 # SECTION 2-3. Objectives 1. Distinguish between accuracy and precision 2. Determine the number of significant figures in measurements 3. Perform mathematical.

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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 = value accepted - value experimental x 100 value accepted Sample Problem A 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 = value accepted - value experimental value accepted = 1.30 g/mL - 1.40 g/mL 1.30 g/mL = - 7.7 % x 100 Video Concept – Click Here

Error in Measurement 1. Skill of the measurer 2. Measuring instruments 3. 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 Visual Concept – Click Here

Sample Problem How 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 5inc. by 142.68  42.7  Less than 5stay17.32  17.3  5, followed by nonzero inc. by 12.7851  2.79  5, not followed by nonzero inc. by 14.635  4.64 Preceded by odd digit  5, not followed by nonzero stays78.65  78.6 Preceded by Even digit Visual Concept – Click Here

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. Express each answer to the correct number of significant figures. 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 10 n, 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 10 4 So: 6.5 x 10 4 km Why? Makes very small or large numbers more workable 60 200 000 000 000 000 000 000 molecules 6.02 x 10 23 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 Notation 1. Addition and subtraction —These operations can be performed only if the values have the same exponent (n factor). a) example: 4.2 × 10 4 kg + 7.9 × 10 3 kg OR

2. Multiplication —The M factors are multiplied, and the exponents are added algebraically. a) example: (5.23 × 10 6 µm)(7.1 × 10 −2 µm) = (5.23 × 7.1)(10 6 × 10 −2 ) = 37.133 × 10 4 µm 2 = 3.7 × 10 5 µm 2 3. Division — The M factors are divided, and the exponent of the denominator is subtracted from that of the numerator. a) example: = 0.6716049383 × 10 3 = 6.7  10 2 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.  Evaluate 1) Check to see that the units are correct. 2) Make an estimate of the expected answer. 3) Expressed in significant figures.

Sample Problem 1. Analyze mass = 3.057 kg density = 2.70 g/cm 3 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/cm 3. D V = m D V = m

3. Compute mass = 3.057 kg x Round using sig figs ------- 1.13 × 10 3 cm 3 4. Evaluate  cm 3, correct units.  correct number of sig figs is three 1 kg 1000 g =3057 g D V = m = 2.70 g/cm 3 = 1132.222... cm 3 Video Concept – Click Here

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

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