Chapter 1 May11

8/6/2019 Chapter 1 May11

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CHAPTER 1CHAPTER 1

INTRODUCTION TO MATTER AND MEASUREMENT


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CONTENTSCONTENTS

1.1 Introduction

1.2 Classification of Matter1.3 Properties of Matter1.4 Units of Measurement 1.5 Uncertainty in Measurement 1.6 Dimensional Analysis


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Learning outcom es:

Able to differentiate between the th ree states

of m a tt er . Able to distinguish between ele m en t s , compou nds and m ix tu res .

Able to distinguish between phy si cal andch e m ical properties.

Able to use and convert different u ni t s of measurement.


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1.1 I n t ro d uct io n

Ch e m is t ry is the study of p rop er t ies ofm a t erials and ch anges th e y u nderg o .

Central role in science and technology.Has a high impact on our daily living , e.g.health and medicine , energy and environment ,

materials and technology and food andagriculture. Able to contribute to problem solving analysis.


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1.2 C lassi fica t io n of M a tt er

Ma tt erPhysical material - anything that has m ass andoccup ies s p a ce .

Classi fica t io ns of M a tt ers

Matter can be classified according to its:Physical state (solid , liquid or gas)Composition (element , compound or mixture)


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Classificationof Matter

Physical State Composition

Gas

Liquid

Solid

Puresubstance Mixture

Element

Compound

Homogeneous

Heterogeneous


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1.2.1 St a t e of M a tt er (Phy si cal

St a t e)G as

no fixed volume/shapeeasy to compress/expandmolecules are far apart move at high speed

often collide


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Co n t : 1.2.1 St a t e of M a tt er

(Phy si cal St a t e)Liq u id

volume independent of containerslightly compressiblemolecules closer than gas

move rapidly but can slide over each other


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Co n t : 1.2.1 St a t e of M a tt er

(Phy si cal St a t e)So lid

defined volume & shape

Incompressiblemolecules packed closely indefinite arrangement/rigid

shape


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1.2.2 Compo si t io n

Pu re Sub s t an ceMatter with fixed composition and distinct properties , E.g H 2O , NaCl

(i) Ele m en t s- simplest form of matter

- cannot be decomposed into simpler substancesby chemical means i.e only one kind of element - can exist as atoms or molecules


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Co n t : 1.2.2 Compo si t io n

114 elements identified

Each given a unique name organized in a PeriodicTable


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(ii) Compou nds- substance composed of atoms of two or more

elements in fixed proportions- can be separated only by chemical means- exist as molecules (H 2O, CO 2)- properties are different from the

elemental properties


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Mix tu reCombination of two or more substances , in which

each substance retains its own chemical identity.

( i) A Homo gene ou s m ix tu re : components uniformly mixed

(one phase) e.g. air also called solutions (gaseous ,

liquid, solid solutions)


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(ii) A H e t er o gene ou s m ix tu re : components are not distributed uniformly (more

than one phase)e.g. sand & rocks

sugar & sand

Separating Mixtures (by physical means):basic techniques: filtration , floatation , crystallization , distillation , extraction and chromatography.


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1.3 P rop er t ies of M a tt er

Properties of matter can be grouped into twocategories:

Phy si cal p rop er t ies : measured and observedwithout changing the composition or identity of asubstance. e.g. color , odor , density , melting point , boiling point.Ch e m ical p rop er t ies : describe how substancesreact or change to form different substances.e.g. hydrogen burning in oxygen.


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Co n t : 1.3 P rop er t ies of M a tt er

Properties of substance can be divided into twoadditional categories:

I n t ensive p rop er t iesDo not depend on the amount of the samplepresent. e.g. temperature , melting point , density.Ex t ensive p rop er t iesDepends on quantity present. e.g. mass , volume.


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1.3.1 Phy si cal and Ch e m ical

Ch angesChanges in matter can involve either chemical or physical changes.Phy si cal ch ange : substance changes physicalappearance but not composition. e.g. changes of state :

liquid gas solid liquidCh e m ical ch ange : substance transform into achemically different substance i.e. identifychanges. e.g. decomposition of water.


