1. LECTURE TOPIC I: Problem Solving in Chemistry
Significant Figures and Scientific Notation
Allowed Digits and “Rounding Off”
Dimensional Analysis and SI Unit Conversions
Density and Percent Problems
(Kotz & Treichel text: )

2. Handling Numbers in Chemistry: The Necessary Skills:
Correct use of significant figures (“sig figs”)
Handling scientific notation
Rounding off computational values
Knowledge of SI, Metric, and English Units
Use of dimensional analysis as problem solving tool

3. Correct use of significant figures (“SF’s”)
All measured values are limited in scope by the
accuracy of the instrument used.
When correctly expressed, all measured values contain
all the digits which can be read directly plus one
estimated digit.
The digits in measured values are described as “significant figures” when they are presented in this
fashion.

4. TEST YOUR MEMORY, Counting SF’s:

5. Correct Number of SF’s:

6. RECALL THE RULES, COUNT ONLY:
1. All non zero digits:
2.35 cm, 3 SF g, 4 SF
2. All Internal zeros:
25.01 oz, 4 SF
3. All ending zeros when the decimal point is
expressed:
2.00 lb, 3 SF in, 2 SF ft, 3 SF

7. DO NOT COUNT AS SIG FIGS:
1. Ending zeros, no decimal point
2000 g, 1 SF
2. All beginning zeros
.0005 g, 1 SF

8. Group Work (Test Yourself!)
These “Group Work Exercises” will occur 2-4 times
during each two hour lecture session...
Your “table-mates” or nearest neighbors will make
up your group... minimum 2, maximum 4, ideal: 3.
Pick a “secretary” to write down your solutions; all
members sign name on top right hand corner of 1st
page; hand in at end of lecture, front desk...
The group work counts 5 points/lecture session!

9. I will present the solution key in class following each
activity, but it will not be in the Topics slides.
The Group Work Solution Keys will be found on-line
at groups.yahoo.com along with the homework and
resources for each topic....
On-line students: “group work activities” are to be used
as “test yourself” exercises as you proceed through each
topic; do the activity and then check out the answers
before proceeding...

10. GROUP WORK 1.1: How many SF’s?

11. SCIENTIFIC NOTATION Scientific Notation is an alternate method of
expressing numerical values in which the original
value is multiplied or divided by ten until there
is only ONE digit to the left of the decimal.
The resulting number is multiplied by 10 raised
to the appropriate power to restore the worth
of the original value.

12. Scientific Notation is used to express very large and
very small values, and to facilitate expression of some
value to correct number of SF’s.
= 2.70 X 10-5 =
10x10x10x10x10
27,000 = 2.70 X 104 = x10x10x10x10
1,000,000 = 1.00 X 106 (3 SF) or 1 X 106 (1 SF)

13. The Methodology:
1. Recall: = = = =100
Equivalent Values: = x 1 = x 100
2. Small Numbers: subtract one from the power of ten
for each right move of decimal:
= x 100 = 2.7 x = 2.7 X 10-5
3. Large Numbers: add one to the power of ten
for each left move of decimal:
27,000 = 27,000 x = 2.7 x = 2.7 X 104

14. In a nutshell:
Add +1 to power of ten for each LEFT MOVE
Add -1 to power of ten for each RIGHT MOVE

16. GROUP WORK 1.2 Place into Scientific Notation:
95,000 (4 SF’s) X 10-4
X 10+5
X 10 +4

17. SF’s IN CALCULATIONS When doing calculations involving measured
values (always the case in science!), you must
limit the number of digits in your results to reflect
the degree of uncertainty introduced by these
values .
You must be familiar with the rules for number of
SF’s or digits allowed and also with the rules for
rounding values down to the allowed number of digits.

18. Calculations Involving SF’s
Multiplication and Division:
The final answer in a computation involving these
operations should have no more SF’s than the
value in the original problem with the least number
of SF’s.
Addition and Subtraction:
The sum of these operations is allowed no more
digits after the decimal than the original value with
the least number of digits after the decimal.

19. Rounding off Answers to Correct # SF’s
If first digit to be dropped is <5, drop it and all following digits, leaving rest of number unchanged.
Round off to 4 SF’s: =
Less than five

20. If the first digit to be dropped is >5, drop it and all
following digits, but increase the last “retained
digit” by one:
Round off to 4 SF’s: =
>5

21. If the first digit to be dropped is exactly five,
no non zero digits following, “even up” the
resulting rounded-off value:
Increase the last retained digit to make it even
if it is odd only.
Round off to 4 SF’s:
= = 23.44
Note:
= = 23.45

22. SF’s in Calculations, Samples:
1.30 in. X .20 in. X in. = in.3 = 7.7 X in.3
3 SF SF SF SF allowed
1.30 in Since one value has
.20 in no digits after decimal,
in none are allowed!
in. = 2962 in.

