How to Balanced an Equation.

1  Balanced Equations, or Stoichiometry:

1.1  A chemical equation is simply a statement of chemical change using chemical symbols. For example, when sulfur is burned in air (one of my favorite chemical reactions, the sulfur is combining with the oxygen in the air to produce an oxide. Let us have a look at this reaction in the form of a chemical equation:

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

1.2  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?

1.3  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.

1.4  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 when combined with one molecular weight of oxygen will 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+O2 -- - SO2

1.5  Let us examine the reaction in some detail.

1.5.1  S - sulfur's atomic weight is about 32.1, so use 32.1 g of this substance + O2 - the atomic weight of O is about 16, the molecular weight of O2 is 32, so 32 g of oxygen would be used) -- - > SO2 (the molar mass is equal to the sum of the previous substances (S + O) moles; a mole of SO2 weighs 64.1 grams).

1.6  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.

2  Let's look at another chemical oxidation reaction, in this case the marvelous reaction of the oxidation of magnesium. When magnesium is oxidized, this reaction occurs:

Mg (magnesium) + O2 (oxygen) -- - > Mg O (magnesium oxide).

2.1  What about the Law of Conservation of Matter? It seems that some of the oxygen was destroyed. This equation is not balanced, and 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 Mg O to make it Mg O2. But, the Law of Definate Proportions would be violated, because magnesium oxide always has the formula Mg O.

When balancing equations, the subscript in the formula may not be changed.

2.2  A skeleton equation is balanced by placing numbers, called coefficients, in front of the formulas of all of the substances in the reaction. Look again at the skeleton equation. By placing the coefficient 2 before Mg O, there are now two oxygen atoms on each side of the equation. The coefficient multiplies all the symbols in the formula immediately after it. This would change the equation to:

Mg + O2 -- - > 2 Mg O.

2.3  However, when looking at the previous equation, there is too much Mg in 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, yeilding the following equation:

2 Mg + O2 -- - > 2 Mg O.

2.4  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."

2.5  The next expression shows how the weights of each of the substances in the balanced equation are indicated:

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

2.6  So, 48.62 g of magnesium combine with 32 g of oxygen to form 80.62 g of magnesium oxide.¹ This weight relationship states 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. Now it will be easy to convert mass into percentage ratios (read on..).

2.7  Let's look at yet another example. Naphthalene (formula C10 H8) 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 H2O. The skeleton equation is:

C10 H8 + O2 -- - > C O2 + H2 O.

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

2.8.1  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 coefficients placed, becomes:

C10 H8 + x O2 -- - > 10 C O2 + 4 H2 O.

2.8.2  Next, we must determine how many oxygen atoms are needed by the hydrogen to completely oxidize to H2 O. 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). The equation becomes:

1 C10 H8 + 12 O2 -- - > 10 C O2 + 4 H2 O.

2.8.3  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.

3  The characteristics of a balanced equation is summarized:

1. A balanced equation obeys the Law of Conservation of Matter.
2. A balanced equation obeys the Law of Definite Proportions.
3. A balanced equations coefficients give the molar proportions of reactants and products involved in the reaction.

• Converting Balanced Equations into Percentage Ratios
• Stoichiometric pyrotechnic compositions are basically balanced equations that are converted into percent ratios from the use of the substance's molecular weights (moles).

3.1  An example of ordinary flash powder:

3 K Cl O4 + 8 Al -- - > 3 K Cl + 4 Al2O3

3.2  To convert this into a percentage ratio, you need to first do some math with the molecular weights:

• 1 mole of K Cl O4 = 138.55
• 138.55 x 3 = 415.65 so, 3 moles of K Cl O4 = 415.65
• 1 mole of Al = 26.98
• 26.98 x 8 = 215.84 so, 8 moles of Al = 215.84
• Next, add the two numbers you obtained together:
• 415.65 [3 moles of K Cl O4] + 215.84 [8 moles of Al] = 631.49

3.2.1  Here is an example of that will determine how much Al will be present:

• First, we substitute our values:

• Next, complete the equation:

• Finally, the percentage obtained (34.17948%) can be used to find the percentage of K Cl O4 present by subtracting it from 100:

Footnotes:

3.3  These units of weight may be grams, kilograms, ounces, pounds, grains, tons, tonnes, etc., just so all three weights are expressed in the same units.