Chemistry Notes Chapter 1 Complete)

-1CHEMISTRY 9701 Chapter 1 : ATOMS, MOLECULES AND STOICHIOMETRY 1.01 Atoms and Molecules 1. Atom is the smallest particle in which an element can be derived without losing its identity. 2. A molecule is a group of atoms (held by covalent bond) that exists under ordinary conditions. E.g. Cl2 , H 2 O , CH 4 1.02 The Atomic Theory of John Dalton 1. All matter consists of indivisible particles called atoms. (False: Atoms can be subdivided into electrons, neutrons and protons.) 2. Atoms of the same element are similar in shape and mass, but differ from atoms of other elements. (Isotopes have different masses.) 3. Atoms cannot be destroyed or created. 4. Atoms combine together to form “compound atoms” or molecules in simple ratio. 1.0 3 Relative Masses of Atoms, Molecules and Ionic Compounds The mass of an atom is compared with the mass of a standard atom of carbon – 12 isotope because carbon is a solid at room temperature and hence, easier to store and transport. The relative mass of a particle =

fff12 fffffffffB fffffffmass ffffffffffffffffffof ffffffffone fffffffffffffparticle fffffffffffffffffffffffffffff mass of one atom of C @12

Relative mass has no unit since it is a ratio. Isotopes are atoms of the same element with different masses. The relative isotopic mass of a particle =

ffffff12 fffffffffB fffffffmass fffffffffffffffffof ffffffffan ffffffffffisotope ffffffffffffffffffffffffffffff mass of one atom of C @12

12 ffffffffB fffffffmass fffffffffffffffffof ffffffffone fffffffffffffatom ffffffffffffffffffof ffffffffan ffffffffffelement fffffffffffffffffffffffff The relative atomic mass of a particle = mass of one atom of C @12

Naturally occurring element has a mixture of isotopes. E.g. Chlorine has 2 isotopes of 35Cl and 37Cl in the ratio of 3:1. The % of isotopes is known as abundance.

3ffff Cl = B 100% = 75% 4 1ffff 37 The abundance of Cl = B 100% = 25% 4 b c ` a relative isotopic mass abundance PffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffB ffffffffffffffffffffffffffffffffffffffffffffffff The relative atomic mass of an element = total abundance The abundance of

35

-2Ar of Chlorine =

` a ` a f75 fffffffffB fffffff35 ffffffffffff+ fffffffff25 fffffffffB fffffff37 fffffffffff

100

= 35.5 Question: K exists as 39.1)

39

K and

40

Let the abundance of Let the abundance of `

a `

K . Calculate the abundance of each isotope. (Ar of K = 39 40

K be x K be 100 @x

a

x 39 + 100 @x 40 39.1 = fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff 100 x = 90 39 40

K = 90% K = 10%

1.0 4 Relative Molecular Mass (Mr) of a Molecule 12 fffffffB fffffffmass ffffffffffffffffffof ffffffffone fffffffffffffmolecule ffffffffffffffffffffffffffffffof ffffffffa fffffsubstance ffffffffffffffffffffffffffffffff The relative molecular mass, Mr = mass of one atom of C @12

Mr is the sum of the Ar of all the atoms shown in the formula. `

2@

a` a

Mr of SO 4 = 32.1 + 16 4

= 96.1

1.0 5 Relative Formula Mass (Fr) of an Ionic Compound 12 fffffffB fffffffmass ffffffffffffffffffof ffffffffa fffffformula fffffffffffffffffffffffffffunit ffffffffffffffof ffffffffa ffffffcompound ffffffffffffffffffffffffffffffffff The relative formula mass of an ionic compound, Fr = mass of one atom of C @12

1.06 The Mole The mole is the amount of substance which contains as many particles (atoms, ions or molecules) as in exactly 12g of Carbon – 12. Molar mass is the Ar of Mr in grams. The unit for molar mass is gmol@1 Avogadro Constant (L) is the no. of atoms that exist in exactly 12g of C – 12. This no. is 6.02 B10 23 ` a mass No Aof mol n = ffffffffffffffffffffffffffff Ar or Mr

No Aof particles = n B L

-3-

Question: a) What is the mass of one mole of aspirin C9 H8 O4 ? (C = 12, H = 1, O = 16) b) How many mole of aspirin are there in i. 146g ii. 1.00g if this substance c) What is the mass in grams of 0.4333 moles of aspirin? d) How many aspirin molecules are there in 1.74g of this substance? e) What is the mass in grams of 1 B10 23 molecules of aspirin? f) How many carbon atoms are there in 1 mole of aspirin? g) How many oxygen molecules are in 1 mole of aspirin? Answer: a)

