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Chapter 1 Introduction: Matter and Measurement.

1.1 The study of chemistry What is chemistry?

Briefly define the following terms:

Matter.

Element.

Atom.

Molecule.

1.2 Classifications of Matter

By state:

Property / State Solid Liquid Gas

Shape

Volume

Compressibility

By composition (see the handout):

Elements

Pure substances

Compounds

Matter

Homogeneous

Mixtures

Heterogeneous

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76900 7.69 x 104

0.0021 2.1 x 10-3

1.3 Properties of Matter

What are properties and where do they come from?

Properties can be Physical or Chemical (does the identity change?).

Properties can also be Extensive or Intensive (does size matter?).

Types of changes: Physical vs Chemical (does the identity change in the process?).

Concepts we need from Math

Scientific Notation

𝑫. 𝑫𝑫 × 𝟏𝟎𝒏

[Just one digit before the decimal × 10𝑛]

𝟏 ≤ 𝑫. 𝑫𝑫 < 𝟏𝟎 ; 𝒏 (the power) can be positive or negative

Standard Format Scientific Notation Q Write in scientific notation. 14723 0.0699 55100000 Q Write as a regular number (standard notation). 3.51 x 10-6 1.8 x 104 8.2301 x 10-3 What do we need to do to compare numbers in scientific notation? Q Which number is larger on each pair? 5.22x10-7 vs 9.87x10-8 ; 9.87x10-3 vs 1.43x102 ; 6.31x10-5 vs 9.87x10-5

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Exponent Rules (See Appendix A.1)

𝑎𝑛 ∙ 𝑎𝑚 = 𝑎𝑛+𝑚 ; 23 ∙ 24 = 27 = 128 ; 𝑎𝑛 ∙ 𝑏𝑛 = (𝑎 ∙ 𝑏)𝑛 ; 32 ∙ 42 = (3 ∙ 4)2 = 144

𝑎𝑛

𝑎𝑚= 𝑎𝑛−𝑚 ;

25

23= 25−3 = 22 = 4 ;

𝑎𝑛

𝑏𝑛= (

𝑎

𝑏)

𝑛

; 43/23 = (4/2)3 = 8

(𝑏𝑛)𝑚 = 𝑏𝑛∙𝑚 ; (23)2 = 26 = 64 ; 𝑏−𝑛 = 1𝑏𝑛⁄ ; 2−3 = 1

23⁄ = 0.125

𝑏0 = 1 ; 𝑏1 = 𝑏 ; 1𝑛 = 1

Simplifying Scientific Notation (×) (2.6 × 105)(9.2 × 10−13) = (2.6)(105)(9.2)(10−13) = (2.6)(9.2)(105−13)

= (2.6)(9.2)(10−8) = 23.92 × 10−8 = (2.392 × 101)(10−8) = 2. 3̅92(101−8) = 𝟐. �̅�𝟗𝟐 × 𝟏𝟎−𝟕 Simplifying Scientific Notation (÷) (1.247 × 10−3) ÷ (2.9 × 10−2) = (1.247 ÷ 2.9)(10−3 ÷ 10−2) = (0.43)(10−3+2) = 0.43 × 10−1

= (4.3 × 10−1)10−1 = 𝟒. 𝟑 × 𝟏𝟎−𝟐

Your turn. Q. Without using your calculator, find:

1. 2.

3.

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Addition and Subtraction To add (or subtract) two numbers in scientific notation, be sure that the exponents in each number are the SAME. In general:

(𝑎 × 10𝑛) + (𝑏 × 10𝑛) = (𝑎 + 𝑏) × 10𝑛 ; (𝑎 × 10𝑛) − (𝑏 × 10𝑛) = (𝑎 − 𝑏) × 10𝑛 Changing the Exponent (smaller)

5.14 × 105 = 51.4 × 104 = 514 × 103 = 5,140 × 102 = 51,400 × 10 = 514,000 Changing the Exponent (larger)

5.14 × 10−5 = 0.514 × 10−4 = 0.0514 × 10−3 = 0.00514 × 10−2 = 0.000514 × 10−1 = 0.0000514 Example

