1. Unit 3 (Chp 1,2,3): Matter, Measurement, & Stoichiometry
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.

2. Chemistry: Quantitative or Qualitative
The study of matter and the changes it undergoes.
(has mass and takes up space)
Ni HCl
nickel hydrochloric acid H
hydrogen
NiCl2
nickel(II) chloride
solid aqueous gas solid crystals
metal solution
Quantitative or
Qualitative

3. Matter H2O CO2 NaCl Atom: simplest particle retaining properties.
Element:
1 or more of the same type of atom. C C H H O O Na C H2 O2 C C
Compound:
different elements bonded.
H2O CO2 NaCl

4. Heterogeneous Mixture
distillation
(boiling)
Matter
cannot separate physically
separate physically
Physical
changes
Pure Substance
Mixture
separate chemically
cannot separate
differences or unevenly mixed
uniform or evenly mixed
filtering
Chemical
changes
Classification of Matter (flowchart handout)
Sugar and Water make Lemonade
Sugar and Water and Fat make Milk
Sugar and Salt and Water and Fat and Iron and Oxygen make Blood
Heterogeneous Mixture
Homogeneous Mixture
Compounds
Elements
salt, baking soda, water, sugar
oxygen, iron, hydrogen, gold
(suspensions/colloids)
(solutions)
NaCl NaHCO3
H2O C12H22O11
O2 Fe
H2 Au

5. d = m V Density: ratio of mass to volume or matter to space occupied
Units: g/mL
g/cm3
kg/L
Why does ice float?
Chemical Property OR
Physical Property
Why?

6. Changes of Matter Physical Changes: Chemical Changes:
do not change the composition of a substance.
temperature, changes of state, amount, etc.
Chemical Changes:
result in new substances.
combustion, oxidation, decomposition, etc.
Name Physical Changes of State
SOLID  LIQUID Melting
LIQUID  GAS Boiling or Evaporation
GAS  LIQUID Condensing
SOLID  GAS Sublimation
GAS  SOLID Reverse Sublimation or Deposition
LIQUID  SOLID Freezing, Solidifying or Crystallizing

7. Chemical Separation:
Compounds can be decomposed into elements.

8. Physical Separation: Filtration:
Separates heterogeneous mixtures (solids from liquids).
Phys or chem?

9. Physical Separation: Distillation:
Separates solution by boiling point differences.
Physical or chemical separation?

10. Physical Separation: Chromatography:
Separates homogeneous mixture by differences in solubility (attractions).
Physical or chemical separation?

11. Scientific Notation Examples: 42000 = 4.2 x 104 0.0508 = 5.08 x 10–2
Power of 10 is the number of places the decimal has been moved.
Examples: = 4.2 x 104
= 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.
Measurements and calculations in chemistry often require the use of very large or very small numbers.

12. Scientific Notation 1. Convert the numbers to scientific notation.
(ii)
(iii) 12002
2.45 x 104
9.85 x 10–4
x 104
2. Convert to standard notation.
(i) 4.2 x 105
(ii) 2.15 x 10-4
(iii) 3 x 10-3
420,000
0.003

13. Metric Prefixes Prefix Symbol Multiplier Examples:
1,000,000,000 GB
1,000,000 MJ
1,000 kg
BASE UNIT: 1 m 1 L g
0.01 cm
0.001 mL
What’s above kilo? Below Micro? µg
(light wavelength) nm
(atoms)
(nuclei)

14. Uncertainty in Measurements
Measuring devices have different uses and different degrees of precision.
(uncertainty)
% Error = |Accepted – Experimental| x100
Accepted

15. Uncertainty
(uncertain)
5.23 cm

16. Significant Figures 5.23 cm 5.230 cm measured digits.
last digit is estimated, but IS significant.
5.23 cm
do not overstate the precision
5.230 cm

17. Significant Zeroes 0.0003700400 grams
All nonzero digits are significant.
Captive Zeroes between two significant figures are significant.
Leading Zeroes at the beginning of a number are never significant.
Trailing Zeroes:
SIG, if at end AND a decimal point.
NOT, if there is no decimal point.
0’s
Exact numbers are infinitely significant. How many boy students in here? Plus girls? How many sig figs?

