Introduction to Chemistry

for Allied Health Sciences

Structure of the Atom

Kirk HunterChemical Technology Department

Texas State Technical College Waco

Structure of the Atom

Subatomic Particles

Name Relative Relative Mass (amu)

ChargeProton 1 +1Neutron 1 0Electron 0 -1

ATOMThe protons and

neutrons are located in the

center in the region called the nucleus.

ATOMElectrons are found around the nucleus in shells or principle energy levels (PEL).

ATOMIC NUMBER

The number of protons in the nucleus of an atom

Z = #p

ATOMIC NUMBER

For a neutral atom, the number of protons is equal to the number of electrons.

#p = #e

MASS NUMBERThe sum of protons and

neutrons in the nucleus of an atom.

A = #p + #n

MASS NUMBERRound the atomic

weight to the nearest whole numberElement Atomic Mass Mass

Number

Li 6.941

Ca 40.078

B 10.81

Au 196.9665

7

41

11

197

Isotopic Notation

A = mass number E = symbol of element

Z = atomic number

EA

Z

Isotopic Notation

Example:Determine the number of protons, electrons and neutrons in 51

23V.

Solution:

# p = 23

# e- = 23

# n = 51 - 23 = 28

Isotopic NotationExample:Determine the number of protons,

electrons and neutrons for the following:

147

N 23892U 23

11Na 20080Hg

#p#e- #n

777

9292146

111112

8080120

Isotopes• Around 20 elements have a fixed

number of neutrons.• Atoms of the same element having

a different number of neutrons are called isotopes.

Isotopes have the same number of protons and a

different number of neutrons.

Isotopes of Hydrogen

11H 2

1H 31H

Protium Deuterium Tritium p+ 1 1 1 n0 0 1 2 e- 1 1 1

Isotopes of Helium

32He 4

2He

p+ 2 2 n0 1 2 e- 2 2

Isotopes of Oxygen

168O 17

8O 188O

p+ 8 8 8 n0 8 9 10 e- 8 8 8

ATOMIC MASS UNIT• Carbon-12 used as the reference

standard.• All atoms are compared to C-12.

1 amu = 1/12 the mass of a C-12 atom.

ATOMIC MASS The average mass of all isotopes of the element.

–exact mass of each isotope

–percent abundance of each isotope in nature

ATOMIC MASSExample:

Determine atomic mass of vanadium.

Isotope exact mass % abundance51V 50.9876 amu 35.64%52V 51.9346 amu 64.36%

At.wt.=(50.9876 *0.3564) + (51.9346 * 0.6436)

= 18.17+33.43

= 51.60 amu

ATOMIC MASSExample:

Determine atomic mass of silicon.

Isotope exact mass % abundance

26Si 25.9854 amu 5.435%28Si 27.9721 amu 76.42%29Si 28.9632 amu 18.15%

At.wt. = (25.9854*0.05435)+(27.9721*0.7642) +(28.9632*0.1815) = 28.05 amu