Chapter 1: Chemistry and Measurement

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Vanessa Prasad-PermaulValencia College

CHM 1045

Properties of Matter

Chemistry: The study of composition, properties, and transformations of matter

Matter: Anything that has both mass & volume

Hypothesis: Interpretation of results

Theory: Consistent explanation of observations

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Conservation of Mass

Law of Mass Conservation: Mass is neither created nor destroyed in chemical reactions.

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Example 1: Conservation of Mass

C(s) + O2(g) CO2(g)

a) 12.3g C reacts with 32.8g O2, ?g CO2

12.3g + 32.8g = 45.1g

a) 0.238g C reacts with ?g O2 to make .873g CO2 0.238g + x = 0.873g = 0.873g-0.238g = 0.635g of O2

a) ?g C reacts with 1.63g O2 to make 2.24g CO2

x + 1.63g = 2.24g = 2.24g - 1.63g = 0.61g C

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Example 1: Conservation of Mass

Exercise 1.1 1.85g of wood is placed with 9.45g of air in a sealed vessel. It

isheated and the wood burns to produce ash and gases. The ashis weighed to yield 0.28g. What is the mass of the gases in thevessel?

1.85g Wood + 9.45g Air heat 0.28g Ash + ? g gases

1.85 + 9.45 - 0.28 = 11.02g of gases

What is the mass of wood that is converted to gas by the end of

the experiment?1.85g of Wood – 0.28g of ash = 1.57g

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Matter

Matter is any substance that has mass and occupies volume.

Matter exists in one of three physical states: solid liquid gas

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Solid

In a solid, the particles of matter are tightly packed together.

Solids have a definite, fixed shape.

Solids cannot be compressed and have a definite volume.

Solids have the least energy of the three states of matter.

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Liquid

In a liquid, the particles of matter are loosely packed and are free to move past one another.

Liquids have an indefinite shape and assume the shape of their container.

Liquids cannot be compressed and have a definite volume.

Liquids have less energy than gases but more energy than solids.

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Gases

In a gas, the particles of matter are far apart and uniformly distributed throughout the container.

Gases have an indefinite shape and assume the shape of their container.

Gases can be compressed and have an indefinite volume.

Gases have the most energy of the three states of matter.

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Phases

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A physical change is a change in the form of matter but not in its chemical identity

A chemical change or a chemical reaction is a change in which one of more kinds of matter are transformed into a new kind of matter or several new kinds of matter

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Physical Properties can be determined without changing the chemical makeup of the sample.

Some typical physical properties are: Melting Point, Boiling Point, Density, Mass, Touch,

Taste, Temperature, Size, Color, Hardness, Conductivity.

Some typical physical changes are: Melting, Freezing, Boiling, Condensation,

Evaporation, Dissolving, Stretching, Bending, Breaking.

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Chemical Properties are those that do change the chemical makeup of the sample.

Some typical chemical properties are: Burning, Cooking, Rusting, Color change,

Souring of milk, Ripening of fruit, Browning of apples, Taking a photograph, Digesting food.

Note: Chemical properties are actually chemical changes

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Classifications of Matter

Matter can be divided into two classes: mixtures pure substances

Mixtures are composed of more than one substance and can be physically separated into its component substances.

Pure substances are composed of only one substance and cannot be physically separated.

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Pure Substances

There are two types of pure substances: Compounds Elements

A compound is a substance composed of two or more elements chemically combined Compounds can be chemically separated into

individual elements. Water is a compound that can be separated

into hydrogen and oxygen. An element cannot be broken down

further by chemical reactions.

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Dalton’s Atomic Theory

Law of Definite Proportions: Different samples of a pure chemical substance always contain the same proportion of elements by mass.

Any sample of H2O contains 2 hydrogen atoms for every oxygen atom

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Mixtures

There are two types of mixtures: homogeneous mixtures heterogeneous mixtures

Homogeneous mixtures have uniform properties throughout. Salt water is a homogeneous mixture.

