1. I. Units of Measurement (p. 33 - 39)
CH. 2 - MEASUREMENT
I. Units of Measurement
(p )

2. A. Number vs. Quantity
Quantity - number + unit
UNITS MATTER!!

3. B. SI Units Quantity Symbol Base Unit Abbrev. Length l meter m Mass m
kilogram kg
Time t
second s
Temp T
kelvin K
Amount n
mole
mol

4. B. SI Units Prefix Symbol Factor mega- M 106 kilo- k 103 BASE UNIT ---
100
deci- d
10-1
centi- c
10-2
milli- m
10-3
micro- 
10-6
nano- n
10-9
pico- p
10-12

5. M V D = C. Derived Units 1 cm3 = 1 mL 1 dm3 = 1 L
Combination of base units.
Volume (m3 or cm3)
length  length  length
1 cm3 = 1 mL
1 dm3 = 1 L
D = M V
Density (kg/m3 or g/cm3)
mass per volume

6. D. Density
Mass (g)
Volume (cm3)

7. Problem-Solving Steps
1. Analyze
2. Plan
3. Compute
4. Evaluate

8. D. Density V = 825 cm3 M = DV D = 13.6 g/cm3 M = (13.6 g/cm3)(825cm3)
An object has a volume of 825 cm3 and a density of 13.6 g/cm3. Find its mass.
GIVEN:
V = 825 cm3
D = 13.6 g/cm3
M = ?
WORK:
M = DV
M = (13.6 g/cm3)(825cm3)
M = 11,200 g

9. D. Density D = 0.87 g/mL V = M V = ? M = 25 g V = 25 g 0.87 g/mL
A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid?
GIVEN:
D = 0.87 g/mL
V = ?
M = 25 g
WORK:
V = M D
V = g
0.87 g/mL
V = 29 mL

10. II. Using Measurements (p. 44 - 57)
CH. 2 - MEASUREMENT
II. Using Measurements
(p )

11. A. Accuracy vs. Precision
Accuracy - how close a measurement is to the accepted value
Precision - how close a series of measurements are to each other
ACCURATE = CORRECT
PRECISE = CONSISTENT

12. B. Percent Error Indicates accuracy of a measurement your value
accepted value

13. B. Percent Error % error = 2.9 %
A student determines the density of a substance to be 1.40 g/mL. Find the % error if the accepted value of the density is 1.36 g/mL.
% error = 2.9 %

14. C. Significant Figures Indicate precision of a measurement.
Recording Sig Figs
Sig figs in a measurement include the known digits plus a final estimated digit
2.35 cm

15. C. Significant Figures Counting Sig Figs (Table 2-5, p.47)
Count all numbers EXCEPT:
Leading zeros
Trailing zeros without a decimal point -- 2,500

16. Counting Sig Fig Examples
C. Significant Figures
Counting Sig Fig Examples
4 sig figs
3 sig figs
3. 5,280
3. 5,280
3 sig figs
2 sig figs

17. C. Significant Figures (13.91g/cm3)(23.3cm3) = 324.103g 324 g
Calculating with Sig Figs
Multiply/Divide - The # with the fewest sig figs determines the # of sig figs in the answer.
(13.91g/cm3)(23.3cm3) = g
4 SF
3 SF
3 SF
324 g

18. C. Significant Figures 3.75 mL + 4.1 mL 7.85 mL 3.75 mL + 4.1 mL
Calculating with Sig Figs (con’t)
Add/Subtract - The # with the lowest decimal value determines the place of the last sig fig in the answer.
3.75 mL mL
7.85 mL
3.75 mL mL
7.85 mL
224 g
+ 130 g
354 g
224 g
+ 130 g
354 g
 7.9 mL
 350 g

19. C. Significant Figures Calculating with Sig Figs (con’t)
Exact Numbers do not limit the # of sig figs in the answer.
Counting numbers: 12 students
Exact conversions: 1 m = 100 cm
“1” in any conversion: 1 in = 2.54 cm

20. C. Significant Figures Practice Problems 5. (15.30 g) ÷ (6.4 mL)
4 SF
2 SF
= g/mL
 2.4 g/mL
2 SF g g
 18.1 g
18.06 g

21. D. Scientific Notation 65,000 kg  6.5 × 104 kg
Converting into Sci. Notation:
Move decimal until there’s 1 digit to its left. Places moved = exponent.
Large # (>1)  positive exponent Small # (<1)  negative exponent
Only include sig figs.

