CH. 2 - MEASUREMENT

I

II

III C. Johannesson

I. Using Measurements

(p. 44 - 57)

CH. 2 - MEASUREMENT

C. Johannesson

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

C. Johannesson

B. Percent Error

Indicates accuracy of a measurement

100literature

literaturealexperimenterror %

your value

accepted value

C. Johannesson

B. Percent Error

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.

100g/mL 1.36

g/mL 1.36g/mL 1.40error %

% error = 2.9 %

C. Johannesson

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

C. Johannesson


Counting Sig Figs (Table 2-5, p.47)

Count all numbers EXCEPT:

Leading zeros -- 0.0025

Trailing zeros without a decimal point -- 2,500

C. Johannesson

4. 0.080

3. 5,280

2. 402

1. 23.50


Counting Sig Fig Examples

1. 23.50

2. 402

3. 5,280

4. 0.080

4 sig figs

3 sig figs

3 sig figs

2 sig figs

C. Johannesson


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) = 324.103g

324 g

4 SF 3 SF3 SF

C. Johannesson


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

+ 4.1 mL

7.85 mL

224 g

+ 130 g

354 g 7.9 mL 350 g

3.75 mL

+ 4.1 mL

7.85 mL

224 g

+ 130 g

354 g

C. Johannesson


Calculating with Sig Figs (con’t) Exact Numbers do not limit the # of sig

figs in the answer.Counting numbers: 12 studentsExact conversions: 1 m = 100 cm “1” in any conversion: 1 in = 2.54 cm

C. Johannesson


5. (15.30 g) ÷ (6.4 mL)

Practice Problems

= 2.390625 g/mL

18.1 g

6. 18.9 g

- 0.84 g

18.06 g

4 SF 2 SF

2.4 g/mL2 SF

C. Johannesson

D. Scientific Notation

Converting into Sci. Notation:

Move decimal until there’s 1 digit to its left. Places moved = exponent.

Large # (>1) positive exponentSmall # (<1) negative exponent

Only include sig figs.

65,000 kg 6.5 × 104 kg

C. Johannesson


7. 2,400,000

g

8. 0.00256 kg

9. 7 10-5 km

10. 6.2 104

mm

Practice Problems

2.4 106 g

2.56 10-3 kg

0.00007 km

62,000 mm

C. Johannesson


Calculating with Sci. Notation

(5.44 × 107 g) ÷ (8.1 × 104 mol) =

5.44EXPEXP

EEEE÷÷

EXPEXP

EEEE ENTERENTER

EXEEXE7 8.1 4

= 671.6049383 = 670 g/mol = 6.7 × 102 g/mol

Type on your calculator:

C. Johannesson

E. Proportions

Direct Proportion

Inverse Proportion

xy

xy

1

y

x

y

x

I

II

III C. Johannesson

II. Units of Measurement

(p. 33 - 39)

CH. 2 - MEASUREMENT

C. Johannesson

A. Number vs. Quantity

Quantity - number + unit

UNITS MATTER!!

C. Johannesson

B. SI Units

Quantity Base Unit Abbrev.

Length

Mass

Time

Temp

meter

kilogram

second

kelvin

m

kg

s

K

Amount mole mol

Symbol

l

m

t

T

n

C. Johannesson

B. SI Units

mega- M 106

deci- d 10-1

centi- c 10-2

milli- m 10-3

Prefix Symbol Factor

micro- 10-6

nano- n 10-9

pico- p 10-12

kilo- k 103

BASE UNIT --- 100

C. Johannesson

C. Derived Units

Combination of base units.

Volume (m3 or cm3) length length length

D = MV

1 cm3 = 1 mL1 dm3 = 1 L

Density (kg/m3 or g/cm3)mass per volume

C. Johannesson

D. DensityM

ass

(g)

Volume (cm3)

Δx

Δyslope D

V

M

C. Johannesson

Problem-Solving Steps

1. Analyze

2. Plan

3. Compute

4. Evaluate

C. Johannesson

D. Density

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

V

MD

C. Johannesson

D. Density

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 = 25 g

0.87 g/mL

V = 29 mLV

MD

I

II

III C. Johannesson

III. Unit Conversions

(p. 40 - 42)

CH. 2 - MEASUREMENT

C. Johannesson

A. SI Prefix Conversions

1. Find the difference between the

exponents of the two prefixes.

2. Move the decimal that many places.

To the leftor right?

C. Johannesson

=


NUMBERUNIT

NUMBER

UNIT

532 m = _______ km0.532

C. Johannesson


mega- M 106

deci- d 10-1

centi- c 10-2

milli- m 10-3

Prefix Symbol Factor

micro- 10-6

nano- n 10-9

pico- p 10-12

kilo- k 103

mo

ve le

ft

mo

ve r

igh

t BASE UNIT --- 100

C. Johannesson


1) 20 cm = ______________ m

2) 0.032 L = ______________ mL

3) 45 m = ______________ nm

4) 805 dm = ______________ km

0.2

0.0805

45,000

32

C. Johannesson

3

3

cm

gcm

B. Dimensional Analysis

The “Factor-Label” Method Units, or “labels” are canceled, or

“factored” out

g

C. Johannesson


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.

C. Johannesson


Lining up conversion factors:

1 in = 2.54 cm

2.54 cm 2.54 cm

1 in = 2.54 cm

1 in 1 in

= 1

1 =

C. Johannesson

1. Dimensional Analysis

How many milliliters are in 1.00 quart of milk?

1.00 qt 1 L

1.057 qt= 946 mL

qt mL

1000 mL

1 L

C. Johannesson


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= 35 cm3

1000 g

1 kg

1 cm3

19.3 g

C. Johannesson


How many liters of water would fill a container that measures 75.0 in3?

75.0 in3 (2.54 cm)3

(1 in)3= 0.191 L

in3 L

1 L

1000 cm3

C. Johannesson


4) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off?

8.0 cm 1 in

2.54 cm= 3.1 in

cm in

C. Johannesson


5) Taft football needs 550 cm for a 1st down. How many yards is this?

550 cm 1 in

2.54 cm= 6.0 yd

cm yd

1 ft

12 in

1 yd

3 ft

CH. 2 - MEASUREMENT

Documents

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