lowest ratio of atoms of CH
3
each element in a compound.
C2H4O
• Molecular formulas:
C2H6 C6H12O3
the total number of atoms of
each element in a compound.
molecular mass = emp. form.
empirical mass
multiple
Calculating Empirical Formulas
Steps (rhyme)
from Mass % Composition
Percent to Mass assume 100 g
Mass to Mole
MM from PT
÷ moles by smallest to CH
4
Divide by Small get mole ratio of atoms
x (if necessary) to get
Times ‘til Whole whole numbers of atoms
75 % C
75 g C
6.2 mol C
1C
25 % H
25 g H
24.8 mol H
4H
1) Percent to Mass
3) Divide by Small
2) Mass to Mole
4) Times ’til Whole
Butane is 17.34% H and 82.66% C by mass.
Determine its empirical formula.
82.66 g C x 1 mol C = 6.883 mol C = 1  1 C x 2
12.01 g C
6.883 mol
=2C
17.34 g H x 1 mol H = 17.20 mol H
1.008 g H
6.883 mol
= 2.499  2.5 H
C2H5 x 2
=5H
If molecular mass is 58 g∙mol–1, what is the
Molecular Formula?
HW p. 113 #43a, 48
molecular mass
empirical mass
58
=2
29.06
2 (C2H5) = C4H10
Calculating Empirical Formulas
Percent to Mass
Mass to Mole
Divide by Small
Times ‘til Whole
Combustion Analysis
• Hydrocarbons with C and H are analyzed
through combustion with O2 in a chamber.
 g C is from the g CO2 produced
Step 1 is
“combustion
 g H is from the g H2O produced to mass”
 g X is found by subtracting (g C + g H)
from g sample
Combustion Analysis
Example 1
When 4-ketopentenoic acid is analyzed by
combustion, a 0.3000 g sample produces
0.579 g of CO2 and
0.142 g of H2O.
The acid contains only C, H, and O.
What is the empirical formula of the acid?
1 mol CO2
1 mol C 12.01 g C
0.579 g CO2 x
x
x
44.01 g CO2 1 mol CO2 1 mol C
?gC
= 0.158 g C
Step 1: “combustion to mass”
1 mol H2O x 2 mol H x1.008 g H
0.142 g H2O x
18.02 g H2O 1 mol H2O 1 mol H
?gH
= 0.0159 g H
0.3000 g sample – (0.158 g C) – (0.0159 g H) =
?gO
= 0.126 g O
1 mol C
0.158 g C x
= 0.0132 mol C = 1.67 C
12.01 g C 0.00788 mol
x3=5C
1 mol H
0.0159 g H x
= 0.0158 mol H = 2 H
1.008 g H 0.00788 mol
x3=6H
1 mol O
0.126 g O x
= 0.00788 mol O = 1 O
16.00 g O 0.00788 mol
x3=3O
C5H6O3
Combustion Analysis
Example 2
A sample of a chlorohydrocarbon with a
mass of 4.599 g, containing C, H and Cl,
was combusted in excess oxygen to yield
6.274 g of CO2 and 3.212 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?
1 mol CO2
1 mol C 12.01 g C
6.274 g CO2 x
x
x
44.01 g CO2 1 mol CO2 1 mol C
?gC
= 1.712 g C
Step 1: “combustion to mass”
1 mol H2O x 2 mol H x1.008 g H
3.212 g H2O x
18.02 g H2O 1 mol H2O 1 mol H
?gH
= 0.3593 g H
4.599 g sample – (1.712 g C) – (0.3593 g H) =
? g Cl
= 2.528 g Cl
1 mol C
1.712 g C x
= 0.1425 mol C = 2 C
12.01 g C 0.07131 mol
1 mol H
0.3593 g H x
= 0.3564 mol H = 5 H
1.008 g H 0.07131 mol
1 mol Cl 0.07131 mol Cl
2.528 g Cl x
= 1 Cl
=
35.45 g Cl 0.07131 mol
C2H5Cl
If the compound has a
MW of 193 g∙mol–1, what MW 193
=3
is the molecular formula? EW 64.51
HW p. 114 #52b
C6H15Cl3
How Many Cookies Can I Make?
• Which ingredient will run out first?
• If out of sugar, you should stop making cookies.
• Sugar is the limiting ingredient, because it will
limit the amount of cookies you can make.
Before
H2
After
O2
Which is
limiting?
2 H2 + O2
Initial: 10
? mol
Change: –10
End: 0 mol
limiting
?7 mol
–5
2 mol
excess
2 H2O
?0 mol
+10
10 mol
Before
H2
After
O2
2 H2 + O2
Initial: 10
? mol
Change: –10
End: 0 mol
?7 mol
–5
2 mol
2 H2O
?0 mol
+10
10 mol
Does limiting mean smallest amount of reactant? No!
O2 is in smallest amount, but…
H2 is in smallest “stoichiometric” amount
Limiting Reactant
• 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 + 3 CuCl2  2 AlCl3 + 3 Cu
54.0 g Al 4.50 mol CuCl2 (Which is limiting?)
54.0 g Al x 1 mol Al x 3 mol CuCl2 = 3.00 mol CuCl
2
26.98 g Al
2 mol Al
(4.50 mol CuCl2) available > needed (3.00 mol CuCl2)
CuCl2 is excess
Al is limiting
Theoretical Yield
theoretical yield: the maximum amount of
product that can be formed
– calculated by stoichiometry
– limited by LR (use LR only to calculate)
limiting
54.0 g Al x 1 mol Al x 3 mol Cu x 63.55 g Cu = 191 g
Cu
26.98 g Al
2 mol Al
1 mol Cu
produced
• different from actual yield (or experimental),
amount recovered in the experiment
HW p. 115 #72
Percent Yield
A comparison of the amount actually obtained
to the amount it was possible to make
Actual
%Yield =
x 100
Theoretical
(calculate using the LR only)
NOT % Error:
% Error = |Accepted – Experimental| x100
Accepted
Percent Yield
Aluminum will react with oxygen gas
according to the equation below
4 Al + 3 O2
2 Al2O3
• In one such reaction, 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?
Percent Yield
4 Al + 3 O2
HW p. 116
#79, 77
2 Al2O3
1mol Al 2 mol Al2O3 101.96 g Al2O3
23.4 g Al x
x
x
26.98 g Al 4 mol Al
1 mol Al2O3
= 44.2 g Al2O3
39.3 g of aluminum oxide are formed.
What is the percentage yield?
39.3 g
%Yield =
44.2 g
x 100
88.9 %