Chemistry: The Study of Change Chang & Goldsby 12 th edition

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Transcription:

Chemistry: The Study of Change Chang & Goldsby 12 th edition modified by Dr. Hahn Chapter 1

Example 1.4 Determine the number of significant figures in the following measurements: (a)478 cm (b)6.01 g end class 8/11 9 am class (c)0.825 m (d)0.043 kg (e)1.310 10 22 atoms (f)7000 ml

Example 1.4 (2) Solution (a) 478 cm -- Three, because each digit is a nonzero digit. (b) 6.01 g -- Three, because zeros between nonzero digits are significant. (c) 0.825 m -- Three, because zeros to the left of the first nonzero digit do not count as significant figures. (d) 0.043 kg -- Two. Same reason as in (c). (e) 1.310 10 22 atoms -- Four, because the number is greater than one so all the zeros written to the right of the decimal point count as significant figures.

Example 1.4 (3) (f) 7000 ml -- This is an ambiguous case.the number of significant figures may be four (7.000 10 3 ), three (7.00 10 3 ), two (7.0 10 3 ), or one (7 10 3 ). This example illustrates why scientific notation must be used to show the proper number of significant figures.

Addition or Subtraction Significant Figures(1) The answer cannot have more digits to the right of the decimal point than any of the original numbers. (use column and least significant column determines significant column of answer) 89.332 +1.1 90.432 3.70 2.9133 0.7867 one significant figure after decimal point round off to 90.4 two significant figures after decimal point round off to 0.79 End 8/11F 10am class 5

Significant Figures (2) Multiplication or Division The number of significant figures in the result is set by the original number that has the smallest number of significant figures. 4.51 3.6666 = 16.536366 = 16.5 3 sig figs 5 sig figs round to 3 sig figs 6.8 112.04 = 0.0606926 = 0.061 5 sig figs 2 sig figs round to 2 sig figs 6

Exact Numbers Significant Figures (3) Numbers from definitions or numbers of objects are considered to have an infinite number of significant figures. The average of three measured lengths: 6.64, 6.68 and 6.70? 6.64 + 6.68 + 6.70 3 = 6.67333 = 6.67 = 7 Because 3 is an exact number 7

Example 1.5 Carry out the following arithmetic operations to the correct number of significant figures: a) 11,254.1 g + 0.1983 g b) 66.59 L 3.113 L c) 8.16 m 5.1355 d) 0.0154 kg 88.3 ml e) 2.64 10 3 cm + 3.27 10 2 cm

Example 1.5 (2) Solution In addition and subtraction, the number of decimal places in the answer is determined by the number having the lowest number of decimal places. In multiplication and division, the significant number of the answer is determined by the number having the smallest number of significant figures. a) 11,254.1 g + 0.1983 g 11,254.2983 g round off to 11,254.3 g b) 66.59 L 3.113 L 63.477 L round off to 63.48 L

Example 1.5 (3) c) 8.16 m 5.1355 = 41.90568 m round off 41.9 m d) 0.0154 kg = 0.000174405436 kg ml round off to 88.3 ml 0.000174 kg ml or 1.74 10 4 kg ml e) First we change 3.27 10 2 cm to 0.327 10 3 cm and then carry out the addition 2.64 cm + 0.327 cm 10 3. Following the procedure in (a), we find the answer is 2.97 10 3 cm. (or use calculator)

Accuracy versus Precision Accuracy how close a measurement is to the true value Precision how close a set of measurements are to each other accurate & precise precise but not accurate not accurate & not precise 11

Precision and Accuracy (continued) Measurements are said to be precise if they are consistent with one another. accurate only if they are close to the actual value.

Dimensional Analysis Method of Solving Problems 1. Determine which unit conversion factor(s) are needed 2. Carry units through calculation 3. If all units cancel except for the desired unit(s), then the problem was solved correctly. given quantity conversion factor = desired quantity given unit desired unit given unit =desired unit 13

Example 1.6 A person s average daily intake of glucose (a form of sugar) is 0.0833 pound (lb). What is this mass in milligrams (mg)? (1 lb = 453.6 g. )

Example 1.6 (2) Strategy The problem can be stated as? mg = 0.0833 lb The relationship between pounds and grams is given in the problem. This relationship will enable conversion from pounds to grams. A metric conversion is then needed to convert grams to milligrams 1 mg = 1 10 3 g. Arrange the appropriate conversion factors so that pounds and grams cancel and the unit milligrams is obtained in your answer.

Example 1.6 (3) Solution The sequence of conversions is pounds grams milligrams Using the following conversion factors 453.6 g 1 lb and 1 mg 1 10 3 g End 9 am class8/14m we obtain the answer in one step:? mg = 0.0833 lb 453.6 g 1 lb 1 mg 1 10 3 g = 3.78 104 mg