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1-1 MEASUREMENT AND SIG FIGS

1-1 MEASUREMENT AND SIG FIGS. 1-2 The number of significant figures in a measurement depends upon the measuring device. Figure 1.9A 32.3 0 C32.33 0 C

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Page 1: 1-1 MEASUREMENT AND SIG FIGS. 1-2 The number of significant figures in a measurement depends upon the measuring device. Figure 1.9A 32.3 0 C32.33 0 C

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MEASUREMENT AND SIG FIGS

Page 2: 1-1 MEASUREMENT AND SIG FIGS. 1-2 The number of significant figures in a measurement depends upon the measuring device. Figure 1.9A 32.3 0 C32.33 0 C

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The number of significant figures in a measurement depends upon the measuring device.

Figure 1.9A

32.30C32.330C

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Rules for Determining Which Digits are Significant

All digits are significant

•Make sure that the measured quantity has a decimal point.•Start at the left of the number and move right until you reach the first nonzero digit.•Count that digit and every digit to it’s right as significant.

Numbers such as 5300 L are assumed to only have 2 significant figures. A terminal decimal point is often used to clarify the situation, but scientific notation is the best!

except zeros that are used only to position the decimal point.

Zeros that end a number and lie either after or before the decimal point are significant; thus 1.030 ml has four significant figures, and 5300. L has four significant figures also.

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Sample Problem 1.7 Determining the Number of Significant Figures

PROBLEM: For each of the following quantities, underline the zeros that are significant figures(sig figs), and determine the number of significant figures in each quantity. For (d) to (f), express each in exponential notation first.

(b) 0.1044 g(a) 0.0030 L (c) 53,069 mL

(e) 57,600. s(d) 0.00004715 m (f) 0.0000007160 cm3

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Rules for Significant Figures in Answers

1. For addition and subtraction. The answer has the same number of decimal places as there are in the measurement with the fewest decimal places.

106.78 mL = 106.8 mL

Example: subtracting two volumes

863.0879 mL = 863.1 mL

865.9 mL - 2.8121 mL

Example: adding two volumes 83.5 mL

+ 23.28 mL

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= 23.4225 cm3 = 23 cm39.2 cm x 6.8 cm x 0.3744 cm

2. For multiplication and division. The number with the least

certainty limits the certainty of the result. Therefore, the answer

contains the same number of significant figures as there are in the

measurement with the fewest significant figures.

Rules for Significant Figures in Answers

Multiply the following numbers:

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Rules for Rounding Off Numbers1. If the digit removed is more than 5, the preceding number increases by 1. 5.379 rounds to 5.38 if three significant figures are retained and to 5.4 if two significant figures are retained.

2. If the digit removed is less than 5, the preceding number is unchanged. 0.2413 rounds to 0.241 if three significant figures are retained and to 0.24 if two significant figures are retained.

3.If the digit removed is 5, the preceding number increases by 1 if it is odd and remains unchanged if it is even.17.75 rounds to 17.8, but 17.65 rounds to 17.6. If the 5 is followed only by zeros, rule 3 is followed; if the 5 is followed by nonzeros, rule 1 is followed: 17.6500 rounds to 17.6, but 17.6513 rounds to 17.7

4. Be sure to carry two or more additional significant figures through a multistep calculation and round off only the final answer.

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Issues Concerning Significant Figures

graduated cylinder < buret ≤ pipet

numbers with no uncertainty

1000 mg = 1 g

60 min = 1 hr

These have as many significant digits as the calculation requires.

be sure to correlate with the problem

FIX function on some calculators

Electronic Calculators

Choice of Measuring Device

Exact Numbers

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Precision and Accuracy Errors in Scientific Measurements

Random Error - In the absence of systematic error, some values that are higher and some that are lower than the actual value.

Precision -Refers to reproducibility or how close the measurements are to each other.

Accuracy -Refers to how close a measurement is to the real value.

Systematic Error - Values that are either all higher or all lower than the actual value.

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Figure 1.10

precise and accurate

precise but not accurate

Precision and accuracy in the laboratory.

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systematic error

random error

Precision and accuracy in the laboratory.Figure 1.10

continued

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Definitions for Components of Matter

Element - the simplest type of substance with unique physical and

chemical properties. An element consists of only one type of atom. It

cannot be broken down into any simpler substances by physical or

chemical means.

Molecule - a structure that consists of two or

more atoms that are chemically bound together

and thus behaves as an independent unit.

Figure 2.1Figure 2.1

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Compound - a substance

composed of two or more elements

which are chemically combined.

Mixture - a group of two or more elements and/or compounds that are physically intermingled.

Definitions for Components of Matter

Figure 2.1

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The total mass of substances does not change during a chemical reaction.

reactant 1 + reactant 2 product

total mass total mass=

calcium oxide + carbon dioxide calcium carbonate

CaO + CO2 CaCO3

56.08g + 44.00g 100.08g

Law of Mass Conservation:

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No matter the source, a particular compound is composed of the same elements in the same parts (fractions) by mass.

Calcium carbonate Calcium carbonate

Analysis by MassAnalysis by Mass(grams/20.0g)(grams/20.0g)

Mass FractionMass Fraction(parts/1.00 part)(parts/1.00 part)

Percent by MassPercent by Mass(parts/100 parts)(parts/100 parts)

8.0 g calcium8.0 g calcium2.4 g carbon2.4 g carbon9.6 g oxygen 9.6 g oxygen

20.0 g20.0 g

40% calcium40% calcium12% carbon12% carbon48% oxygen 48% oxygen

100% by mass100% by mass

0.40 calcium0.40 calcium0.12 carbon0.12 carbon0.48 oxygen 0.48 oxygen

1.00 part by mass1.00 part by mass

Law of Definite (or Constant) Composition:

Figure 2.2

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If elements A and B react to form two compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers.

Example: Carbon Oxides A & BCarbon Oxide I : 57.1% oxygen and 42.9% carbonCarbon Oxide II : 72.7% oxygen and 27.3% carbon

Assume that you have 100g of each compound. In 100 g of each compound: g O = 57.1 g for oxide I & 72.7 g for oxide II

g C = 42.9 g for oxide I & 27.3 g for oxide II

g O

g C=

57.1

42.9= 1.33

= g O

g C

72.7

27.3= 2.66

2.66 g O/g C in II

1.33 g O/g C in I

2

1=

Law of Multiple Proportions: