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. Matter H2O CO2 NaCl Atom: Element: same type of atom (1 or more) H2 O2
simplest particle with properties of element
Element:
same type of atom (1 or more) C C H H O O Na C H2 O2 C C
Compound:
different atoms bonded
H2O CO2 NaCl
molecule

3. Heterogeneous Mixture
chromatography
distillation
Matter
(boiling)
separate physically
cannot 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

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

5. 0.0003700400 grams Significant Digits Box & Dot nonzero
trailing 0’s if decimal
grams
Nonzero digits
Captive Zeroes (between two sig digs)
Leading Zeroes (at the beginning of a number) are NEVER significant
Trailing Zeroes:
Significant ONLY if there’s a decimal
0’s
Exact numbers are infinitely significant. How many boy students in here? Plus girls? How many sig figs?
What if 500 is significant to the 10’s digit?
5.0 x 102

6. Sigs Digs in Operations
+ or –
round answers to keep the fewest decimal places
round answers to keep the fewest significant digits =
5.68
5.7
x or ÷
6.40 x =
12.8 13

7. 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 (Z), but…
can have different mass numbers.

8. 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

9. 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)(%) …

10. 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.

11. 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)

12. Ions Cations metals lose e’s (+) charge (metal) ion Anions nonmetals
gain e’s
(–) charge
(nonmetal)ide
Ions

13. 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

14. Nick and his Polyatomic Ions
Nick the
Camel ate
Clam
Crepes for
Supper in
Phoenix with
Mandy.
NO31–
(nitrate)
CO32–
(carbonate)
ClO31–
(chlorate)
CrO42–
(chromate)
SO42–
(sulfate)
PO43–
(phosphate)
MnO41–
(permanganate)
#consonants = O’s
#vowels = charge

15. Naming Polyatomic Ions & Acids
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
perchlorate
ClO4–
chlorate
ClO3–
chlorite
ClO2–
hypochlorite
ClO–
perchloric acid
HClO4
chloric acid
HClO3
chlorous acid
HClO2
hypochlorous acid
HClO
sulfate
SO42–
sulfite
SO32–
sulfuric acid
H2SO4
sulfurous acid
H2SO3

16. Reaction Types Synthesis Decomposition Single Replacement
A + B → AB 2 → 1
Decomposition
AB → A + B 1 → 2
Single Replacement
AB + C → A + CB
Double Replacement
AB + CD → AD + CB
Zn in acid releases H2 gas
Combustion
CxHy + O2 → CO2 + H2O 16

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

18. (groups of 6.022x1023 particles)
Using Moles
Moles are the bridge from
the particle (micro) scale to
the real-world (macro) scale.
(from PT)
(equation sheet)
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

19. 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
mol-to-mol ratio
2 mol B
1 mol A
g B
1 mol B
molar mass B
g B
mol B 19

20. 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

21. Types of Formulas Empirical Formulas: Molecular Formulas: CH3 C2H4O
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

22. Calculating Empirical Formulas
from Mass % Composition
Steps (rhyme)
Percent to Mass
Mass to Mole
Divide by Small
Times ‘til 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

23. C2H5 C4H10 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.
If molecular mass 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 (C2H5) =
C4H10

24. 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?

25. 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
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
0.142 g H2O x
? g H
= g H
g sample – (0.158 g C) – ( g H) =
? g O
Then:
Steps 2-4 to get empirical formula
= g O

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

27. Find the Limiting Reactant
Convert reactant A to reactant B
Compare B available to B needed
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?)
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
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 27

28. Theoretical Yield
theoretical yield: the maximum possible 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
HW p. 115 #72

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