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Stoichiometry 1: Composition Stoichiometry.

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Teaches how to handle Calculations in Stoichiometry 1. It helps the student overcome one of the biggest challenges in Chemistry. Good for Students in high school and the first year in university. Highly simplified, written in simple language and lucid style for easy comprehension.

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STOICHIOMETRY
1

CHEMICAL FORMULAS

and

COMPOSITION STOICHIOMETRY



By


Temple C. Eke

,BACKGROUND
The chemical formula of a substance shows its chemical composition.
This represents the elements present, and the ratio in which the atoms of the
elements are combined.
Composition stoichiometry describes the quantitative relationships
among elements in compounds.

THE MOLE AND MOLAR MASS
The mole is the amount of a substance (element or compound) that
contains the same number of particles (atoms, molecules, and ions) as atoms
present in of the Carbon-12 isotope. Generally, 1 mole of any substance
contains particles. This number is known as Avogadro’s number,
. Thus, .
The molar mass, of a substance is the mass (in grams) of 1 mole of
the substance. The molar mass of a compound is given by the sum of the
relative atomic masses of the elements in its formula, each taken the number
of times the element occurs in the formula. The molar mass of a compound is
also known as its formula weight.
Let us consider some illustrative examples on finding the molar masses of
compounds.

Example 1: Molar Masses From Chemical Formulae.
Calculate the molar mass of each of the following substances:
(a)
(b)
(c)
(d)




Solution:
The relative atomic mass of each element is multiplied by the number
of times it occurs in the formula. The results are then added to give the molar
mass of the substance.

(a) contains 2 Oxygen (O) atoms.

, Molar mass of = (16 2) = 32

(b) contains 1 Ca atom, 1 C atom, and 3 O atoms.
Molar mass of = 40 + 12 + (16 3)
= 40 + 12 + 48 = 100

(c) contains 2 Al atoms and three O atoms.
Molar mass of = (27 2) + (16 3)
= 54 + 48 =102.

(d) contains 1 Fe atom, 1 S atom, 11 O atoms (4 + 7),
and 14 H atoms (7 2).
Molar mass of = 56 + 32 + (16 11) + (1 14)
= 56 + 32 + 176 + 14 = 278.

The number of moles present in a given mass, can be found as follows:

Let the number of moles present in grams be .
i.e mole = grams (1)

But 1 mole of the substance has a mass equal to the molar mass, .
i.e 1 mole = grams (2)

By taking the ratio , we have




Thus,
Hence,


.
In a similar manner, the number of particles, present in a given number of
moles can be found as follows:

Let the number of particles present in moles be .
i.e particles = moles (3)

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