Heterogeneous mixtures do not have uniform properties throughout. Sand and water is a heterogeneous mixture.

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Example 2: Matter

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Which of the following represents a mixture?

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Accuracy, Precision, and Significant Figures in Measurement

Accuracy is how close to the true value a given measurement is.

Precision is how well a number of independent measurements agree with one another.

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Example 8: Accuracy & Precision

Which of the following is precise but not accurate?

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Significant Figures are the total number of digits in the measurement.

The results of calculations are only as reliable as the least precise measurement!!

Rules exist to govern the use of significant figures after the measurements have been made.

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Rules for Significant Figures:

Zeros in the middle of a number are significant

Zeros at the beginning of a number are not significant

Zeros at the end of a number and following a period are significant

Zeros at the end of a number and before a period may or may not be significant.

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Example 4: Significant Figures

How many Significant Figures ?a) 0.000459 = 3

b) 12.36 = 4

c) 36,450 = 4

d) 8.005 = 4

e) 28.050 = 5

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Rules for Calculating Numbers: During multiplication or division, the

answer can’t have more significant figures than any of the original numbers.

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a) 218.2 x 79 = 17237.8 = 1.7 x 104

a) 12.5 / 0.1272 = 94.33962264150943 = 94.3

b) 0.2895 x 0.29 = 0.083955 = 0.084

c) 32.567 / 22.98 = 1.417188859878155 = 1.417

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-During addition or subtraction, the answer can’t have more digits to the right of the decimal point than any of the original numbers.

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a) 218.2 + 79 = 297.2 = 297

b) 12.5 - 0.1272 = 12.3728 = 12.4

c) 0.2895 + 0.29 = 0.5795 = 0.58

d) 32.567 - 22.98 = 55.547 = 55.55

e) 185.5+2.224 = 187.724 = 187.7


Rules for Rounding Numbers:

If the first digit removed is less than 5 round down (leave # same)

If the first digit removed is 5 or greater round up

Only final answers are rounded off, do not round intermediate calculations

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Example 7: Rounding and Significant Figures

Round off each of the following measurements

a) 3.774499 L to 4 sig. figs. = 3.774L

b) 255.0974 K to 3 sig. figs. = 255K

c) 55.265 kg to 4 sig. figs. = 55.27kg

d) 1.2151ml to 3 sig. figs. = 1.22ml

e) 1.2143g to 3 sig. figs. = 1.21g

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Exercise 1.3Give answers to the following arithmetic setups. Round to the correct number of significant figures:

a) 5.61 x 7.891 = 4.864671 = 4.9 9.1

b) 8.91 - 6.435 = 2.475 = 2.48

c) 6.81 – 6.730 = 0.08 = 0.08

d) 38.91 x (6.81-6.730) = 38.91 x 0.08 = 3.1128 = 3

Scientific Notation

Changing numbers into scientific notation Large # to small # Moving decimal place to left, positive

exponent123,987 = 1.23987 x 105

Small # to large # Moving decimal place to right, negative

exponent0.000239 = 2.39 x 10-4

How to put into calculator

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Example 3: Scientific Notation

Put into or take out of scientific notation

a) 1973 = 1.973 x 103

b) 5.5423 x 10-4 = 0.00055423

c) 0.775 = 7.75 x 10-1

d) 3.55 x 107 = 35,500,000

e) 8500 = 8.5 x 103

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Measurement and Units

Physical Quantity Name of Unit Abbreviation Mass kilogram kg

Length meter m Temperature kelvin K

Amount of substance mole mol Time second s

Electric current ampere A Luminous intensity candela cd

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SI Units


Factor Prefix Symbol 1,000,000,000 = 109 giga G

1,000,000 = 106 mega M 1,000 = 103 kilo k 100 = 102 hecto h 10 = 101 deka da 0.1 = 10-1 deci d