` a

c)

b) (i)

b) (ii)

d)

e)

146g No Amole = fffffffffffffff 180 = 0.81mol

mass g 1 mole = ffff`ffffffffffaffffffffffffffffffffffffffff`fffffffffffafff 9 12 + 8 + 4 16 ` a mass ffffffffffffffffffg ffffffff 1 mole = 180 mass = 180g ` a

1.00g No Amole = fffffffffffffffff 180 = 0.006mol

1.74g 23 No Amole = fffffffffffffffffB 6.02 B10 180 = 0.01mol

mass g 0.4333 mol = ffffffffffffffffffffffffff 180 mass = 77.99g

f)

g) 23

23

No Aparticles = 6.02 B10 B 9 = 5.42 B10

No Aparticles = 6.02 B10 B 4

24

= 1.204 B10

24

1.07 Molecules and Moles Concentration of an aqueous solution may be expressed as: i. Mole of solute per dm³ of solution ( moldm @3 ) @3 ii. Mass of solute per dm³ of solution ( gdm ) b

No Aof mole =

c

M B Vol cm3 fffffffffffffffffffffffffffffffffffffffffffff 1000

Question: How many moles of water in 1 dm³? Answer:

@3

M = moldm

Mass = No Amole B Molar mass 23 ffff1fffffB10 fffffffffffffffffffffffff = B 180 23 6.02 B10 = 29.9mol

-43

1dm = 1000cm3 = 1000g 1g Density of water = fffffffffff3fff 1cm

` a

Mr of H2 O = 1 2 + 16 @1

= 18gmol

= 1gcm@3

` a

mass g No Amole = ffffffffffffffffffffffffff Mr 1000g ffffffff = fffffffffffffffffffff@1 18gmol = 55.56mol

When a given solution is diluted, the no. of moles of solute remains

M1 V1 = M1 V 2 Where M1 = Initial molarity M 2 = Finalmolarity after addition of water unchanged after dilution. V1 = Initialvolume of water V2 = Finalvolume of water 1.08 Moles of Gases 1 mole of gas at s.t.p. occupies 22.4dm³ 1 mole of gas at r.t.p. occupies 24dm³ S.T.P. = Standard temperature and pressure f

0 ° C and 1atmpressure 273K and 1atmpressure

g

R.T.P. = Room temperature and pressure f

25 ° C and 1atmpressure 295K and 1atmpressure

1.09 Determination of Ar or Mr

g

-5-

In a mass spectrometer, bombarding the vaporized with high energy electrons from the hot wire produces positive ions:  Positive ions are needed so that it can be deflected by the magnetic field  The ions are accelerated by an electric field into the magnetic field, which causes the ions to be deflected from their straight line  The ions are accelerated to avoid doubly charged ions ( X2 + ) being formed  The extent of deflection depends on the mass to charge ratio (m/e) of an ion  A lighter ion is deflected more than heavier one with the same charge  A doubly charged ion is deflected more than singly charged ion of the same mass In a mass spectrometer, the field strength is steadily changed so that particles of increasing mass arrive one after the other at the detector. A vacuum is maintained inside the mass spectrometer since any air molecules present may hinder the movement of ions. A chart record will print a mass spectrum.

-635

+

Cl

37

+

Cl

How many lines in a spectrum represent how many isotopes. Most ions will have charge +1. Hence m/e ratio = mass of ion The mass spectrum of an element provides the following information: 1. No. of isotopes present – from the no. of peaks or lines 2. Relative isotopic mass and hence the identity of the isotope from the m/e 3. Relative abundance of each isotope – from the height of each peak and hence, 4. The relative atomic mass of an element Question: The mass spectrum of Mg as shown 24

Mg

+

25

Mg

+

26

Mg

+

a) What are the isotopes of Mg? b) What are the species responsible for the peaks? c) Calculate the Ar of Mg Answer:

a)

24

b)

24

c)

25

26

Mg , Mg , Mg +

25

+

26

Mg , Mg , Mg

+

` a

` a

` a

24 8 + 25 1 + 26 1 Relative abundance = fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff 10 = 24.3

-7Question: The Ar of Rb is 85.6. Rb exists as spectrum of Rb. Answer:

85

Rb and

= 85.6 ` a ` a 85 fffffffffffxffffffff+ ffffff87 ffffffffffff100 fffffffffffff@x fffffffffffffff 85.6 = 100 8560 = 85x + 8700 @87x x = 70

