USING SCIENTIFIC MEASUREMENTS

SECTION 2-3

Objectives

1. Distinguish between accuracy and precision

2. Determine the number of significant figures in measurements

3. Perform mathematical operations involving significant figures

4. Convert measurements into scientific notation

5. Distinguish between inversely and directly proportional relationships

Accuracy and Precision Accuracy – the closeness of measurements to

the correct or accepted value of the quantity measured

Precision – the closeness of a set of measurements of the same quantity made in the same way.

Visual Concept – Click Here

Percent Error Calculated by subtracting the experimental

value from the accepted value, dividing the difference by the accepted value, and then multiplying by 100

Percent error = valueaccepted - valueexperimental x 100

valueaccepted

Sample ProblemA student measures the mass and volume of a substance and calculates its density as 1.40 g/mL. The correct, or accepted, value of the density is 1.30 g/mL. What is the percentage error of the student’s measurement?

Percent error

= valueaccepte

d

- valueexperimental

valueaccepte

d

= 1.30 g/mL- 1.40

g/mL 1.30 g/mL

= - 7.7 %

x 100

x 100

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Error in Measurement

1. Skill of the measurer2. Measuring instruments3. Conditions of measurement

Significant Figures Consists of all digits known with

certainty, plus one final digit

6.75 cm

certain

estimated

Total of 3 significant figures, to two places after the decimal.

52.8 mL

**Read from bottom of meniscus.

Rules for determining significant zeros

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Sample ProblemHow many significant figures are in each of the following measurements?a) 28.6 g

3 s.f.

b) 3440. cm 4 s.f.

c) 910 m 2 s.f.

d) 0.046 04 L 4 s.f.

e) 0.006 700 0 kg 5 s.f.

Rounding

Greater than 5 inc. by 1 42.68 42.7

Less than 5 stay 17.32 17.3 5, followed by nonzero inc. by 1 2.7851 2.79 5, not followed by nonzero inc. by 1 4.635 4.64 Preceded by odd digit 5, not followed by nonzero stays 78.65 78.6

Preceded by Even digit

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Addition/subtraction with significant figures

The answer must have the same number of digits to the right of the decimal point as there are in the measurement having the fewest digits to the right of the decimal point.

35. 1+ 2.3456

So : 37.4

37.4456

Multiplication/Division with significant figures The answer can have no more significant

figures than are in the measurement with the fewest number of significant figures.

3.05 g ÷ 8.470 mL =0.360094451 g/mL

3 s.f.

4 s.f. Should be 3 s.f.

So : 0.360 g/mL

Sample Problems

Carry out the following calculations. Expresseach answer to the correct number of significantfigures.a) 5.44 m - 2.6103 m =

Answer: 2.84 m

b) 2.4 g/mL x 15.82 mL =Answer: 38 g

Conversion Factors and Significant Figures There is no uncertainty exact

conversion factors. Significant figures are only for

measurements, not known values!

Scientific Notation Numbers are written in the form M x

10n, where the factor M is a number greater than or equal to 1 but less than 10 and n is a whole number. 65 ooo km M is 6.5 Decimal moved 4 places to left

X 104

So: 6.5 x 104 kmWhy? Makes very small or large numbers more workable60 200 000 000 000 000 000 000

molecules6.02 x 1023 molecules Visual Concept – Click

Here

Extremely small numbers – negative exponent Ex: 0.0000000000567 g 5.67 x 10-11 g M should be in significant figures

Mathematical Operations Using Scientific Notation1. Addition and subtraction —These

operations can be performed only if the values have the same exponent (n factor).

a) example: 4.2 × 104 kg + 7.9 × 103 kg

OR

2. Multiplication —The M factors are multiplied, and the exponents are added algebraically.

a) example: (5.23 × 106 µm)(7.1 × 10−2 µm)= (5.23 × 7.1)(106 × 10−2)= 37.133 × 104 µm2 = 3.7 × 105 µm2

3. Division — The M factors are divided, and the exponent of the denominator is subtracted from that of the numerator.

a) example:

= 0.6716049383 × 103

= 6.7 102 g/mol

Using Sample Problems Analyze

Read the problem carefully at least twice and to analyze the information in it.

Plan Develop a plan for solving the problem.

Compute Substitute the data and necessary conversion factors.

Evaluate1) Check to see that the units are correct.2) Make an estimate of the expected answer.3) Expressed in significant figures.

Sample Problem

1. Analyzemass = 3.057 kgdensity = 2.70 g/cm3

volume = ?2. Plan

conversion factor : 1000 g = 1 kg

Calculate the volume of a sample of aluminum that has a mass of 3.057 kg. The density of aluminum is 2.70 g/cm3.

DV

=m

DV =

m

3. Computemass = 3.057 kg x

Round using sig figs ------- 1.13 × 103 cm3

4. Evaluate cm3, correct units. correct number of sig figs is three

1 kg

1000 g = 3057

g

DV =

m=

3057 g2.70

g/cm3

= 1132.222 . . . cm3

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Direct Proportions Two quantities if

divided by the other gives a constant value If one doubles so does

the other y = kx

Inverse Proportions Two quantities who’s product is a

constant If one doubles, the other is cut in half xy = k

Visual Concept – Click Here