0.01 = 10-2 centi c 0.001 = 10-3 milli m

0.000,001 = 10-6 micro µ 0.000,000,001 = 10-9 nano n

0.000,000,000,001 = 10-12 pico p

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*

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* Important

Some prefixes for multiples of SI units

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Exercise 1.4Express the following quantities using an SI prefix

and abase unit. For instance, 1.6 x 10-6m = 1.6m. A

quantity suchas0.000168g could be written 0.168mg or 168g.

a) 1.84 x 10-9 m = 1.84 nm (nanometer)b) 5.67 x 10-12 s = 5.67 ps (picosecond)c) 7.85 x 10-3 g = 7.85 mg (milligram)

d) 9.7 x 103 m = 9.7 km (kilometer)e) 0.000732 s = 0.732 ms (millisecond)

= 732us (microsecond)f) 0.000000000154 m = 0.154nm (nanometer)

= 154pm (picometer)

Changes in Physical State

Most substances can exist as either a solid, liquid, or gas.

Water exists as a solid below 0 °C; as a liquid between 0 °C and 100 °C; and as a gas above 100°C.

A substance can change physical states as the temperature changes.

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Solid Liquid

When a solid changes to a liquid, the phase change is called melting.

A substance melts as the temperature increases.

When a liquid changes to a solid, the phase change is called freezing.

A substance freezes as the temperature decreases.

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Liquid Gas

When a liquid changes to a gas, the phase change is called vaporization.

A substance vaporizes as the temperature increases.

When a gas changes to a liquid, the phase change is called condensation.

A substance condenses as the temperature decreases.

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Solid Gas

When a solid changes directly to a gas, the phase change is called sublimation.

A substance sublimes as the temperatureincreases.

When a gas changes directly to a solid, the phasechange is called deposition.

A substance undergoes deposition as the temperature decreases.

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Temperature

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Diagram of the various phases of temperature change

Temperature

Temperature Conversions:The Kelvin and Celsius scales have equal size units (a change of 1oC is equivalent to a change of 1K)

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100 K100 oC180oF

Temperature Conversions: Celsius (°C) — Kelvin (K) temperature

conversion:Kelvin (K) = t°C x 1K + 273.15K

1oC Fahrenheit (°F) — Celsius (C) temperature

conversions: there are exactly 9oF for every 5oC. Knowing that 0oC = 32oF

tF = tC x 9oF + 32 5oC

tC = 5oC x (toF – 32) 9oF

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Example 9: Temp. Conversions

Carry out the indicated temperature conversions:

a) –78°C = ? K = (-78oC x 1K/1oC) +273.15K = 195.15 = 195K

b) 158°C = ? °F = (158oC x 9oF/5oC)+32oF = 316.4 = 316oF

c) 373.15 K = ? °C = (373.15K x 1oC/1K)– 273.15K = 100K

d) 98.6°F = ? °C = 5oC/9oF x (98.6oF – 32oF) = 37oC

e) 98.6°F = ? K = (37oC x 1K/1oC) +273.15K = 310.15 = 310K

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Exercise 1.5A person with a fever has temperature of

102.5oF.What is this temperature in oC? A cooling

mixtureof dry ice and isopropyl alcohol has a

temperatureof -78 oC. What is the temperature in kelvins?

a) oC = 5oC x (oF – 32 ) = 0.555 x (102.5 – 32) = 39.2oC

9oF

b) K = oC + 273.15 = -78 + 273.15 = 195 K

Volume

Volume: how much three-dimensional space a substance (solid, liquid, gas) or shape occupies or contains often quantified numerically using the SI derived unit (m3) the cubic meter.

The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.

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http://en.wikipedia.org/wiki/SI_derived_unit

Volume

units of Volume:

m3 or cm3 (cc)

Traditionally chemists use liter (L)

1cm3 = 1cc = 1mL48


Density: relates the mass of an object to its volume.

Density = mass / Volume D = m / VV = m / Dm = V D

Density decreases as a substance is heated because the substance’s volume increases.