87

Rb . Draw a predicted mass

Relative Abundance

Ar

85

87

7

85

Rb

+

87

3

Rb = 70% Rb = 30%

85

Rb

+

87

1.10 Mass Spectrum of Molecules HCl A mass spectrum of a compound provides the following information: 1) The identity of fragment ions – from the m/e ratio of the ion 2) The identity of the molecular ion and hence the Mr of the compound – from the highest m/e value 3) Relative abundance of fragment ions or molecular ions of each peak Relative abundance 35

+

Cl

H

35

+

H Cl

+

37

+

Cl

37

+

H Cl

m/e 1

35

36

37

38

E.g.: A molecule of chlorine gas is subjected to a mass spectrometer The mass spectrum obtained is Relative abundance 35

+

Cl

35

+

Cl2

37

+

Cl

35

37

+

Cl @ Cl

-8-

37

+

Cl2

m/e 35

37

Molecule Molecular species m/e Relative height

70

35

35

35

Cl @ Cl 35 + Cl2

72

37

37

Cl @ Cl 35 37 + + Cl @ Cl

70

f

3ffff 3ffff B 4 4

74 37

Cl @ Cl 37 + Cl2

72

74

g 3ffff 1ffff

1ffff 1ffff B 4 4

9 = fffffff 16

B B2 4 4 6 = fffffff 16

1 = fffffff 16

9

6

1

Ratio

Question: Draw a mass spectrum when a Br 2 molecule is subjected to a mass spectrometer. Calculate the relative abundance of the molecular species. Answer: Molecule

79

Br 2

Molecular Species

79

+

m/e

158

79

79

Br 2

f

1fff 1fff 1ffff B = 2 2 4

Relative Abundance Ratio

1 Relative Abundance 79

2

81

Br Br

+

81

Br Br

81

Br 2

81

81

+

Br Br

+

Br 2

160

162

g

1fff 1fff 1ffff B = 2 2 4

2

1

1fff 1fff 1 B B 2 = fff 2 2 2

-979

1

81

+

Br 2

+

Br 2

m/e 158

160

162

Draw a mass spectrum when a BrCl molecule is subjected to a mass spectrometer. Calculate the relative abundance of the molecular species. Answer: Molecule

79

Molecular Species

79

m/e

35

79

Br Cl 35

+

79

Br Cl

f

Ratio

81

37

+

116

g

f

3ffff 1fff 6 B B 2 = fff 4 2 8

35

+

81

Br Cl 116

g

f

1fff 1ffff 2 B B 2 = fff 2 4 8

3

81

35

Br Cl

81

Br Cl

114

Relative Abundance

37

Br Cl

37

g

3

f

g

1fff 1ffff 2 B B 2 = fff 2 4 8 1

Relative Abundance 79

4 3

79

37

+

Br Cl 35

35

81

+

, Br Cl

+

Br Cl

81

1

37

+

Br Cl

m/e 114

116

+

Br Cl 118

1fff 3ffff 6 B B 2 = fff 2 4 8

1

37

Br Cl

118

1.11 Empirical and Molecular Formula  Empirical shows the simplest whole number ratio of the atoms of the different element in the compound  Molecular formula shows the actual number of atoms of each element present in one molecule of compound

- 10 b

c

b

c

n empirical formula = molecular formula n empirical formula = Mr 1.12 Calculations of Mr from Ideal Gas Equation Ideal gas equation given by:

PV = nRT

`

a

b

c

P = Pressure Pa V = Volume m3

mass n = no Aof mole = fffffffffffffff Mr @1 @1 R = gas constant = 8.31 Jk mol ` a T = Temperature K

mass PV = fffffffffffffffRT Mr mass RT Mr = ffffffffffffffffffffffffff PV RT = density ffffffffff P density = gm@3

b

c

@3

volume cm3 Q m3 B10 3

b

c

@3

dm Q m3 B10 `

a

° C Q K + 273

Question: 0.4g of a gas has a volume of 227cm³ at 27 ° C and a pressure of 100kPa. Calculate the Mr of the gas. Answer:

Mass

= 0.4g @6

Volume = 227 B10 m3 Temperature = 27 + 273 = 300K 3 Pressure = 100 B10 Pa

mass RT Mr = ffffffffffffffffffffffffff PV fffffffff0.4 fffffffffffB fffffff8.31 fffffffffffffffB fffffff300 ffffffffffffffffffffff = 3 @6 100 B10 B 227 B10 997.2 = fffffffffffffffff 22.7 = 43.9

Question: A gas has a density of g/dm³ at 373 ° C and at a pressure of 200kPa. Calculate the Mr of the gas.