23. Group Work 1.3 Calculate and round off to correct # SF’s:
g g = ?
32.35 cm X cm X cm = ?
54.01 lb lb + 1, lb =?

24. METRIC AND SI UNITS Prefixes you should know:
M, 1,000,000 (106) X basic unit “mega”
k, 1, (103) X basic unit “kilo”
d, 1/ (10-1) X basic unit “deci”
c, 1/ (10-2) X basic unit “centi”
m, 1/ (10-3) X basic unit “milli”
, 1/1,000, (10-6) X basic unit “micro”
n, 1/1,000,000, (10-9) X basic unit “nano”
p, 1/1,000,000,000,000 (10-12) X basic unit “pico”

25. Common SI/Metric/ English Conversions (Know!)
1. MASS
1000 g = 1 kg mg = 1 g
(1 g = 10-3 kg ) (1 mg = 10-3 g)
1 lb = g
16 oz avoir = 1 lb
SI / English
gateway

26. 2. LENGTH:
1 m = 100 cm = 103 mm m = 1 km
(1 cm = 10-2 m) (1 m = 10-3 km) (1 mm = 10-3 m)
1 m = 109 nm = 1012 pm
2.540 cm = 1 inch
SI/ ENG
gate
1 yd = 3 ft = 12 in yd = 1 mi

27. 3. VOLUME:
1 L = 1000 mL = 1000 cm3
1 qt = L= mL
SI / ENG
GATE
1 qt = 2 pt gal = 4 qt
1pt = 16 fl oz

28. Unit Conversion Using Dimensional Analysis
Three Steps Involved:
a) Recognizing and stating the question
b) Recognizing and stating relationships
c) Multiplication of Initial value by appropriate
conversion factors to get desired value

29. Dimensional Analysis and Mass Conversion
Problems
A typical goal weight for a 5’ 6” female might be
135 pounds. What is this weight in kilograms (kg) ?
1. State the Question: (Use following format):
Original value, old unit = ? Value, new unit
135 lb = ? kg

30. 2. State the relationship(s) between the old unit
Eng / SI conversion
135 lb = ? kg
2. State the relationship(s) between the old unit
and the new unit:
1 lb = g
103 g = 1 kg
Eng / SI mass
gateway
Problem pathway: lb  g  kg

31. 135 lb = ? kg
Problem pathway: lb  g  kg
3. Multiply the old value, unit by the appropriate
conversion factor(s) to arrive at the new value, unit:
1 lb = g lb = 1 = g
453.6 g lb
103 g = 1 kg g = 1 = 1 kg
1 kg g

32. 135 lb = ? kg
Problem pathway: lb  g  kg
3. Setup and Solve:
Old value, unit X factor 1 X factor 2 = new value, unit
135 lb X g X 1 kg = kg
1 lb g
Calculator answer
Not done yet!

33.  SF’s All defined conversions within a system are exact;
The “1” in all conversion relationships is exact
4 SF
3 SF
Exact
135 lb X g X 1 kg = kg
1 lb g
Exact
Exact,
 SF’s
# SF’s allowed, 3
kg = kg (Answer!)

34. 2. LENGTH:
1 m = 100 cm = 103 mm m = 1 km
(1 cm = 10-2 m) (1 m = 10-3 km) (1 mm = 10-3 m)
1 m = 109 nm = 1012 pm
2.540 cm = 1 inch
SI/ ENG
gate
1 yd = 3 ft = 36 in yd = 1 mi

35. Dimensional Analysis and Length Conversion
Problems
Orange light has a wavelength of 625 nm. What would
this length be in centimeters?
1. State the question:
625 nm = ? cm
2. State the needed relationships:
109 nm = 1 m
1 m = 102 cm
Pathway: nm  m  cm

36. 3. Setup and solve:
625 nm X 1 m X cm = X cm
109 nm m
= X cm
= X 10-5 cm

37. Group Work 1.4
A football field is 100. yd long. What length is this in
meters? (1 m = in)
In water, H2O, the bond length between each H and O is
94 pm. What is this length in millimeters?

