1 Chapter 2 Measurements in Chemistry Standard measuring device Standard scale gram (g)
2 Reliability of Measurements Accuracy closeness to true value Precision reproducibility Example: 98.6 o F 98.5 o F 98.7 o F average: precision: 0.1 o F o F + 0.1 o F
3 Ch. 2.1 Measurement Systems English & metric
4 Ch. 2.2 Metric System Units base unit =
5 base unit = mass vs weight
base unit = 6
7 10cm x 10cm x 10cm 1 L = 1000 ml
8 Table 2.1 Common Metric System Prefixes You must know all of these for exams and quizzes.
9 Ch. 2.3 Exact and Inexact Numbers Exact no uncertainty direct count defined equivalency Inexact any measurement has a degree of uncertainty
10 Ch. 2.4 Uncertainty in Measurement and Significant Figures # of significant figures = all certain digits + one estimated digit 2 sig. fig. 3 sig. fig.
Guidelines for Determining Significant Figures 1. All nonzero digits are significant. 2. Zeros may or may not be significant. a. Leading zeros not significant 0.00729 0.0000052 sig fig sig fig b. Confined zeros always significant 50.003 0.00601 sig fig sig fig c. Trailing zeros are significant if a decimal point is present in the number 63.050 0.09010 sig fig sig fig 11
12 Trailing zeros in numbers that do not contain a decimal point: 30600 g ambiguous To remove ambiguity, use prefixes or scientific notations (Ch 2.6) 30.600 kg 5 sig fig 30.6 kg 3 sig fig Stoker text pg 43 100 o F = 3 sig fig 30 o C = 2 sig fig
Ch. 2.5 Significant Figures and Mathematical Operations Rounding off Numbers These rules are listed on page 36 of your lab manual: 1. If the first nonsignificant figure is less than 5, drop it and all other nonsignificant figures. 2. If the first nonsignificant figure is more than 5 or is 5 followed by digits other than all zeros, drop all nonsignificant figures and increase the last significant figure by 1. 3. If the first nonsignificant figure is 5 alone or is 5 followed by only zeros, drop all nonsignificant figures and increase the last significant figure by 1 if it is odd but leave it alone if it is even. (Round Even Rule) 13
14 Examples of rounding off numbers 7.248 round to 2 sig fig less than 5, round down = 10.532 round to 2 sig fig a 5 followed by nonzeros, round up = 524.650 round to 4 sig fig a 5 followed by zero, round even = 34.235 round to 4 sig fig a 5 alone, round even =
15 Operational Rules 1. multiplication and division least number of sig fig in any of the measurements 12.45 x 5.1 = (63.495) calculator answer sig fig sig fig correct answer = sig fig
16 2. addition and subtraction common number of digits to the right of the decimal point 8.30 + 3.1246 (11.4246) calculator answer correct answer to the common hundredth place
17 Mixed mode calculation Always watch the significant figures of intermediate answers (52.70 0.359) x 28.4332 = 1488.222121 calculator answer = correct answer 52.70-0.359 1 sig fig for intermediate answer Round only the final answer!