18. + or – x or ÷ Significant Figures 3.48 + 2.2 = 5.68 5.7 6.40 x 2.0 =
round answers to keep the fewest decimal places
round answers to keep the fewest significant figures =
5.68
5.7
x or ÷
6.40 x =
12.8 13

19. Significant Figures WS 1s 1. How many sig figs are in each number?
250.0
4.7 x 10–5 4 2 2 4
2. Round the answer to the correct sig figs.
(i) x 23.46
(ii) 123/3
(iii)
(iv) x 2.0 – 3
809 40
35.1 2

20. Warm Up (for QUIZ!!!) Review WS 1s # 1, 3, 10
Complete WS 1a #1, 2, 8, 9, 10

21. 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.
√ H2O O H H H
X H2O2 H O O
elemental formulas (composition) of pure compounds cannot vary.

22. 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.
__H2 + __O __H2O 2 2
Why we balance equations.
Balancing Equations!!!

23. C Symbols of Elements 12 6 Mass Number = p’s + n’s 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?

24. H H H Isotopes 1 2 1 3 1 element: same or different mass:
why?
same # of protons (& electrons), but different # of neutrons 1 H 2 1 H 3 1 H
protium
deuterium
tritium

25. Average Atomic Mass Avg. Mass = (Mass1)(%) + (Mass2)(%) …
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%.
amu = 1/12 mass of C amu = X10^-24 grams
(6.015)(0.0750) + (7.016)(0.925) = 6.94 amu
Avg. Mass = (Mass1)(%) + (Mass2)(%) …

26. isotopes separated by difference in mass
Mass Spectrometry
atomized,
ionized
magnetic field
element sample
isotopes separated by difference in mass
~75%
~25%
WS 2a
(35)(~0.75) + (37)(~0.25) = ?

27. Molecular (Covalent) Compounds
Covalent compounds contain nonmetals that “share” electrons to form molecules.
(molecular compounds)

28. 7 Diatomic Molecules “H-air-ogens”
Seems like all gases, why not noble gases? (have octet already)
These seven elements occur naturally as molecules containing two atoms.

29. 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
N2O5: ________________
Where are the less electro negative elements(direction)? The more?
dinitrogen pentoxide
CuSO4∙5H2O
copper(II) sulfate pentahydrate
(ionic & covalent)

30. Ions Cations metals lose e’s positive (+) (metal) ion Anions nonmetals
gain e’s
negative (–)
(nonmetal)ide
Ions

31. Ionic Bonds
Attraction between +/– ions formed by metals & nonmetals transferring e–’s.

32. 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.
Do Al and SO4 on board
Pb4+ O2–
Pb2O4
PbO2

33. Naming Ionic Compounds
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
Anion: Write nonmetal name with –ide
OR the polyatomic anion name. (–ate, –ite)
Iron(II) sulfide, FeS
Magnesium sulfate, MgSO4
“Naming”

34. Common Polyatomic Ions
Name
Symbol
Charge
*ammonium
NH4+ 1+
*acetate
(ethanoate)
C2H3O2– (CH3COO–) 1–
*hydroxide
OH–
*perchlorate
ClO4–
*chlorate
ClO3–
chlorite
ClO2–
hypochlorite
ClO–
bromate
BrO3–
iodate
IO3–
*nitrate
NO3–
nitrite
NO2–
cyanide
CN–
*permanganate
MnO4–
*bicarbonate
(hydrogen carbonate)
HCO3–
*carbonate
CO32– 2–
*sulfate
SO42–
sulfite
SO32–
*chromate
CrO42–
dichromate
Cr2O72–
*phosphate
PO43– 3–
Common Polyatomic Ions
* these 12 will be on
Quiz 1
- all 20 Polyatomic Ions
will be on Quiz 2
WS 2d

35. “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

36. 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–
WS 2e

37. Anatomy of a Chemical Equation
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

38. Anatomy of a Chemical Equation
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
Reactants appear on the left side of the equation.