38. 3. VOLUME:
1 L = 1000 mL = 1000 cm3
1 qt = L= mL
SI / ENG
GATE
1 qt = 2 pt gal = 4 qt
1pt = 16 fl oz

39. Dimensional Analysis and Volume Conversion
Problems
What is the volume in cm3 occupied by one half gallon
( .50 gal) of Sunkist Orange Juice? What is this value in
in3?
1. State First Question: gal = ? cm3
2. Relationships:
1 gal = 4 qt
1 qt = mL
1 mL = 1 cm3
Pathway: gal  qt  mL  cm3

40. Pathway: gal  qt  mL  cm3
3. Setup and Solve:
.50 gal X 4 qt X mL X 1 cm3 = cm3
1 gal qt mL
= X 103 cm3
= X 103 cm3

41. What is the volume in cm3 occupied by one half gallon
( .50 gal) of Sunkist Orange Juice? What is this value in
in3? ( .50 gal = 1.9 X 103 cm3)
1. State the Question: X 103 cm3 = ? in3
2. State the relationships:
2.540 cm = 1 in
( cm)3 = ( 1 in)3
16.39 cm3 = 1 in3
Pathway: cm3  in3

42. Pathway: cm3  in3
3. Setup and solve:
1.9 X 103 cm3 X in3 = in3
16.39 cm3
= X 102 in3
= X 102 in3
Calculator: enter 1.9, then “E” button, then “3”

43. Dimensional Analysis and Density Conversions
“Density” is a handy conversion factor which relates
the mass of any object or solution or gas to the amount
of volume it occupies.
It is a physical property of matter, and can be obtained
for all elements, most compounds and common
solutions in a reference handbook (or on line).
Denser matter “feels heavier” and accounts for the
properties of “floating” and “sinking”

44. Common Density Values gases Dry air: 1.2 g/L at 25 oC, 1 atm pressure
Water: g/mL at 0 oC
g/ mL at 4.0 oC
g/mL at 25 oC
Sea Water: g/mL at 15 oC
Antifreeze: g/mL at 20oC
liquids
Magnesium: g/cm3
Aluminum: g/cm3
Silver: g/cm3
Gold: g/cm3
solids

45. SOLVING DENSITY PROBLEMS
1. Density itself can be calculated from experimental
values:
Density = mass of object, solution, substance
volume occupied (mL, cm3, L)
Volume can be determined in several ways:
a) direct measurement, liquid
b) liquid displacement, solid
c) measurement of dimensions, calculation

46. Volume by Displacement:
A liquid in which the solid does not dissolve is
suitable for this technique

47. Calculation, Volume:
V = l X w X ht (rectangle)
V = e (cube)
V =  r2 ht (cylinder)
radius = diameter / 2; all dimensions in same units w
diameter
height ht l

48. SOLVING DENSITY PROBLEMS
2. In all other cases, where conversion of mass to volume
or volume to mass is desired, density is used as a
“conversion factor”:
D, Al = 2.70 g /cm g Al = 1 cm3 Al
2.70 g Al = 1 = 1 cm3 Al
1 cm3 Al g Al
“conversion factors”

49. Sample, Obtaining Density Value
A “cup” is a volume used by cooks in US. One cup is
equivalent to 237 mL. If one cup of olive oil has a mass
of 205 g, what is the density of the oil in g/mL and in
oz avoir / in3?
1. State the question:
D, olive oil = ? g/ mL = ? oz avoir / in3
2. State formula:
Density = mass, g
volume, mL

50. 3. Substitute values and solve:
D = mass = g = g = g
volume mL mL mL
Second question: What is this density factor expressed
as oz avoir/ in3?
1. State Question:
.865 g = ? oz avoir
mL in3

51. .865 g = ? oz avoir
mL in3
2. State relationships:
453.6 g = 1 lb
1 lb = 16 oz avoir
1 mL = 1 cm3
2.540 cm = 1 in
( cm)3 = (1 in)3
16.39 cm3 = 1 in3
Pathway: g  lb  oz avoir; mL  cm3  in3

52. .865 g = ? oz avoir
mL in3
Pathway: g  lb  oz avoir; mL  cm3  in3
3. Setup and Solve:
.865 g X 1 lb X 16 oz avoir X 1 mL X cm3
1 mL g lb cm in3
= oz avoir
in3
= oz avoir

53. Density as a Conversion Factor:
Peanut oil has a density of .92 g/mL. If the recipe calls
for 1 cup of oil (1 cup = 237 mL), what mass of oil, in
lb, are you going to use?
State question:
237 mL oil =? lb oil
State relationship:
( D, oil= .92 g / mL) mL oil = .92 g oil
453.6 g = 1 lb
Pathway: mL  g  lb