(7.2 × 105) + (5.1 × 104) = (72 × 104) + (5.1 × 104) = (72 + 5.1) × 104 = 77.1 × 104 = 𝟕. 𝟕𝟏 × 𝟏𝟎𝟓 Example (3 × 10−2) − (2.2 × 10−3) = (3 × 10−2) − (0.22 × 10−2) = (3 − 0.22) × 10−2 = 𝟐. 𝟕𝟖 × 𝟏𝟎−𝟐 Example (backwards) (3 × 10−2) − (2.2 × 10−3) = (30 × 10−3) − (2.2 × 10−3) = (30 − 2.2) × 10−3 = 27.8 × 10−3 = 𝟐. 𝟕𝟖 × 𝟏𝟎−𝟐

Your turn. Q. Without using your calculator, find:

1. 2.

3.

Why do we need to know this if the calculator does it for us?

Avoid writing just whatever the calculator displays on the screen without thinking… If your answer is outrageously wrong, you won’t get points for it… Why?

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Q. Estimate the answer to:

1.4 Units of Measurement

What is the difference between a qualitative and a quantitative observation?

Which are the two “ingredients” a measurement must possess?

The Equation of the Line – Length

The slope of a line is an equivalency (or ratio).

L[cm]

L [in]

P1 (0cm, 0in)

P2 (1in, 2.54cm)

x1 = 0 x2 = 1y1 = 0

y2 = 2.54

b = 0

L(cm) = mL(in) + b

m =y2 - y1

x2 - x1

b = 0cm

1in - 0in

2.54cm - 0cm==

L(cm) = L(in)

in

cm2.54

in

cm2.54

1 in = 2.54 cm

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Example: How many cm are there in 5 inches? How many in are there in 100 cm? Factor label method. Temperature

Units of Measurement Système International d’Unités (an agreement). A different base unit is used for each quantity (Check Table 1.4). Prefixes from the Metric System (Why do we use them in the first place?). Memorize Micro to Giga! See the Handout. Table 1.5. Using Prefixes:

1 km = 1000 m

1 cm = 0.01 m or 1 m = 100 cm

1 mm = 0.001 m or 1 m = 1000 mm

1 μm = 1x10-6 μm or 1 m = 106 μm

1 nm = 1x10-9 nm or 1 m = 109 μm

L(cm) = L(in)in

cm2.54

L(cm) = 5(in) L(cm) = 12.7cm2.54cm

1 in

100cm2.54cm

1 in= 39.4 in

1 in = 2.54 cm

T[°F]

T [°C]

P1 (0°C, 32°F)

P2 (100°C, 212°F)

x1 = 0 x2 = 100

y1 = 32

y2 = 212

b

T(°F) = mT(°C) + b

m

b =

100°C - 0°C

212°F - 32°F==

T(°F) =

1.8°F

°C

32°F - 1.8°F

°C0°C = 32°F

1.8°F

°CT(°C) + 32°F

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Measuring Length 1 m = 100 cm (1 m ≈ 1 yard ) 1 inch = 2.54 cm (Come from equations…)

What is the difference between mass and weight? Which one do we measure with a beam balance and a top

balance (or analytical balance)?

Define Temperature (at the microscopic level):

What is the difference between absolute and empirical temperature scales?

Empirical Absolute

𝑇(℃)

100 ℃=

𝑇(℉)−32 ℉

100 ℉=

𝑇(𝐾)−273 𝐾

100 𝐾

Volume 1 m3 = 1000 dm3 1 dm3 = 1000 cm3 = 1 L (1 L ≈ 1.06 qt) 1 cm3 = 1 mL

Density

𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =𝑀𝑎𝑠𝑠 (𝑔)

𝑉𝑜𝑙𝑢𝑚𝑒 (𝑚𝐿 𝑜𝑟 𝑐𝑚3)

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1.5 Uncertainty in Measurements

Why is uncertainty ever present? How do we account for it?

See the handout on significant figures to get acquainted with the rules you need to follow when working with

numbers that have uncertainty and exact numbers; also, follow the rounding rules given.

