Introduction to Chemistry

1. 1/ Atoms, Molecules and Stoichiometry
• Background: Matter is a substance that has inertia and occupies
physical space and consists of various types of particles, each with
mass and size.
• The particles are the electron, the proton and the neutron.
• Combinations of these particles form atoms.
• A combination of atoms forms a molecule.
• Atoms and/or molecules can join together to form a compound.

2. Chemical reactions consume
and produce materials
knowledge of relative atomic
and molecular masses.
Stoichiometry: calculate the
quantities of these materials
e.g titrations enable the
determination of the
concentration of a sample

3. 1.1/ Relative Masses of Atoms and Molecules
• atomic masses are based on 12C as the standard and 12C has a mass of
12 atomic mass units (a.m.u).
• All other atoms are given masses relative to the 12C standard.
• masses of atoms are compared using an instrument called a mass
spectrometer.

4. The instrument operates as follows:
• Atoms or molecules are passed into a beam of electrons moving at
high speed
• electrons convert the atoms or molecules being analysed into
positively charged ions
• The positive ions pass through an electric field which accelerates
them towards a magnetic field.
• Accelerating ions create their own magnetic field n interaction
between the two magnetic fields results in the ion being deflected
and the amount of deflection for each ion depends on its mass.
• Comparison of the areas where the ions hit the detector plate
enables the accurate determination of their relative masses.

5. • Example: When the masses of 12C and 13C are compared in a mass
spectrometer their ratio is found to be:

6. 1.1.1/ Relative atomic masses for elements with
isotopes
• though carbon -12 was assigned a mass of exactly 12amu, its mass number as shown in the
periodic table is 12.01 or is not 12.
• since carbon is found in nature as a mixture of isotopes 12C, 13C and 14C, having different
numbers of neutrons but the same number of protons. The atomic mass of 12.01 is actually
an average value
• Which is calculated by taking into account the relative abundances of the three isotopes.
• Natural carbon consists of 98.89% of 12C, 1.11% of 13C and the amount 14C is negligible
• The relative atomic mass is obtained as follows:
• 98.89% of 12 amu + 1.11% of 13.003
= (0.9889 x 12 amu) + (0.011 x 13.003 amu) = 12.01 a.m.u
• 12.01 is the atomic weight of carbon.

7. 1.1.2/ Relative atomic and molecular masses
from mass spectra
• The mass spectrometer is also used for determining the isotopes of a
natural element.
• When a sample of a naturally occurring element is analysed in a mass
spectrometer, a mass spectrum is obtained.
• a plot of the number of atoms against mass number.
• If the mass spectrum shows a number of peaks, then they are due to
the isotopes of the element.
• spectrum enables the determination of the masses of each isotope
and their relative abundances.

8. • Below is a spectrum of natural neon:
• Therefore natural Neon has three
• isotopes:
20Ne, 21Ne and 22Ne.
• their relative abundances are
91%, 0.2% and 8.8% respectively.
• The relative atomic mass of Neon is :
• (0.91x20) + (0.002x 21) + (0.088 x 22) = 20.18

9. 1.1.3/ Relative Molecular Mass
• mass spectrum of a compound shows a variety of peaks that
correspond to the various fragments of the molecule.
• The relative mass of the heaviest particle shown on a mass spectrum
is that of the unfragmented particle with a positive charge.
• This mass is taken as the relative molecular mass of the compound,
the sum of all the atomic masses of the elements making up the
molecule.

10. • Examples:

11. 1.1.4/ Empirical and molecular formulae of
compounds
• the simplest whole number ratio of the different types of atoms that
make up a compound whereas the molecular formula represents the
exact formula.
• Two methods can be used to determine the formula of a compound.
These involve the use of composition by mass data and combustion
data.
A weighed sample is either decomposed into its constituent elements
(mass spectrometry) or it is reacted with oxygen to produce CO2, H2O
and N2.

12. • E.g: An organic substance is composed of carbon, hydrogen and nitrogen. When
0.1156 grams of the compound was burnt in oxygen, 0.1638g of carbon dioxide
CO2 and 0.1676g of Water H2O were collected. Given that the molar mass of the
organic compound is 62.12 deduce its empirical and molecular formulae.
• Step 1: Determine the number of atoms (in moles) of each element present
• the fraction of carbon in terms of mass present in CO2 is
= (atomic mass of carbon / molar mass of CO2)
• Step 2: Determine the mass of elements in original sample

13. Step 3: Determine the moles of elements in original sample Step 4: Determine the ratios of elements in original sample

14. 1.2/ Stoichiometric Calculations