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Chapter 2 Measurements in Chemistry Standard measuring device Standard scale

gram (g)

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Reliability of Measurements Accuracy closeness to true value Precision reproducibility Example:

98.6oF 98.5oF 98.7oF

average: oF + 0.1oF

precision: 0.1oF

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Ch. 2.1 Measurement Systems

“English” & metric

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Ch. 2.2 Metric System Units

base unit =

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base unit = mass vs weight

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base unit =

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10cm x 10cm x 10cm 1 L = 1000 mL

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Table 2.1 Common Metric System Prefixes

You must know all of these for exams and quizzes.

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Ch. 2.3 Exact and Inexact Numbers Exact • no uncertainty • direct count • defined equivalency

Inexact • any measurement has a degree of uncertainty

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Ch. 2.4 Uncertainty in Measurement and Significant Figures # of significant figures = all certain digits + one estimated digit

2 sig. fig. 3 sig. fig.

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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

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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 100oF = 3 sig fig 30oC = 2 sig fig

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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)

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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 =

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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

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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

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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!

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Ch. 2.6 Scientific Notation A x 10n Exponent Coefficient

1000 = 1 x 10 x 10 x 10 = 1 x 103 0.01 = 1 = 1 = 1 x 10-2 10 x 10 102 75,000 = 7.5 x 104

decimal point is moved ___ places 0.0000056 = 5.6 x 10-6

decimal point is moved ___ places

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Significant Figures and Scientific Notation 37,000 = 3.7000 x 104 5 sig fig 3.700 x 104 sig fig 3.70 x 104 sig fig 3.7 x 104 sig fig 4 x 104 sig fig

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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

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sample calculation: convert 0.0156 L to mL from table 2.1: 1 mL = 10-3 L which is equal to: 1000 mL = 1 L (preferred) conversion factors: 1000 mL 1L . 1 L 1000 mL 0.0156 L x 1000 mL = mL 1 L

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Table 2.2 Equalities and Conversion Factors

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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

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Ch. 2.9 Density demo Pb vs styrofoam

mass m density = __________ = _______

volume V usual units: g/cm3 solids g/mL liquids g/L gases

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Table 2.3 Densities of Selected Substances Density of water at 4oC = g/mL This is the only density you need to memorize.

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Fig. 2.9

A penny floats on liquid mercury (Hg)

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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

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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/cm3 Aluminum Al 2.70 g/cm3 Zinc Zn 7.13 g/cm3 Silver Ag 10.5 g/cm3 Lead Pb 11.3 g/cm3

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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

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Ch. 2.10 Temperature Scales Figure 2.10

Conversions Between Temperature Scales oF = 9 (oC) + 32 oC = 5 (oF – 32) 5 9

K = oC + 273.15 oF – 32 = oC 180 100

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Sample calculation: Convert 3650oC into oF oF – 32 = oC 180 100 oF – 32 = 3650 180 100 oF = (3650 x 180/100) + 32

oF = sig fig

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Ch. 2.11 Heat Energy and Specific Heat

Temperature vs Heat Energy

Intensive Extensive Property 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 1oC. Specific heat is the quantity of heat energy, in cal, necessary to raise the temperature of 1 g of a substance by 1oC.

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Table 2.4 Specific Heats of Selected Common Substances

You must KNOW the specific heat of water = 1 cal/g • oC

(this is a defined quantity and has no uncertainty)

demo: burn $ bill

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Specific heat (c) = cal . g x oC cal = c x g x oC heat energy = specific heat x mass x temperature change heat energy (absorbed or released) = c x m x ∆t

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Sample calculation: 5.5 g of Au at 25oC is heated by 10 cal of heat. What is the final temperature? cAu = 0.031 cal/g • oC heat energy (absorbed or released) = c x m x ∆t 10 cal = 0.031 cal/g • oC x 5.5 g x ∆t

∆t = 59oC tfinal = tinitial + ∆t

= + 59oC

=

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Exp 3 Determination of the specific heat of a metal hot metal + cold water tmetal decreases twater increases tfinal is heat loss of metal = heat gain of water c1 x m1 x ∆t1 = c2 x m2 x ∆t2

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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 4.1 Cal/g Fat 9.3 Cal/g Protein 4.1 Cal/g

Hambuger 3.6 Cal/g Green beans 0.38 Cal/g Apples 0.59 Cal/g

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Sample calculation: One ounce of cereal gives 112 Cal of energy on oxidation. How many kg of water can be heated from 20oC to 30oC 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.)

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 oC) x (30oC – 20oC) m = g = kg