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Density

What is the density of glass (in mL) if a sample weighing 26.43 g has a volume of 12.40 cm3?

d = ?m = 26.43 gV = 12.40 cm3 = 12.40 mLd = m = 26.43 g = 2.13145 = 2.131

g/mL V 12.40 mL

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Density

What is the volume of an unknown solution if the mass is 12.567 g and the density is 14.621 g/mL ?

d = m/V V x d = m V = m/d

V = 12.567 g / 14.621 g/mL = 0.85952 mL

12.567g x 1mL = 0.85952 mL 14.621g

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Density

What is the mass of an unknown solution if the

volume is 20.2 mL and the density is 2.613 g/mL?

d = m/V m = d x V

m = 2.613g x 20.2 mL = 52.7826 = 52.8 g

mL

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Exercise 1.6A piece of metal wire has a volume of 20.2 cm3

and amass of 159 g. What is the density of the metal?

D = m = 159 g = 7.87128712 = 7.87 g /cm3

V 20.2cm3

We know that the following metals have the following

densities. Which metal is the wire made of?

Mn = 7.21 g/cm3

Fe = 7.87 g/cm3

Ni = 8.90 g/cm3

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Exercise 1.7

Ethanol (grain alcohol) has a density of 0.789 g/cm3.

What volume (mL) of ethanol must be poured into

a graduated cylinder to equal 30.3 g?

d = m/V V x d = m V = m / d

V = 30.3 g x 1 cm3 = 38.4cm3

0.789 g

Dimensional Analysis & Units

Dimensional-Analysis method uses a conversion factor to express the relationship between units.

Original quantity x conversion factor = equivalentquantity

Example: express 2.50 kg lb.Conversion factor: 1.00 kg = 2.205 lb

2.50 kg x 2.205 lb = 6.00 lb 1.00 kg

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Exercise 1.8The oxygen molecule (O2) consists of two

oxygenatoms a distance of 121 pm apart. How manymillimeters (mm) is this distance?

121 pm x 10-12 m x 1mm = 1.21 x 10-7 mm

1 pm 10-3

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Exercise 1.9A large crystal is constructed by stacking small

identicalpieces of crystal. A unit cell is the smallest piece

fromwhich a crystal can be made. A unit cell of a

crystal ofgold metal has a volume of 67.6 A3. What is the

volumein dm3?

67.6 A3 x 10-10 m x 10 dm = 6.76 x 10 -26 dm3

1 A3 1 m

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Exercise 1.10Using the following definitions, obtain theconversion factor for yards to meters. How

manymeters are there in 3.54 yd?1 in = 2.54cm (exactly) 1 yd = 36in (exactly)

1 yd x 1 in x 1 cm = 1.093613298 = 1.094 yd/m

36 in 2.54 cm 10-2 m

3.54 yd x 1 m = 3.24 m 1.094 yd

Conversions

a) 1.267 km m cm1.267km x 1000m x 100cm = 126700cm = 1.267

x 105

1km 1m

b) 0.784 L mL0.784L x 1000mL = 784L

1L

c) 3.67 x 105 cm in 3.67 x 105cm x 1in = 144488.1889in = 1.44 x

105in 2.54cm

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Conversions

d) 79 oz g79oz x 28.35g = 2239.65g = 2.2 x 103g

1oz

e) 9.63 x 10-3 yd ft9.63 x 10-3yd x 1m x 1km x 0.62137mile x

5280ft 1.0936yd 1000m 1km

1mile= 0.0289ft

f) 23.5 cm2 m2

23.5cm2 x 1m2 = 0.235m2

100cm2

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Conversions

g) 1.34 x 1012 pm m1.34 x 1012pm x 1m = 1.34 x 1024m

10-12pm

h) 4.67 x 10-7 nm pm4.67 x10-7nm x 1m x 10-12pm = 4.67 x 10-12pm

10-9nm 1m

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Documents

Chapter 1: Chemistry and Measurement