stoichiometry

 

In 12 gr carbon there are 6*10exp23  (exactly 6.02205*10exp23) carbon Atoms. It’s the figure of Avogadro and is one Mole.

In periodic system of Elements you’ll find the element C, Carbon. The figure 12.011 is the molare mass of element C.

12.011 gr = 6*10exp23 Atoms of element C = 1 Mole.

In the periodic tabel of Elements above you’ll find the molare mass of each element  on the third line.

Examples: Helium, He, an inert gas, ist molare mass weighs 4.003 gr that means 1 mole of Helium weighs 4.003 gr.

Oxygen, O, has a molare mass of 15.999 gr. Be Aware Oxygen does not exist as an atome. O2 most important component of the air is a molecule.

One mole O2 weighs 31.998 gr (about 32 gr).

 

Avogadros number is significant all over chemistry, especially in stoichiometry determinig the reaction equations.

Thanks of the avogadros number chemical formulas can be defined knowing the percentage composition of compounds.

And also the concentration of a solution and determination of base and acids.

 

Reaction equation

Reactants  ===>  products

example:

hydrogen an oxygen form water. 2H2 + O2 ===> H2O

It means: 2 moles of hydrogen, H2,  and 1 mole of Oxygen, O2,  yield water, H2O.

There are reactions they run by itselves, like the one above. Energy is released, it is called an exothermic reaction.

Other reactions energy (heat) is needed or a catalyst, a compound of an element, which helps run on.

When energy or a catalyst ist needed we speak of an endothermic reaction.

Photosynthesis, an example of an endothermic biochemical reaction.

6 CO2 + 6 H2O ===> C6H12O6 (glucose) + 6 O2

 

Determining of reaction equations:

The number of Atoms of the reactants must match with those of the products. Let’s controle the reaction of photosynthesis

reactants: we get 6 C, 18 O and 12 H

products: we get 6 C, 18 O and 12 H

Other example: Elementary sodium, Na, is given in water. You get caustic soda, NaOH and hydrogen, H2. It is an exothermic raction, runnig by itself.

What’ s the reaction equation?

We note the reactants and the products: Na + H2O ===> NaOH + H2

On the productside there are 3 H-Atoms, 1 H is missing on the left,  we try to equalize Na,H2O and NaOH with an additional mole.

Multiplying each compound and element exept hydrogen with 2, you get: 2 Na + 2 H2O ===> 2NaOH + H2

Let’s see if it matchs:

reactants: we get 2 Na, 4H and 2O

products: we get 2 Na, 4H and 2O

 

Determination of empirical chemical formulas due percentage composition

From a compound the following percentage compositions are known:

Carbon, C 45.9%, hydrogen, H 3.21% and bromine, Br 50.89%

Dividing each element by its molare mass you get:

C 45.9/12.011 = 3.82, H 3.21/1.0079 = 3.18, Br 50.89/79.904 =  0.636

And then dividing the results of C and H by the smallest result of Br:

C 3.82/0.636 = 6, H 3.18/0.636 = 5

The empirical formula found is C6H5Br

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