22. D. Scientific Notation Practice Problems 7. 2,400,000 g 8. 0.00256 kg
9. 7  10-5 km
 104 mm
2.4  106 g
2.56  10-3 kg km
62,000 mm

23. D. Scientific Notation Calculating with Sci. Notation
(5.44 × 107 g) ÷ (8.1 × 104 mol) =
Type on your calculator:
EXP EE
EXP EE
ENTER
EXE
5.44 7
8.1 ÷ 4 =
= 670 g/mol
= 6.7 × 102 g/mol

24. Chemistry Binder Organization

25. III. Unit Conversions (p. 40 - 42)
CH. 2 - MEASUREMENT
III. Unit Conversions
(p )

26. A. SI Prefix Conversions
Symbol
Factor
mega- M
106
kilo- k
103
BASE UNIT
---
100
deci- d
10-1
move left
move right
centi- c
10-2
milli- m
10-3
micro- 
10-6
nano- n
10-9
pico- p
10-12

27. M V D = C. Derived Units 1 cm3 = 1 mL 1 dm3 = 1 L
Combination of base units.
Volume (m3 or cm3)
length  length  length
1 cm3 = 1 mL
1 dm3 = 1 L
D = M V
Density (kg/m3 or g/cm3)
mass per volume

28. D. Density
Mass (g)
Volume (cm3)

29. Problem-Solving Steps
1. Analyze
2. Plan
3. Compute
4. Evaluate

30. D. Density V = 825 cm3 M = DV D = 13.6 g/cm3 M = (13.6 g/cm3)(825cm3)
An object has a volume of 825 cm3 and a density of 13.6 g/cm3. Find its mass.
GIVEN:
V = 825 cm3
D = 13.6 g/cm3
M = ?
WORK:
M = DV
M = (13.6 g/cm3)(825cm3)
M = 11,200 g

31. D. Density D = 0.87 g/mL V = M V = ? M = 25 g V = 25 g 0.87 g/mL
A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid?
GIVEN:
D = 0.87 g/mL
V = ?
M = 25 g
WORK:
V = M D
V = g
0.87 g/mL
V = 29 mL

32. Homework
P. 54
Practice problems 1 & 2

33. Conversion Factors Problems
Dimensional Analysis
Conversion Factors
Problems

34. B. Dimensional Analysis
A tool often used in science for converting units within a measurement system
Conversion Factor
A numerical factor by which a quantity expressed in one system of units may be converted to another system

35. B. Dimensional Analysis
The “Factor-Label” Method
Units, or “labels” are canceled, or “factored” out

36. B. Dimensional Analysis
Steps:
1. Identify starting & ending units.
2. Line up conversion factors so units cancel.
3. Multiply all top numbers & divide by each bottom number.
4. Check units & answer.

37. B. Dimensional Analysis
Lining up conversion factors:
= 1
1 in = 2.54 cm
2.54 cm cm
1 =
1 in = 2.54 cm
1 in in

38. B. Dimensional Analysis
How many milliliters are in 1.00 quart of milk? qt mL
1.00 qt
1 L
1.057 qt
1000 mL
1 L
= 946 mL 

39. B. Dimensional Analysis
You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3. lb
cm3
1.5 lb
1 kg
2.2 lb
1000 g
1 kg
1 cm3
19.3 g
= 35 cm3

40. B. Dimensional Analysis
How many liters of water would fill a container that measures 75.0 in3?
in3 L
75.0 in3
(2.54 cm)3
(1 in)3
1 L
1000 cm3
= 1.23 L

41. B. Dimensional Analysis
5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off? cm in
8.0 cm
1 in
2.54 cm
= 3.2 in

42. B. Dimensional Analysis
6) Taft football needs 550 cm for a 1st down. How many yards is this? cm yd
550 cm
1 in
2.54 cm
1 ft
12 in
1 yd
3 ft
= 6.0 yd

43. B. Dimensional Analysis
7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire? cm
pieces
1.3 m
100 cm
1 m
1 piece
1.5 cm
= 86 pieces