39. Anatomy of a Chemical Equation
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
Products appear on the right side of the equation.

40. Anatomy of a Chemical Equation
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
What are the states and symbols? s, l, g, aq, ↓, ↑
States (s, l, g, aq) written in parentheses next to each compound

41. Anatomy of a Chemical Equation
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
Subscripts show how many atoms of each element

42. 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.

43. Reaction Types

44. Combination
2 → 1
A + B → AB
Demo: MgO
2 Mg(s) O2(g)  2 MgO(s)

45. 1 → 2 Decomposition AB → A + B 2 NaN3(s)  2 Na(s) + 3 N2(g)
Sodium azide, heated decompose rapid and exo, hot N2 gas expands (50 milliseconds) then cools and deflates
(50 milliseconds!)
2 NaN3(s)  2 Na(s) N2(g)

46. Replacement Reactions (or “Displacement”)
Single Replacement
AB + C → A + CB
video clip
AgNO3(aq) + Cu(s)  Ag(s) + CuNO3(aq)
Double Replacement
AB + CD → AD + CB
Video Clip: Formation of Silver Crystals
Pb(NO3)2(aq) + KI(aq)  PbI2(s) + KNO3(aq)
Demo: PbI2 46

47. 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)

48. Formula Weights

49. Formula Weight (FW) Molecular Weight (MW) Ca: 1(40.08 amu)
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…
Usually go to at 2 dec places from PT
Ca: 1(40.08 amu)
+ Cl: 2(35.45 amu)
amu
C: 2(12.01 amu)
+ H: 6(1.008 amu)
30.07 amu

50. Percent Composition (# of atoms)(AW) % element = x 100 (FW)
One can find the percent by mass of a compound of each element in the compound by using this equation.

51. Percent Composition So the percentage of carbon in ethane (C2H6) is…
(2)(12.01)
(30.07)
24.02
30.07 =
x 100
= 79.88% C

52. Moles

53. Avogadro Constant 6.022 x 1023 1 mole of 12C has a mass of 12 g
So we don’t mix 2 Hs with 1 O, we mix 2 moles of Hs with 1 mole of Os.

54. Mole Relationships
One mole of atoms, ions, or molecules contains the Avogadro constant of those particles 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 ?

55. 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
amu is for an atom, g/mol is a mole of atoms (1 amu = 1/12 of a C-12 atom or × g)

56. (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

57. Using Moles = 6.022 x 1023 particles
What is the mass of 1 mole of copper(II) bromide, CuBr2?
How many moles are there in 112 g of copper(II) bromide, CuBr2?
How many particles present in each of the questions #1 & #2 above?
(63.55) + 2(79.90) = g
1 mol CuBr2
g CuBr2
112 g CuBr2 x
= mol CuBr2
6.022 x 1023 particles
1 mol
0.501 mol x
= 3.02 x 1023 particles

58. Finding Empirical Formulas

59. 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 = emp. form.
empirical mass multiple

60. 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

61. Calculating Empirical Formulas
The compound para-aminobenzoic acid (PABA in your sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%).
Determine the empirical formula of PABA.

62. 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
HW p. 113 #43a, 48 58
29.06
molecular mass
empirical mass
= 2
2 (C4H10) =
C4H10

63. Calculating Empirical Formulas
Percent to Mass
Mass to Mole
Divide by Small
Times ‘till Whole 63

64. 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
Is this chemical or physical separation?