Remember this…!! The Last digit written in a measurement is the number that is considered uncertain.

54.07 g A mass between 54.06 g and 54.08 g (±0.01 g) 54.071 38 g A mass between 54.071 37 g and 54.071 39 g (±0.000 01 g) Unless it is explicitly given, consider the uncertainty as ± 1 at the relative position of the last significant figure.

Q. In each measurement determine: A. # of sig. figs. B. the uncertainty. a. 256.7 lb d. 0.00317 g b. 51 mL e. 0.0200 g c. 150 m f. 720. in

Q. Round up each number to the marked digit. 1. 𝟕, �̅�𝟗𝟏 2. 𝟏𝟓, �̅�𝟒𝟑 3. 𝟗𝟗, �̅�𝟔𝟏 4. 𝟑, �̅�𝟓𝟎 5. �̅�𝟐𝟎 Problem 1 Round to the correct number of significant figures. 1) 0.219 × 0.012310 = 2) 14.8 + 1.223 + 102.11 = 3) 520 − 1.79 − 10.2 =

4) 1.534 + 3.2

15.44=

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Problem 2 What is the answer using the proper number of significant figures?

Accuracy versus Precision, what is the difference? (More about this topic will be addressed at the lab.)

1.6 Dimensional Analysis Factor-Label Method:

(𝑺𝒕𝒂𝒓𝒕𝒊𝒏𝒈 𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚) × (𝑪𝒐𝒏𝒗𝒆𝒓𝒔𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓) = 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚

𝒈𝒊𝒗𝒆𝒏 𝒖𝒏𝒊𝒕 ×𝒅𝒆𝒔𝒊𝒓𝒆𝒅 𝒖𝒏𝒊𝒕

𝒈𝒊𝒗𝒆𝒏 𝒖𝒏𝒊𝒕= 𝒅𝒆𝒔𝒊𝒓𝒆𝒅 𝒖𝒏𝒊𝒕 ; 𝟕𝟐. 𝟎 𝒊𝒏 (

𝟐. 𝟓𝟒 𝒄𝒎

𝟏 𝒊𝒏) = 𝟏𝟖𝟑 𝒄𝒎

Treat units as constants in an algebraic expression.

3𝑎 × 4𝑎 = 12𝑎2 ; 3 𝑖𝑛 × 4 𝑖𝑛 = 12 𝑖𝑛2

16𝑎3

4𝑎2= 4𝑎 ;

16 𝑓𝑡3

4 𝑓𝑡2= 4 𝑓𝑡

Be careful! Keep in mind that “you cannot add apples and oranges”:

2𝑎 + 4𝑏 = 2𝑎 + 4𝑏 ; 2 𝑙𝑏 + 4 𝑘𝑔 = 2 𝑙𝑏 + 4 𝑘𝑔 If you want to do it, first you need to convert one type of unit into the other:

1 𝑘𝑔 = 2.20 𝑙𝑏

2 𝑙𝑏 + 4 𝑘𝑔 (2.20 𝑙𝑏

1𝑘𝑔) = 2 𝑙𝑏 + 8̅. 8 𝑙𝑏 = 1̅1 𝑙𝑏 → 10 𝑙𝑏

2 𝑙𝑏 (1𝑘𝑔

2.20 𝑙𝑏) + 4 𝑘𝑔 = 0. 9̅0𝑘𝑔 + 4 𝑘𝑔 = 4̅. 90 𝑘𝑔 → 5 𝑘𝑔

Conversion factor Conversion factor

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Q. What are the units of "𝑋“ in the following equations? 𝑔𝑟𝑎𝑚𝑠[=](𝑋)(𝑚𝐿) 𝐽𝑜𝑢𝑙𝑒𝑠[=](𝑋)(𝑔𝑟𝑎𝑚𝑠)(℃) Systematic Approach to Problem Solving 1. Read the problem carefully; you MUST understand the question: What are you trying to find?

– Identify all quantities given (#’s and units). – Identify all relationships (explicit and implicit) in the problem.

