An equation is simply a statement of chemical change using chemical symbols. For example, when sulfur is burned in air (one of my favourite chemical reactions ;-), it is combining with the oxygen in the air to produce an oxide. Let's look at this reaction in the form of a chemical equation:

S (sulfur) + O2 (oxygen) -----|> SO2 (sulfur dioxide).

Examine the equation closely. Is it consistent with the Law of Conservation of Matter? In other words, are there equal numbers of each type of atom on each side of the equation? There is. This equation is therefore said to be balanced. An equation is meaningless unless it is balanced. When an equation is balanced, it is said to be stoichiometric.

This equation tells us more than merely that sulfur combines with oxygen to produce sulfur dioxide. It tells us that one atomic weight's worth, of sulfur withh combinr with one molecular weight's worth, of oxygen to produce one mole of sulfur dioxide. The molecular or atomic weight's worth of something is properly called a mole. If the units of grams are used, this would be:

S (sulfur's atomic weight is about 32.1, so you would use 32.1 grams of this subtance) + O2 (the atomic weight of O is about 16, so therefore the molecular weight of O2 is 32, so 32grams of this would be used) -----|> SO2 (this substance's mole is equal to the sum of the previous subtances moles; a mole of this would therefore weigh 64.1 grams).

In other words, this equation tells us that 1 mole of sulfur combines with 1 mole of oxygen to form 1 mole of sulfur dioxide.

Let's look at another chemical oxidation reaction, in this case the marvelous reaction of the oxidation of magnesium (oh joy! ;-). When magnesium is oxidized, it forms an oxide (duh ;-). This reaction is indicated as follows:

Mg (magnesium) + O2 (oxygen) -----|> MgO (magnesium oxide).

What about our beloved Law of Conservation of Matter now? Do you see that we have apparently destroyed some of our beloved oxygen? This equation is not balanced. It is called a skeleton equation, for it indicates only the names of the substances involved.

This equation would be balanced if we could put a 2 after the O of the MgO to make it MgO2. But this would violate the Law of Definate Proportions, because magnesium oxide always has the formula MgO. In balancing equations, the subscript in the formulas may NOT be changed.

A skeleton equation is balanced by placing numbers, called coefficients, in front of the formulas of the substances in the reaction. Look again at our skeleton equation. By placing the coefficient 2 in front of our lovely MgO, we would have two oxygen atoms on each side of the equation, for the coefficient multiplies all the symbols in the formula immediately after it. This would change our equation to this:

Mg + O2 -----|> 2 MgO.

However, looking at this equation, it is apparent we have too much Mg at the right side of the equation. This can be remedied by placing another coefficient of 2 in front of the Mg at the left side of the equation, giving us the following:

2 Mg + O2 -----|> 2 MgO.

Now the equation is balanced. We have 2 magnesium atoms and 2 oxygen atoms on each side of the equation. The balanced equation now reads 2 moles of magnesium combine with 1 mole of oxygen to produce 2 moles of magnesium oxide. The following expression shows how the weights of each of the substances in the balanced equation may be indicated:

2 Mg [2 x 24.31] + O2 [2 x 16] -----|> 2 MgO [ 2 x (24.31 + 16)].

So, 48.62 units of weight of magnesium combine with 32 units of weight of oxygen to form 80.62 units of weight of magnesium oxide. These units of weight may be grams, kilograms, ounces, pounds, grains, tons, tonnes, etc., just so long as all three weights are expressed in the same units. This weight relationship also tells us that magnesium and oxygen combine in a ratio of 48.62 parts by weight of magnesium to 32 parts by weight of oxygen. Similarly, 80.62 parts by weight of magnesium oxide are formed for every 32 parts of oxygen or every 48.62 parts of magnesium. It is quite easy to convert this into percentage ratios (read on..).

