New Way Chemistry for Hong Kong A-Level Book 11 Atomic Structure 1.1The Atomic Nature of Matter...

New Way Chemistry for Hong Kong A-Level Book 11

Atomic StructureAtomic Structure

1.11.1 The Atomic Nature of MatterThe Atomic Nature of Matter

1.21.2 The Experimental Evidence ofThe Experimental Evidence of

Atomic StructureAtomic Structure

Chapter 1Chapter 1


Dalton’s atomic theory (1808)

John Dalton proposed his Dalton’s atomic theory.

1.1 The atomic nature of matter (SB p.2)


Main points of Dalton’s atomic theory

1.1 The atomic nature of matter (SB p.2)

5. Atoms of different elements combine to form a compound. The numbers of various atoms combined bear a simple whole number ratio to each other.

1. All elements are made up of atoms.

2. Atoms can neither be created nor destroyed.

3. Atoms of the same element are identical. They have the same mass and chemical properties.

4. Atoms of different elements are different. They have different masses and chemical properties.


1.2 The experimental evidence of atomic structure (SB p.3)

Discovery of electrons

- A beam of rays came out from the cathode (-) and hit the anode (+).

Cathode Ray Discharge Tube



The beam was composed of negatively charged fast-moving particles called ‘electrons’.Deflected in the

electric field

Deflected in the magnetic field



Gold foil scattering experiment

- performed by Ernest Rutherford

Location of protons?



He bombarded a thin gold foil with a beam of fast-moving -particles (+ve charged)

Observation:

-most -particles passed through the foil without deflection

-very few -particles were scattered or rebounded back


Interpretation of the experimental results

- Most of the atom is empty space.

- The mass concentrated at the center of an atom called ‘nucleus’.

- The nucleus is positively charged.- The positively charged particle is called ‘proton’.



Chadwick’s atomic model

- the presence of neutrons

- proved by James Chadwick

Mass of atom > Total mass of protons


Chadwick’s atomic model


ProtonElectron

Neutron


Characteristics of sub-atomic particles1.3 Sub-atomic particles (SB p.6)

Sub-atomic particle

Proton Neutron Electron

Symbol p or n or e- or

Location in atom Nucleus Nucleus Surrounding the nucleus

Actual charge (C)

1.6 x 10-9 0 1.6 x 10-9

Relative charge +1 0 -1

Actual mass (g) 1.7 x 10-24 1.7 x 10-24 9.1 x 10-28

Approximate relative mass (a.m.u.)

1 1 0

H11

n10

e0-1


Isotopes1.4 Atomic number, mass number and isotopes (SB p.8)

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.

Cl has 2 isotopes: Cl-35 and Cl-37

Isotopes Relative abundance

Cl-35 75%

Cl-37 25%


1.8 Mass spectrometer (SB p.20)



Mass spectrometer

A highly accurate instrument!


Mass spectrometer consists of 6 parts:1.8 Mass spectrometer (SB p.20)


Mass spectrum of Cl2:1.8 Mass spectrometer (SB p.21)

m/e ratio Corresponding ion

35 35Cl+

37 37Cl+

70 35Cl─35Cl+

72 35Cl ─ 37Cl+

74 37Cl ─37Cl+


Mass spectrum of CH3Cl:

m/e ratio Corresponding ion

35 35Cl+

37 37Cl+

50 12CH3─35Cl+

51 13CH3 ─ 35Cl+

52 12CH3 ─37Cl+



Relative atomic mass

The relative atomic mass of an element is the weighted average of the relative isotopic masses of its natural isotopes on the carbon-12 scale.

1.9 Relative isotopic, atomic and molecular masses (SB p.23)



What is the relative atomic mass of Cl?

The relative abundances of Cl-35 and Cl-37 are 75.77 and

24.23 respectively

Relative atomic mass of Cl

=

= 35.48

24.23)(75.5524.23)(3775.77)(35


Relative molecular mass


The relative molecular mass is the relative mass of a molecule on the carbon-12 scale.



What is the relative molecular mass of CH3Cl?

Relative molecular mass of CH3Cl

=

= 50.540)2(123

40)(522)(51123)(50



Check Point 1-6Check Point 1-6

(a) The mass spectrum of lead is given on the right. Given that the relative atomic mass of lead is 207.242, calculate the relative abundance of the peak at m/e 208.

Answer

Let x be the relative abundance of the peak at m/e 208.

(204 1.5 + 206 23.6 + 207 22.6 + 208x) (1.5 + 23.6 + 22.6 + x) = 207.242

x = 52.3

The relative abundance of the peak at m/e 208 is 52.3.



Check Point 1-2 (cont’d)Check Point 1-2 (cont’d)

(b)The mass spectrum of dichloromethane is given on the right. Calculate the relative molecular mass of dichloromethane.

Answer

Let y be the relative molecular mass of dichloromethane.

y = (84 94 + 85 3.0 + 86 59 + 87 2.2 + 88 13 + 89 2.5 + 90 0.8 ) (90 + 3.0 + 59 + 2.2 + 13 + 2.5 + 0.8)

= 85.128

The relative molecular mass of dichloromethane is 85.128.


The END

New Way Chemistry for Hong Kong A-Level Book 11 Atomic Structure 1.1The Atomic Nature of Matter...

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