2. Sketch a strategy (conceptual map) to solve the problem (you don’t need to write it!).

– Each step involves an equation or conversion factor (you may need to work backwards…) 3. Solve the problem following the outlined strategy.

– Check that units cancel properly. 4. Check your final answer (two ways):

– Perform a rough estimate (simple problems) – Double check your calculations (more complex)

You will not get points for answers that are impossible/make no sense. Remember: You must double check what your calculator displays as an answer… Using Density in Calculations

𝑑 =𝑚

𝑉 (𝑟𝑒𝑙𝑎𝑡𝑒𝑠 𝑡𝑤𝑜 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠) ; 𝑉 =

𝑚

𝑑= 𝑚𝑑−1 ; 𝑚 = 𝑑𝑉

Olive oil has a density of 0.92 g/mL. Conversion factors:

𝑑 =0.92 𝑔

1 𝑚𝐿 𝑜𝑟 𝑑−1 =

1 𝑚𝐿

0.92 𝑔

Q. a. What is the volume of 50̅0 𝑔 of olive oil? b. What is the mass of 20̅0 𝑚𝐿 of olive oil?

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Problem 3 Convert 1.76 yd to centimeters. (1 m = 1.094 yd) Problem 4 Mercury has a density of 13.534 g/mL –it is the densest liquid at room temperature. How many quarts will 195.5 lbs (mass of the average man) of mercury occupy? (453.592 g = 1 lb; 1 qt = 946.353 mL) Problem 5 Dextrose is a form of glucose (sugar). Dextrose 5% in water is injected into a vein through an IV to replace lost fluids and provide carbohydrates to the body; therefore, it is used to treat hypoglycemia (low blood sugar), insulin shock, or dehydration. It also given for nutritional support to patients who are unable to eat because of illness, injury or other medical condition. An order reads: “Infuse 1.00 𝐿 of dextrose 5% in water over 12 hours”. With a tubing set of drop factor of 10̅ (the drop factor is the number of drops per milliliter), what is the rate in drops per minute the tubing should be set up to?

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Problem 6 You have a patient on a Dopamine drip ordered at 10 mcg/kg/min (mcg stands for micrograms). The drip comes

as 400 mg/250 mL D5W (D5W stands for a 5% solution of dextrose in water). The patient weighs 75 kg. What

is the rate in ml/hr? Round your answer to the nearest tenth. Converting Squared/Cubed units To convert squared or cubed units, we have to square or cube the entire conversion factor (# and unit). Problem 7 How many cubic centimeters are there in a box that measures 2.23 in by 4.713 in by 16.564 in? Problem 8 A 2.50 cm x 2.50 cm square piece of platinum (Pt) has a mass of 1.656 g. Pt has a density of 21.45 g/cm3. What

is the thickness of the Pt sheet in mm?

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Problem 9 A copper refinery produces a copper ingot weighing 150 lb. If the copper is drawn into wire whose diameter is

7.50 mm, how many feet of copper can be obtained from the ingot? The density of copper is 8.94 g/cm3.

Assume that the wire is a cylinder with volume V = πr2h, where r is its radius and h is its height or length.

(1 lb = 453.59 g ; 1 in = 2.54 cm (exactly!) ; 1 ft = 12 in ; 1 m = 100 cm ; 1 m = 1000 mm)

Density (a word of caution…) Volumes are not necessarily additive (i.e. the final density may not be the addition of the individual densities).

𝑑𝐺𝑜𝑙𝑑 = 19.3 𝑔

𝑐𝑚3 ; 𝑑𝐸𝑎𝑟𝑡ℎ = 5.52

𝑔

𝑐𝑚3

Problem 10 Brass is an alloy (mixture) principally used for decoration due to its bright gold-like appearance. It is made of copper and zinc, but their proportions can be varied to create a range of brasses with varying properties. Consider a sample of brass that weighs 9.85g and has a volume of 1.13 cm3. The densities of copper and zinc are 8.94 g/cm3 and 7.14 g/cm3, respectively. If the total volume of the piece of brass is the sum of the volumes of the copper and zinc that it contains, calculate the percentage of copper (by mass) in the sample.