gc-s004-scientificnotation

Henry R. Kang (1/2010)

General Chemistry

Lecture 4

Scientific Notation


Contents

• Scientific NotationRules of scientific notation

Arithmetic operations

• Significant FiguresRules of significant numbers

• Dimensional Analysis

• Problem Solving


Scientific Notation


Scientific Notation

• Scientific notation is designed to deal with very large and very small numbers. The number of gold atoms in 197.0 g of pure gold nugget.

602,200,000,000,000,000,000,000 = 6.022×1023

The mass of a single gold atom in grams0.000,000,000,000,000,000,000,327 = 3.27×10-22

• General expression N × 10n

N represents a number, its absolute value is between 1 and 10.1 ≤ |N| < 10

The exponent n is a positive or negative integer.


Rules of Scientific Notation

• Decimal point placement Value >> 1

Move decimal to left after the first non-zero integer.

The exponent is positive.• The value of the exponent is equal to the number of places that the

decimal point has moved to left.

Example: • 568.762 5.68762×102

Value << 1 Move decimal to right after the first non-zero integer.

The exponent is negative.• The value of the exponent is equal to the number of places that the

decimal point has moved to right.

Example: • 0.0000007548 7.548×10-7


Arithmetic Operations of Scientific Notation - Addition

• Example: 4.31×104 + 3.9×103 Write each quantity with the same

exponent n 4.31×104 + 0.39×104

Combine N1 and N2 4.70×104

The exponent, n, remains the same


Arithmetic Operations of Scientific Notation - Subtraction

• Example: 4.3×106 − 3.96×107 Write each quantity with the same

exponent n 0.43×107 − 3.96×107

Combine N1 and N2 -3.53×107

The exponent, n, remains the same


Arithmetic Operations of Scientific Notation - Multiplication

• Multiplication = (N1×N2) × 10 (n1+n2)

Multiply N1 and N2

Add exponents n1 and n2

Adjust the resulting N, if N>10

Example: (4.2 × 10-7) × (7.5 × 103)

(4.2 × 7.5) × (10-7+3) = 31.5 × 10-4

= 3.15 × 10-3


Arithmetic Operations of Scientific Notation - Division

• Division = (N1/N2) × 10 (n1− n2)

Divide N1 and N2

Subtract exponent n2 from n1

Adjust the resulting N, if N<1

Example: (3.5 × 104) ÷ (5.0 × 109)

(3.5 ÷ 5.0) × 104–9 = 0.70 × 10-5

= 7.0 × 10-6


Significant Figures


Significant Figures

• Significant figures are the meaningful digits in a measured or calculated quantity.

• The number of significant figures that a measuring device can provide is depended on the resolution (division, mark, scale, etc.) of the device. Top-loading balance can measure mass to 1/10, 1/100, or

1/1000 of a gram, depending on the model & manufacturer.

Analytical balance can read to 1/10000 of a gram.

Buret can measure volume to 1/100 of a mL.

• The last digit is understood to be uncertain, which is estimated.


Rules of Determining Significant Figures• Any digit that is not zero is significant

4.2183 m : 5 significant figures

• Zeros between nonzero digits are significant 80.0054 lb : 6 significant figures

• Zeros to the left of the first nonzero digit are not significant 0.0000349 g : 3 significant figures

• For numbers contain decimal point; all the trailing zeros count as significant figures 3.400 cm : 4 significant figures

0.0200500 kg : 6 significant figures

• For numbers do not contain decimal point; the trailing zeros may or may not be significant 6800 mi : 4, 3, or 2 significant figures

Use scientific notation to avoid this ambiguity.


Examples of Significant Figures

• Determine the number of significant figures in the following measurements:

1. 328 cm Answer: 3

2. 6.01 g Answer : 3

3. 0.805 m Answer : 3

4. 0.043 kg Answer : 2

5. 1.510×1022 atoms Answer : 4

6. 5000 mL Answer : 1, 2, 3, or 4

7. 5.0×103 mL Answer : 2

8. 5.000×103 mL Answer : 4

9. 0.030200 m3 Ans: 5

10. 6.40×104 molecules Ans: 3


Rules of Significant Figures: Addition

• In addition and subtraction, the answer cannot have more digits to the right of the decimal point then either of the original numbers.

89.453+ 2.5 one significant figure after decimal point

91.953 round off to 92.0

3.72 two significant figures after decimal point

- 2.91730.8027 round off to 0.80


Rules of Significant Figures: Multiplication

• In multiplication and division, the number of significant figures in the final product or quotient is determined by the original number that has the smallest number of significant figures (s.f.).

4.51 × 3.6666 = 16.536366 = 16.5 3 s.f. 5 s.f. Round to 3 s.f.

6.8 / 112.04 = 0.0606926 = 0.061 2 s.f. 5 s.f. Round to 2 s.f.


Rules of Significant Figures: Exact Number• Exact numbers are a number obtained from definition or by

counting numbers of objects. Examples: 12 inches to a foot (definition), 8 coins in a bottle (counting), 5

measurements of the mass of a penny coin (counting).

• Exact numbers are considered to have an infinite significant figures. This means that the exact number does not involve in determining the

resulting significant figures; other numbers determine the final number of significant figures.

Example: The average of three measured lengths; 6.64, 6.68 and 6.70?

6.64 + 6.68 + 6.70 = 6.67333 = 6.67 ( ≠ 7)

3

3 is an exact number, which does not involve in determining the significant figures.


Examples of Significant Figures• Carry out the following arithmetic operations to the correct

number of significant figures:

1. 2.568 × 5.8 / 4.186In one step:

2.568 × 5.8 / 4.186 = 3.558146202 ← round off to 3.6

In two steps:

14.8944 / 4.186 = 15 / 4.186 = 3.583373149 ← round off to 3.6

1. 5.41 – 0.398 = 5.012 ← round off to 5.01

2. 3.38 – 3.01 = 0.37Note that the result has one less significant figure.

1. 4.18 – 58.16 × (3.38 – 3.01)Underline the least significant digits:

4.18 – 58.16 × (3.38 – 3.01) = 4.18 – 58.16 × 0.37

= 4.18 – 21.5192 = -17.3392 ← round off to -17


Examples (Continued)

• Carry out the following arithmetic operations to the correct number of significant figures:

1. 254 g + 0.1983 g = 254.1983 g ← round to 254 g

Adding a large number and a small number, which is below the uncertainty of the large number, will not change the magnitude of the large number.

1. 66.59 L – 3.113 L = 63.477 L ← round to 63.48 L

2. 8.16 m × 5.1355 = 41.90568 m ← round off to 41.9 m

3. 0.0154 kg ÷ 88.3 mL = 0.000174405436 kg/mL ← round to 0.000174 kg/ml ( or 1.74 ×10-4 kg/mL)

4. 2.64×103 cm + 3.27×102 cm = 2.64×103 cm + 0.327×103 cm

= 2.967×103 cm ← round to 2.97×103 cm


Dimensional Analysis


Dimensional Analysis (Factor-Label Method)

• Dimensional analysis is a technique for converting a quantity from one unit to another. A computational method in which units are included with

numbers. Conversion factors are used such that source units will be

cancelled to obtain destination unit.

• General relationship for quantity (Source quantity) × (Conversion factor)

= Destination quantity

• General relationship for units (Source unit) × (Conversion factor) = Destination unit


Problem Solving by Dimensional Analysis• All units are included in calculation.

• Decide which equality is needed.

• Arrange the equality into a correct conversion factor for canceling units.

The end result should contain the desired units only.

• Example: How many mL are in 0.580 L? Equality: 1 L = 1000 mL

Conversion factor: (1000 mL) / (1 L)

There are two ways to arrange the conversion ratio; the correct conversion factor will give the correct unit for the destination quality.

1 L1 L1000 mL

1000 mL0.580 L 0.580 L× ×

= 580. mL = 0.000580 L2/mL


Example 1 of Dimensional Analysis

• A sample of nitrogen gas in a glass bulb weighs 185 mg. What is this mass in SI base unit of mass (kg)?

• Answer: Equalities: Conversion factors:

1 g = 1000 mg (1 g) / (1000 mg)

1 kg = 1000 g (1 kg) / (1000 g)

185 mg × (1 g / 1000 mg) × (1 kg / 1000 g)

= 1.85×10-4 kg



• Convert the speed of sound from 767 miles/hour to meters/second.

Equalities: Conversion factors: 1 mi = 1609 m (1609 m) / (1 mi)

1 hour = 60 min (1 hour) / (60 min)

1 min = 60 s (1 min) / (60 s)

1 mi1609 m

= 343 m/s

× × 60 s1 min

60 min

1 hour×767 mi/hour



• The oxygen molecule consists of two oxygen atoms (O2) a distance of 121 pm apart. How many millimeters is this distance?


1 pm = 10-12 m (10-12 m) / (1 pm)

1 m = 1000 mm (1000 mm) / (1 m)

121 pm × (10-12 m / 1 pm) × (1000 mm / 1 m)

= 1.21×10-7 mm



• The world’s oceans contain approximately 1.35×1021 L of water. What is this volume in km3?


1 km = 1000 m (1 km) / (1000 m)

1 m = 100 cm (1 m) / (100 cm)

1 cm3 = 1 mL (1 cm3) / (1 mL)

1 L = 1000 mL (1000 mL) / (1 L)

1.35×1021 L × (1000 mL / 1 L) × (1 cm3 / 1 mL)

× (1 m / 100 cm)3 × (1 km / 1000 m)3

= 1.35×1024 cm3 × (1 m3 / 106 cm3) × (1 km3 /109 m3)

= 1.35×1018 m3 × (1 km3 /109 m3)

= 1.35×109 km3


Problem Solving

• Read the question carefully Understand the information given Know “what you are asked to solve?”

• Find the appropriate equation Relate the given information and the unknown Use dimensional analysis

• Check your answer Sign (+ or -), units, and significant figures

• Judge whether the answer is reasonable Make a “ball-park” estimate


• The average daily intake of glucose (a form of sugar) of a person is 1.30 ounce (oz). What is this mass in milligram (mg)? (16 oz = 1 lb and 1 lb = 453.6 g)

• Answer: This is a unit conversion problem by

converting Oz → lb → g → mg

1.30 oz × (1 lb / 16 oz) × (453.6 g / 1 lb)

× (1000 mg / 1 g) = 3.69× 104 mg

Example 1 of Problem Solving


• An average adult has 5.2 L of blood. What is the volume of blood in pints (pt)? 1 L = 1.06 qt and 1 qt = 2 pt

• Answer: This is a unit conversion problem by

converting L → quart → pint

5.2 L × (1.06 qt / 1 L) × (2 pt / 1 qt)

= 11 pt



• The density of silver is 10.5 g/cm3. Convert the density to unit of kg/m3.

• Answer: This is a unit conversion problem by converting

g→ kg (1 kg = 1000 g)

cm3 → m3 (1 m = 100 cm)

10.5 g/cm3 × (1 kg / 1000 g) × (100 cm / 1 m)3

= 10500 kg/m3

= 1.05×104 kg/m3


gc-s004-scientificnotation

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