18 Ch. 2.6 Scientific Notation A x 10 n Exponent Coefficient 1000 = 1 x 10 x 10 x 10 = 1 x 10 3 0.01 = 1 = 1 = 1 x 10-2 10 x 10 10 2 75,000 = 7.5 x 10 4 decimal point is moved places 0.0000056 = 5.6 x 10-6 decimal point is moved places
19 Significant Figures and Scientific Notation 37,000 = 3.7000 x 10 4 5 sig fig 3.700 x 10 4 sig fig 3.70 x 10 4 sig fig 3.7 x 10 4 sig fig 4 x 10 4 sig fig
Ch. 2.7 Conversion Factors Obtained from equalities and used in dimensional analysis (factor unit method) 12 in. = 1 ft 12 in. = 1 1 ft = 1 1 ft 12 in. example: Convert 2.5 ft into inches 2.5 ft x 12 in. = in. 1 ft information x conversion = information given factor sought 20
21 sample calculation: convert 0.0156 L to ml from table 2.1: which is equal to: 1 ml = 10-3 L 1000 ml = 1 L (preferred) conversion factors: 1000 ml 1L. 1 L 1000 ml 0.0156 L x 1000 ml = ml 1 L
Table 2.2 Equalities and Conversion Factors 22
23 ft to m not given in Table 2.2 but 1.00 m = 39.4 in. and we know 1 ft = 12 in. ft in. m 38 ft x 12 in. x 1 m = (11.57360406) 1 ft 39.4 in. correct answer: m
24 Ch. 2.9 Density demo Pb vs styrofoam density = mass volume = m V usual units: g/cm 3 solids g/ml liquids g/l gases
25 Table 2.3 Densities of Selected Substances Density of water at 4 o C = need to memorize. g/ml This is the only density you
26 Fig. 2.9 A penny floats on liquid mercury (Hg)
27 Density is an Intensive Property Intensive Properties are independent of the sample size and represent qualities that identify the substance: density b.p. m.p. Extensive Properties depend on the amount of matter: mass volume length
28 demo: Determination of the density and identity of an unknown silver cube mass: volume: density: identity of metal: Densities of Selected Silver Metals Beryllium Be 1.85 g/cm 3 Aluminum Al 2.70 g/cm 3 Zinc Zn 7.13 g/cm 3 Silver Ag 10.5 g/cm 3 Lead Pb 11.3 g/cm 3
29 Density as a Conversion Factor Example: A patient s urine sample has a density of 1.02 g/ml. How many grams of urine are eliminated on a day in which 1250 ml is excreted? information x conversion = information given factor sought 1250 ml x 1.02 g = (1275) g calculator answer ml correct answer = g
30 Ch. 2.10 Temperature Scales Figure 2.10 Conversions Between Temperature Scales o F = 9 ( o C) + 32 o C = 5 ( o F 32) 5 9 K = o C + 273.15 o F 32 = o C 180 100
31 Sample calculation: Convert 3650 o C into o F o F 32 = o C 180 100 o F 32 = 3650 180 100 o F = (3650 x 180/100) + 32 o F = sig fig
32 Ch. 2.11 Heat Energy and Specific Heat Temperature vs Heat Energy Intensive Property Extensive Property Common Unit of Heat Energy: calorie (cal) One cal is the amount of heat energy needed to raise the temperature of 1 g of water by 1 o C. Specific heat is the quantity of heat energy, in cal, necessary to raise the temperature of 1 g of a substance by 1 o C.
33 Table 2.4 Specific Heats of Selected Common Substances You must KNOW the specific heat of water = 1 cal/g o C (this is a defined quantity and has no uncertainty) demo: burn $ bill
34 Specific heat (c) = cal. g x o C cal = c x g x o C heat energy = specific heat x mass x temperature change heat energy (absorbed or released) = c x m x t
Sample calculation: 5.5 g of Au at 25 o C is heated by 10 cal of heat. What is the final temperature? c Au = 0.031 cal/g o C heat energy (absorbed or released) = c x m x t 10 cal = 0.031 cal/g o C x 5.5 g x t t = 59 o C t final = t initial + t = + 59 o C = 35
36 Exp 3 Determination of the specific heat of a metal hot metal + cold water t metal decreases t water increases t final is heat loss of metal = heat gain of water c 1 x m 1 x t 1 = c 2 x m 2 x t 2
37 Other common units for heat energy The joule (J) 1 cal = 4.184 J The dietary Calorie (must know) 1 Cal = 1000 cal = 1 kcal Average basic metabolic rate: 2000 Cal/day 3500 Cal/lbs to gain or lose body weight Carbohydrates Fat Protein Hambuger Green beans Apples 4.1 Cal/g 9.3 Cal/g 4.1 Cal/g 3.6 Cal/g 0.38 Cal/g 0.59 Cal/g
Sample calculation: One ounce of cereal gives 112 Cal of energy on oxidation. How many kg of water can be heated from 20 o C to 30 o C by burning the cereal? (The specific heat of water, the equation, and the necessary conversion factors for this question will not be given on exams and quizzes.) 38 Heat energy released by cereal and absorbed by the water: 112 Cal = 112 kcal = cal heat energy = c x m x t m = heat energy c x t m = cal. 1 cal/(g o C) x (30 o C 20 o C) m = g = kg