1. LECTURE 2 CHM 151 ©slg TOPICS: SF's IN CALCULATIONS
SI, METRIC, ENGLISH UNITS
DIMENSIONAL ANALYSIS
UNIT CONVERSIONS
DENSITY

2. 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.

3. 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.

4. 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

5. 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

6. 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

7. 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.

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

9. Group Work #1, Answers: 24.569 g - .0056 g = ? 24.569 g - .0055 g
“ No more digits after decimal than value with
least number of digits after decimal”

10. 32.35 cm X cm X cm = ?
32.35 cm X cm X cm = “ ” cm3
allowed SF’s: = X 102
= 5.4 X 102 cm3
You may also use 540 cm3 which is also 2 SF’s

11. 54.01 lb lb + 1, lb =? lb lb
+ 1, lb
1, lb
= 1, lb

12. 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”

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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!

19.  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!)

20. 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

21. 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

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

23. Group Work
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?

24. A football field is 100. yd long. What length is this in
meters? (1 m = in)
1. State question: yd = ? m
2. State relationships:
1 yd = 3 ft
1 ft = 12 in
39.37 in = 1 m
Pathway: yd  ft  in  m

25. Pathway: yd  ft  in  m
3. Setup and Solve:
100. yd X 3 ft X 12 in X 1 m = m
1 yd ft in
= 91.4 m

26. In water, H2O, the bond length between each H and O is
94 pm. What is this length in millimeters?
1. State the question:
94 pm = ? mm
2. State the needed relationships:
1012 pm = 1 m
1 m = 103 mm
Pathway: pm  m  mm

27. Pathway: pm  m  mm
3. Setup and solve:
94 pm X 1 m X mm = X mm
1012 pm m
= X mm
= X 10-8 mm

28. 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

29. 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

30. 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

31. 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

32. 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”

33. 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”

34. Common Density Values 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
Magnesium: g/cm3
Aluminum: g/cm3
Silver: g/cm3
Gold: g/cm3

35. SOLVING DENSITY PROBLEMS
1. When the Density factor itself is desired:
Density = mass of object, etc (g)
volume occupied (mL, cm3, L)
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

36. 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

37. 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

38. .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

39. .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

40. 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:
1 mL oil = .92 g oil
453.6 g = 1 lb
Pathway: mL  g  lb
SETUP AND SOLVE AS GROUP WORK!

41. 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