Mass

= 1.7g 3

= 1dm @3 = 1 B10 m3 Temperature = 27 + 373 = 400K 3 Pressure = 200 B10 Pa Volume

RT Mr = density ffffffffff P Mass B RB T Mr = fffffffffffffffffffffffffffffffffffffffffffffffffffffffff Volume B Pressure 1.7 Bfffff8.31 fffffffffffffffB fffffff400 fffffffffffffffffff Mr = fffffffffffffffff@3 3 1 B10 B 200 B10

Question: Calculate the volume occupied when 1g of ice is heated to 323K at 1 atm pressure.

- 11 PV = nRT mass ` a` a PV = fffffffffffffff R T Mr `

a ` a

101000Pa B V =

V = 1.48 B10

@3

f

g fff1ffffffff

189

8.31 B 323K

@3

dm

1.13 Calculations using Combustion Data The molecular formula of hydrocarbon can be determined by combustion in excess oxygen to form CO 2 and H2 O . ` a

d

e

Cx Hy g + x +

` a y ` a yffff ffff O 2 Q xCO 2 g + H 2 O l 4 2

Under room temperature or s.t.p. the water produced is a liquid. Hence water is negligible compared with the volumes of C x H y , O 2 and CO 2 gases. Equal volumes gases, under the same conditions of temperature and pressure, contain equal numbers of molecules or moles. (Avogadro’s Law) Question: 10cm³ of a gaseous hydrocarbon required 20cm³ of oxygen gas for complete combustion. 10cm³ of CO 2 was produced in the combustion. All gases were measured under the same condition. Calculate the relative molecular formula and hence the relative molecular mass. Answer:

d

e

` a y `a y C x H y g + x + ffff O2 Q xCO 2 g + ffffH2 O l 4 2 ` a

10cm³

20cm³

X=1

10cm³

Y X + fffff= 2 4 Y fffff = 2 @1 4 Y=4

Molecular formula = CH4 ` a Mr = 12 + 4 16

Ratio: 1 2 1 O 2 gas can be absorbed completely by soda lime or KOH when the gaseous products of combustion are passed by over soda lime or KOH, there is a decrease in volume of gas = volume of CO 2 Question: 20cm³ of a gaseous hydrocarbon was mixed with 150cm³ of oxygen. The mixture was sparked so that the hydrocarbon was completely burnt. The gaseous products had a total volume of 130cm³, when this product was passed over soda lime, the volume of the product decreased to 90cm³. Deduce the molecular formula of the hydrocarbon.

- 12 d

e

` a ` a y `a y C x H y g + x + ffff O2 Q xCO 2 g + ffffH2 O l 4 2

20cm³

150cm³ 130cm³ - 40cm³ = 90cm³X = 2

20cm³ Ratio

1

60cm³ 3

40cm³ 2

Question: 25cm³ of C2 H 6 was mixed with 35cm³ of oxygen gas. The mixture was electrically sparked so that ethane gas burnt in oxygen. Calculate:a) The volume of gases left after combustion at r.t.p. b) What is the total change in volume after the reaction.

1.14 Percentage Composition E.g. Calculate the percentage composition by mass of benzene C6 H 6 .

=

` a ffff12 ffffffffffff6 fffffffffffff ` a B 100%

Percentage Composition % of Hydrogen H =

` a ffffff1 ffffffff6 fffffffffffffff ` a B 100%

`

a

` a

Percentage Composition % ofCarbon C

`

a

` a

12 6 + 6 = 92.3%

12 6 + 6 = 7.7%

Y X + fffff= 3 4 Y fffff = 3 @2 4 Y=4

- 13 -

1.15 Yield The mass of product formed in a chemical reaction is called the yield. Theoretical yield: Yield of a product calculated from the chemical equation. Actual yield: The yield actually obtained in an experiment or in an industrial process. It will be much less due to incomplete reaction or product loss during the reaction. Percentage yield: Is a measure of the efficiency of the reaction. Percentage yield =

fffffffActual ffffffffffffffffffffffyield ffffffffffffffffffffffff B 100% Theoretical yield

Question: In an experiment, 23.0g of bromobenzene was obtained from 20g sample of benzene. Calculate the percentage yield.

C6 H 6

+

Benzene 20g

Br

Q

C6 H5 Br

+

HBr

Bromobenzene

Actual yield = 23g

20 = ffffffff 78 20 Thereoticalyield ofC6 H5 Br = ffffffff 78 a 20 ` Mass ofC6 H5 Br = ffffffffB 72 + 5 + 79.9 78 = 40g No Aof mole C6 H6

Percentage yield

23g = ffffffffffffB 100% 40g = 58%