chemistry notes chapter 1 complete)

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- 1 - CHEMISTRY 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. , , 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. Relative mass has no unit since it is a ratio. Isotopes are atoms of the same element with different masses. Naturally occurring element has a mixture of isotopes. E.g. Chlorine has 2 isotopes of and in the ratio of 3:1.

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Page 1: Chemistry Notes Chapter 1 Complete)

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CHEMISTRY 9701

Chapter 1 : ATOMS, MOLECULES AND STOICHIOMETRY

1.01 Atoms and Molecules1. 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. ,

,

1.02 The Atomic Theory of John Dalton1. 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 CompoundsThe 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.

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

Naturally occurring element has a mixture of isotopes.E.g. Chlorine has 2 isotopes of and in the ratio of 3:1.

The % of isotopes is known as abundance.

The abundance of =

The abundance of =

Question: K exists as and . Calculate the abundance of each isotope. (Ar of K = 39.1)

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Let the abundance of be Let the abundance of be

= 90%= 10%

1.0 4 Relative Molecular Mass (Mr) of a Molecule

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

Mr of

1.0 5 Relative Formula Mass (Fr) of an Ionic Compound

1.06 The MoleThe 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

Avogadro Constant (L) is the no. of atoms that exist in exactly 12g of C – 12. This no. is

Question:a) What is the mass of one mole of aspirin ? (C = 12, H = 1, O = 16)b) How many mole of aspirin are there in

i. 146gii. 1.00g if this substance

c) What is the mass in grams of 0.4333 moles of aspirin?

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d) How many aspirin molecules are there in 1.74g of this substance?e) What is the mass in grams of 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) b) (i) b) (ii)

c) d) e)

f) g)

1.07 Molecules and MolesConcentration of an aqueous solution may be expressed as:i. Mole of solute per dm³ of solution ( )ii. Mass of solute per dm³ of solution ( )

Question:How many moles of water in 1 dm³?

Answer:

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

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1.08 Moles of Gases 1 mole of gas at s.t.p. occupies 22.4dm³1 mole of gas at r.t.p. occupies 24dm³S.T.P. = Standard temperature and pressure

R.T.P. = Room temperature and pressure

1.09 Determination of Ar or Mr

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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 ( ) 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.

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 lines2. Relative isotopic mass and hence the identity of the isotope from the m/e3. Relative abundance of each isotope – from the height of each peak and hence,4. The relative atomic mass of an element

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Question:The mass spectrum of Mg as shown

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

Answer:a)

b)

c)

Question:The Ar of Rb is 85.6. Rb exists as and . Draw a predicted mass spectrum of Rb.

Answer: Relative Abundance

7

3

85 871.10 Mass Spectrum of Molecules

HCl

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A mass spectrum of a compound provides the following information:1) The identity of fragment ions – from the m/e ratio of the ion2) The identity of the molecular ion and hence the Mr of the compound – from the highest m/e value3) Relative abundance of fragment ions or molecular ions of each peak

Relative abundance

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

m/e35 37 70 72 74

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

Answer:Molecule

Molecular Species

MoleculeMolecular species

m/e 70 72 74Relative height

Ratio 9 6 1

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m/e 158 160 162

Relative Abundance

Ratio 1 2 1

Relative Abundance

2

1

m/e

158 160 162

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

Answer:Molecule

Molecular Species

m/e 114 116 116 118

Relative Abundance

Ratio 3 1 3 1

Relative Abundance

4

3

1

m/e114 116 118

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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

1.12 Calculations of Mr from Ideal Gas Equation

Ideal gas equation given by:

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

Answer:

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

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

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1.13 Calculations using Combustion DataThe molecular formula of hydrocarbon can be determined by combustion in excess oxygen to form and .

Under room temperature or s.t.p. the water produced is a liquid. Hence water is negligible compared with the volumes of and 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 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:

10cm³ 20cm³ 10cm³

Ratio: 1 2 1

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

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.

20cm³ 150cm³

130cm³ - 40cm³ = 90cm³

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20cm³ 60cm³ 40cm³

Ratio 1 3 2

Question:25cm³ of 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 CompositionE.g. Calculate the percentage composition by mass of benzene .

1.15 YieldThe 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:

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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.

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

Benzene Bromobenzene

20g Actual yield = 23g