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1.4 U ni t s of M eas u re m en t

SI Units1960 : All scientific units use Systme International dUnits (SI Units).Seven base units :

Physical Quantity Name o nit bbreviationMass Kilogram Kg

Length Meter mTime Second s (sec)Electric current mpereTemperature Kelvin K Luminous intensity Candela cd

mount o substance Mole mol

1.4 U ni t s of M eas u re m en t


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1.4.1 Leng th and Mass

SI base unit of length : meter (m)1 m = 1.0936 yards

Mass :A measure of the amount of material in anobject.SI base unit of mass : kilogram (kg)

1 kg = 2.2 pounds


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1.4.2 T e mp era tu re

Temperature is a measure of hotness or coldnessof an object 3 temperature scales are currently in use:

(i) OF (degrees Fahrenheit)(ii) OC (degrees Celsius)(iii) K (Kelvin)

Scientific studies commonly usedCelsius and Kelvin scales


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Co n t : 1.4.2 T e mp era tu re

K elvin (SI U ni t )Based on properties of gases

0 K is the lowest temperature that can beattained theoretically (absolute zero)0 K = -273.15 rC


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Temperaturescale

Properties of water at sea level

Freezing point Boiling point

Fahrenheit, F 32 212

Celcius, C 0 100

Kelvin, K 273.15 373.15


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Co n t : 1.4.2 T e mp era tu re

Te mp era tu re co nversi o nsK = 0 rC + 273.15

rC = K - 273.15

3295

325

9

r!r

r!r

F C

C F


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1.4.3 Vo lum e

SI u nit of volume = (unit of length) 3 = m 3

Generally , chemists work with much smaller

volumes:cm 3 , mL or cc1 cm 3 = 1 mL = 1 v 10 -6 m3

1000 cm3

= 1 L* N ote: liter ( ) is not an SI unit1 dm 3 = 1 v 10 -3 m3


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1.4.4 D ensi ty

W idely used to characterize substances.Defined as mass divided by volume, d = mass (m)

volume (V)Unit : g/cm 3

Varies with temperature because volume changeswith temperature.

Can be used as a conversion factor to change massto volume and vice versa.Common units :

g/m for l iquid , g/cm 3 for so l id , g/ for gas .


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1.5 U n cer t ain ty in Meas u re m en t

Obj e ct ives

i. Determine the number of significant figures ina measured quantity.

ii. Express the result of a calculation with theproper number of significant figures.


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Co n t : 1.5 U n cer t ain ty in

Meas u re m en tTwo typ es of n umb ers:

(i) Exa ct n umb ers - those that have defined

values or integers resulting from countingnumbers of objects. e.g. exactly 1000g in akilogram , exactly 2.54 in an inch.(ii) I nexa ct n umb ers - those that obtainedfrom measurements and require judgement.Uncertainties exist in their values.Note : Uncertainties always exist in measuredquantities.


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1.5.1 P re c isi o n and Accu ra cy

P re c isi o n - how well measured quantities agree

with each other.

Accu ra cy - how well measured quantities agreewith the true value .


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Co n t : 1.5.1 P re c isi o n and

Accu ra cy

Good precisionGood accuracy

Good precisionPoor accuracy

Poor precisionGood accuracy

Poor precisionPoor accuracy


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1.5.2 S igni fican t F ig u res

Measured quantities (inexact) are generallyreported in such a way that the last digit is the

first uncertain digit. (2.2405g) All certain digits and the first uncertain digit arereferred to as significant figures.

Ru les:(i) Non-zero numbers are always significant

e.g. 2.86 : has three significant figures.


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Co n t : 1.5.2 S igni fican t F ig u res

(ii) Zeros between non-zero numbers are alwayssignificant. E.g. 1205 has four significant figures.

(iii) Zeros before the first non-zero digit are not significant. E.g. 0.003 : has one significant figure.

(iv) Zeros at the end of a number after a decimalplace are significant.. E.g. 0.0020 : has twosignificant figures.


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(v) Zeros at the end of a number before a decimalplace are ambiguous.

E.g . 100: has one significant number unlessotherwise stated. If it is determined from countingobjects , it has three significant figures.

Method - Scientific notation removes the ambiguity of knowing how many significant figures a number

possesses.


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Example:

(i) 225 , 2.25 v 10 2 : th ree signi fican t f ig u res ( s .f. ) .(ii) 10.004 , 1.0004 v 10 4 : five s .f.(iii) 0.0025 , 2.5 v 10 -3 : two s .f.(iv) 0.002500 , 2.500 v 10 -3 : fou r s .f.(v) 14 100.0 , 1.41000 v x 10 4 : six s .f.(vi) 14100 , 1.4100 v 10 4, 1.41 v 10 4, 1.410 v 10 4 :could have th ree , fou r or five s .f . - need knowledge .