54. Solution:
237 mL oil =? lb oil
Pathway: mL  g  lb
237 ml oil X g oil X 1 lb = lb oil
1 mL oil g
= lb oil
Density conversion factor

55. GROUP WORK 1.5: Setup and solve as Conversion Type Problem
A gold coin is 2.75 cm in diameter and .50 cm thick.
If the density of gold is 19.3 g / cm3, what is the mass
of the coin in grams?
Note:
Volume =  r2 ht
r = d / 2
ht = “thickness”
D, Au = 19.3 g/cm g Au = 1 cm3 Au

56. Percent Composition by Mass
“Percent composition” is a convenient way to
describe a mixture or solution or compound in terms
of mass of the part contained in 100 mass units of the
whole:
“This solution is 15% salt” means that for every 15 g of
salt there is 100 g of solution or:
15 g salt = 100 g solution
“The brass alloy is 15 % tin and 45% copper”
100 g alloy = 15 g tin = 45 g copper

57. Like Density, Percent Composition is:
calculated from experimental values
used as a conversion factor.
Calculation of % by mass (g):
% Part = g part X 100%
g whole
A brass alloy weighing g was found to contain
g of copper. What is the % of Cu in the alloy?
%Cu = g Cu X 100% = % Cu
g alloy
= g Cu in 100 g alloy

58. Percent as a Conversion Factor
What mass in grams of a brass alloy would contain
25.00 g Cu if the alloy is 43.16% Cu?
Question: g Cu = ? g brass
Relationship: g Cu = 100 g brass
Setup and solve:
25.00 g Cu X g brass = g brass
43.16 g Cu
% factor

59. Other Percent Relationships, Solutions:
Generally, chemists use percent in terms of “mass, g
per 100 g of the whole”.
However, for solutions, % is sometimes given in terms of g solute per volume in mL of the solution:
“ 5.00% solution of salt, 5.00 g salt / 100 mL solution”
5.00 g solute salt = 100 mL solution
You must carefully note how problem defines % !

60. If a 5.00 % salt solution is made up to contain 5.00 g
of salt for every 100 mL solution, how many mL of
this solution would contain 19.0 g of the salt?
Question: g salt = ? mL solution
Relationship: g salt = 100 mL solution
19.0 g salt X 100 mL soltn = mL solution
5.00 g salt

61. Density/Percent Solution Problems
Automobile batteries are filled with sulfuric acid, which
is a solution of liquid H2SO4 in water.
What is the mass (in grams) of the H2SO4 in mL of
the battery acid solution, if the density of the solution
is g /mL and the solution is 38.08% H2SO4 by mass?

62. 1.“Describing the Scene”:
Analysis of Problem
1.“Describing the Scene”:
Automobile batteries are filled with sulfuric acid, which
is a solution of liquid H2SO4 in water.
2. “Stating the Question”:
What is the mass (in grams) of the H2SO4 in mL of
the battery acid solution
3.“Giving the Relationships”:
if the density of the solution is g /mL and the
solution is 38.08% H2SO4 by mass?

63. Automobile batteries are filled with sulfuric acid, which
is a solution of liquid H2SO4 in water.
What is the mass (in grams) of the H2SO4 in mL of
the battery acid solution, if the density of the solution
is g /mL and the solution is 38.08% H2SO4 by mass?
Question: mL soltn = ? g H2SO4
Relationships: g soltn = 1 mL soltn
38.08 g H2SO4 = 100 g soltn
Pathway: mL soltn  g soltn  g H2SO4

64. Question: 500. mL soltn = ? g H2SO4
Relationships: g soltn = 1 mL soltn
38.08 g H2SO4 = 100 g soltn
Pathway: mL soltn  g soltn  g H2SO4
Setup and solve:
500. mL soltn X density factor X % factor = g H2SO4
500. mL soltn X g soltn X g H2SO4
1 mL soltn g soltn
= g H2SO4 = 245 g H2SO4

65. Group Work 1.6
Automobile batteries are filled with sulfuric acid, which
is a solution of liquid H2SO4 in water.
How many mL of the acid solution would contain g
H2SO4, if the density of the solution is g /mL and the
solution is 38.08% H2SO4 by mass?
Question: g H2SO4 = ? mL soltn
Relationships: g soltn = 1 mL soltn
38.08 g H2SO4 = 100 g soltn
Pathway: g H2SO4  g soltn  mL soltn D %