Let's look at yet another example. Naphthalene (formula C10H8) is a substance commonly used to produce large orange fireball effects in movies. It burns with oxygen under optimum conditions to form carbon dioxide and hydrogen monoxide (water). The skeleton equation is:

C10H8 + O2 -----|> CO2 + H2O.

Let us balance this skeleton equation using the "even numbers" technique described in the previous examples.

1. First, we balance the equation for the carbon and hydrogen. Since there is 10 carbons on the left side, there must therefore be 10 carbons on the right side (of course, combined with O2). Also, there is 8 hydrogens on the left side, so therefore there must be 8 on the right side. The equation, with the coefficients placed, becomes this:

w C10H8 + x O2 -----|> 10 CO2 + 4 H2O.

This gives us an equal number of carbon and hydrogen on both sides. Now, we must balance the equation to work O2-wise.

2. First, we must determine how many oxygen atoms are needed by the hydrogen to completely oxidize to H2O. This number is 4 oxygen atoms, or 2 O2s. For complete combustion of the C into CO2, 20 oxygen atoms, or 10 O2s are needed. Therefore, the equation needs a total of 12 O2s (2 O2 for the hydrogen + 10 O2 for the carbon). Thus, the equation becomes so:

1 C10H8 + 12 O2 -----|> 10 CO2 + 4 H2O.

This equation reads: 1 mole of naphthalene combine with 12 moles of diatomic oxygen to produce 10 moles of carbon dioxide and 4 moles of hydrogen monoxide (water). The weight proportions involved are:

Reactants: (Naphthalene: 120 + 8 = 128; Oxygen: 12(32) = 384); 128 + 384 = 512.

Products: (Carbon Dioxide: 10(12 + 32) = 440; Water: 4(2 + 16) = 72); 440 + 72 = 512.

The charactaristics of a balanced equation may be summarized as follows:

1: It obeys the Law of Conservation of Matter.

2: It obeys the Law of Definite Proportions.

3: Its coefficients give the molar proportions of reactants and products involved in the reaction.

Symbols, formulas, and equations all have definite quantitative meanings. We are now ready to look at some numerical applications based upon these ideas.

Stoichiometric pyrotechnic compositions are basically balanced equations that are converted into percent ratios from the use of the substance's molecular weights (moles). An example is that of ordinary flash powder:

3 KClO4 + 8 Al -----|> 3 KCl + 4 Al2O3

As you can see, a ratio of 3 moles of potassium perchlorate to 8 moles of aluminium powder are needed for a complete, balanced reaction. To convert this into a percentage ratio, you need to do some math with the molecular weights:

1 mole of KClO4 = 138.55

138.55 x 3 = 415.65

Therefore, 3 moles of KClO4 = 415.65

1 mole of Al = 26.98

26.98 x 8 = 215.84

Therefore, 8 moles of Al = 215.84

Next, you add the two numbers you obtained together:

415.65 [3 moles of KClO4] + 215.84 [8 moles of Al] = 631.49.

Then, you do this, to obtain the percentage:

(X times 100) divided by Y = percentage for comp, [where X equals either the 3 moles of the Al or the 8 moles of KClO4, and the Y equals the sum of both moles added together (see examples below). I hope you know some algebra!]

Here is an example of this equation in action. This will determine how much Al must be present.

First, we substitute our values:

(215.84 times 100) divided by 631.49 = ?%

Then, we finish the equation:

(215.84 times 100 = 21584) divided by 631.49 = ?%; 21584 divided by 631.49 = 34.17948%

The percentage thus obtained (34.17948%) can be used to extrapolate the percentage of KClO4 that must be present, by simply subtracting it from 100:

100 minus 34.17948 = 65.82052%

Thus, the composition, stoichiometrically, would be ~65.8% KClO4, ~34.2% Al. As you can see, the traditional 70/30 flash is a little overoxidized.

You can use this equation to easily determine a basis for most of your pyrotechnic compositions.

I hope this helps. If you have any questions, feel free to ask.