65. Combustion Analysis
When 4-ketopentenoic acid is analyzed by combustion, a g sample produces g of CO2 and g of H2O. The acid contains only C, H, and O. What is the empirical formula of the acid?
C-C-C=C-C-OH C5H6O3
O O

66. 1 mol CO2
44.01 g CO2
1 mol C
1 mol CO2
12.01 g C
1 mol C
0.579 g CO2 x x x
? g C
= g C
1 mol H2O
18.02 g H2O
2 mol H
1 mol H2O
1.008 g H
1 mol H x x
0.142 g H2O x
? g H
= g H
g sample – (0.158 g C) – ( g H) =
? g O
= g O

67. 1 mol C
12.01 g C
mol C
0.158 g C x = =
1.67 C
mol
x 3 = 5 C
1 mol H
1.008 g H
mol H
g H x = =
2 H
mol
x 3 = 6 H
1 mol O
16.00 g O
mol O
0.126 g O x = =
1 O
mol
x 3 = 3 O
C5H6O3

68. A sample of a chlorohydrocarbon with a mass of 4
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?

69. 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
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

70. 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
HW p. 114 #52b
C6H15Cl3

71. Stoichiometry: mol-to-mol ratios
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

72. Stoichiometric Calculations
Rxn: A(aq) + 2 B(aq)  C(aq) + 2 D(aq)
molar mass A
g A
mol A
Coefficients of balanced equation
???
g A
1 mol A OR
1 mol A
g A OR
mol-to-mol ratio
2 mol B
1 mol A
1 mol A
2 mol B
g B
1 mol B
molar mass B
g B
mol B 72

73. 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)
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

74. 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) HCl(aq)  MgCl2(aq) + H2(g)
mole ratio B/A
molar mass B 1
mol HCl 1
mol H2
2.016
g H2 x x
18.0 g HCl x
36.46
g HCl 2
mol HCl
mol H2 1
g of A
molar mass A
= ____ g H2
0.498 g H2

75. 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.

76. Before After 2 H2 + O2 2 H2O H2 O2 Initial: ? mol ? mol ? mol Change:
Which is limiting?
2 H O H2O
Initial: ? mol ? mol ? mol
Change:
End:
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? 10 7
–10 –5
+10
0 mol
2 mol
10 mol
limiting
excess

77. Before After 2 H2 + O2 2 H2O H2 O2 Initial: ? mol ? mol ? mol Change:
End: 10 7
–10 –5
+10
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
2 mol
10 mol
Does limiting mean smallest amount of reactant?
No!
O2 is in smallest amount, but…
H2 is in smallest “stoichiometric” amount

78. Limiting Reactant Al(s) + CuCl2(aq) 2 3 3 Cu(s) + AlCl3(aq) 2
Solid aluminum metal is reacted with aqueous copper(II) chloride in solution.
If g Al are reacted with mol CuCI2, which is the limiting reactant?
How much product will be produced?
Al(s) + CuCl2(aq) 2 3 3
Cu(s) AlCl3(aq) 2
demo
DEMO – g CuCl2-2H2O in 200mL H2O makes M CuCl2, then…
Add 0.100g Al to 10mL CuCl2 for Al as limiting, then…
Add 0.200g Al more to CuCl2 as limiting and Al as excess

79. 2 Al(s) + 3 CuCl2(aq) 3 Cu(s) + 2 AlCl3(aq)
Limiting Reactant
1 mol Al
26.98 g Al
3 mol Cu
2 mol Al
0.030 g Al x x
= mol Cu
3 mol Cu
3 mol CuCl2
mol CuCl2 x
= mol Cu
2 Al(s) + 3 CuCl2(aq) Cu(s) + 2 AlCl3(aq)
Al is limiting
HW p. 115 #72

80. Theoretical Yield
theoretical yield: the maximum amount of product that can be formed
calculated by stoichiometry
(using LR only)
This is different from the actual yield, the amount one actually produces and measures (or experimental)
1 mol Al
26.98 g Al
3 mol Cu
2 mol Al
0.030 g Al x x
= mol Cu

81. (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

82. Percent Yield 4 Al + 3 O2 2 Al2O3 Aluminum will react with oxygen gas
according to the equation below
4 Al + 3 O 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?

83. Percent Yield 4 Al + 3 O2 2 Al2O3 39.3 g 44.2 g %Yield = x 100
HW p. 116 #79
Percent Yield
4 Al + 3 O Al2O3
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 of aluminum oxide are formed.
What is the percentage yield?
39.3 g
44.2 g
88.9 %
%Yield = x 100