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1.5.3 S igni fican t F ig u res in

Cal cu la t io ns1.5.3.1 Addi t io n (+ ) and Subt ra ct io n (- )

Result must be reported to the least number of d ecimal pla ces .E.g. 20.4 g - 3.322 g = 17.1 gOther Examples:The final answer should have thesame uncertainty , with the greatest uncertainty.

(i) 325.24 (uncertainty = s 0.01)21.4 (uncertainty = s 0.1)

+ 145 (uncertainty = s 1)

491.64 Answer : 492


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Cont:1.5.3.1 Addition (+) and

Subtraction (-)O ther Examples :

(ii) 12.25 + 1.32 + 1.2 = 14.771.2 has the greatest uncertainty ( s 0.1) @ theanswer must be rounded to one digit to theright of the decimal point. Answer : 14.8

(iii) 13.7325 - 14.21 = -0.4775, Answer: -0.48


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1.5.3.2 Multiplication ( v ) and

Division (z

)R esult must be to the least number of significant figures .E .g. 6.221 cm v 5.2 cm = 32 cm 2

T o round off the final calculated answer so that ithas the same number of significant figures as theleast certain number.

O ther Example:(i) 1.256 v 2.42 = 3.03952T he least certain/precise number is 2.42 p 3significant figures(s.f.). @ T he answer must berounded to the 3 s.f.: 3.04


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Cont: 1.5.3.2 Multiplication ( v )

and Division (z

)O ther Examples :

(ii) 16.231 2.20750 = 7.352661T he least precise number is 16.231 (5 s.f.).@ Answer is 5 s.f. : 7 .352 7

(iii) (1.1)(2.62)(13.5278) 2.650 = 14.712121T he least precise number is 1.1 (2 s.f.).@ Answer must be rounded to 2 s.f. : 15


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1.5.3.3 Rules for Rounding O ff

Numbers(i) When the figures immediately following the

last digit to be retained is less than 5 , the last

digit unchanged.e.g. 6.4362 to be rounded off to four significant figures : 6.436(ii) When the figure immediately following the last digit to be retained is greater than 5 , increase thelast retained figure by 1.e.g 6.4366 to be rounded off to four significant figures : 6.437


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Cont: 1.5.3.3 Rules for RoundingO

ff Numbers(iii) When the figure immediately following the last digit to be retained is 5 , the last figure to beretained is increased by 1 , whether it is odd oreven.e.g. 2.145 becomes 2.15 if three significant figuresare to be retained.

(iv) When a calculation involves an intermediateanswer , retain at least one additional digit past thenumber of significant figures.


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1.6 D im ensi o nal Anal ysis

Obj e ct ive: To be able to convert different measurement units by using dimensional analysis.Dim ensi o nal Anal ysis is the algebraic processof changing from one system of units to another.Conversion factors are used.

A conversion factor is a fraction whose numerator

and denominator are the same quantity expressedin different units.Given units are being multiplied and divided togive the desired units.


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Co n t : 1.6 D im ensi o nal Anal ysis

Desired unit = given unit v conversion factorconversion factor

In dimensional analysis , always ask threequestions:

(i) What data are given?(ii) What quantity do we need?(iii) What conversion factors are available totake us from what are given to what we need?

)unitgiven(

)unitdesired(


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Exa mp le 1

Q uantity 1 in. = 2.54 cm yields two conversionfactors

2.54 cm and 1 in.1 in. 2.54 cm

Convert 5.08 cm to in. and 4.00 in. to cm

5.08 cm v 1 in. = 2.00 in.2.54 cm

4.00 in. v 2.54 cm = 10.2 cm1 in.


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Exa mp le 2

Convert 6.23 ft 3 to the appropriate SI unit.ft3 to m 3 and 3.272 ft = 1m

(1 ft ) 3 = (1m) 3

(3.272ft) 3

@ 6.23 ft 3 = 6.23 ft 3 v (1m) 3 = 0.1 7 8 m 3

(3.272ft) 3


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Exer c ise 1.1

A persons average daily intake of glucose is0.0833 pound. W hat is this mass in milligrams?( 1 lb = 453.6 g)

lb1

g6.453

Answer: 3.78 x 10 -4 mg

lb g mg

0.0833 lb x x =g1

mg1000


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END of CHAPTER 1

Chapter 1 May11

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