 # Understanding Chemical Equations And Formulae

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### #1 PyroNitrate

PyroNitrate

Posted 15 August 2003 - 04:52 AM

I found this very helpful being a noob. Here it is for all you to read

Balanced Equations, or Stoichiometry:
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.

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). 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.

### #2 Stuart

Stuart

Posted 15 August 2003 - 08:11 AM

Good find. Mabye it would be better to put it into the User Guide section fro other n00bs.

### #3 Richard H

Richard H

Posted 15 August 2003 - 12:32 PM

I will pin in this forum so it always appears 'at the top'. I agree that this post is useful for beginners.

### #4 karlfoxman

karlfoxman

Resident Maltese shell builder

• General Public Members
•    • 1,139 posts

Posted 14 July 2005 - 09:10 AM

just found this, it fairly simple but may help newbies understand a little more about the reactions taking place. remove if you feel its a waste...

http://www.chemsoc.o...omposition.html

k

### #5 Phoenix

Phoenix

Posted 17 July 2005 - 09:06 PM

I get the impression that the site's author doesn't really have much background in firework making. The site is full of small technical inaccuracies and incorrectly spelled firework terms. It's not dangerous, just mildly amusing.

### #6 Shake

Shake

Posted 22 July 2008 - 09:44 AM

Time to pull down my pin up girl poster and hang my periodic table in the garage ....or is that going to look too mad scientist-ish?
Fireworks are paint brushes for the night sky...with a few bloody huge BOOMS thrown in for good measure!

### #7 Arthur Brown

Arthur Brown

Posted 22 July 2008 - 05:12 PM

Excellent guide! However in pyro it's not always the simple reactions that are important, and sometimes the reactions are pressure and temperature dependent. Some producet especially stars are under oxidised and only function correctly when at speed in air.
http://www.movember.com/uk/home/

Keep mannequins and watermelons away from fireworks..they always get hurt..

### #8 Potassium chlorate

Potassium chlorate

Posted 05 February 2009 - 06:38 PM

The stochiometric formula isn't necessarily the formula in reality. There is, alas, a lot of trial and error.
"This salt, formerly called hyperoxymuriate of potassa, is
used for sundry preparations, and especially for experimental
fire-works."

Dr. James Cutbush

### #9 Guest_anyka_*

Guest_anyka_*
• • Guests

Posted 24 August 2010 - 12:57 PM

The stochiometric formula isn't necessarily the formula in reality. There is, alas, a lot of trial and error.

So true!
i.e
Aluminum powder and potassium perchlorate are basically the 2 only components of the pyro industry flash powder. It has stability and power but look here at the balanced equation really is The balanced equation for the reaction is:

3 KClO4 + 8 Al → 4 Al2O3 + 3 KCl
Yet almost all pryrotechnicians use a 70/30 mixture.

However, a ratio of 2 mass units potassium perchlorate to 1 mass unit Dark Pyro Aluminum is closer to stoichiometric, and would surely produce a louder report.

### #10 Atom Fireworks

Atom Fireworks

Posted 24 August 2010 - 03:29 PM

So true!
i.e
Aluminum powder and potassium perchlorate are basically the 2 only components of the pyro industry flash powder. It has stability and power but look here at the balanced equation really is The balanced equation for the reaction is:

3 KClO4 + 8 Al → 4 Al2O3 + 3 KCl
Yet almost all pryrotechnicians use a 70/30 mixture.

However, a ratio of 2 mass units potassium perchlorate to 1 mass unit Dark Pyro Aluminum is closer to stoichiometric, and would surely produce a louder report.

It does, infact it makes it so much sharper and the percussion from this mix is unreal, NOT FOR Non-experienced pyros.

### #11 alany

alany

Posted 24 August 2010 - 03:32 PM

I disagree that stoichiometric flash would necessarily produce louder reports. It may, or may not? I am unsure how much study has been performed on optimal flash report compositions. In fact I am curious just how much we really know about how something as simple as binary flash actually works.

Reports are produced by motion of air, the flash reaction is ultimately "gasless" if truly stoichiometric, so I strongly suspect air inside the composition and unreacted oxygen from the reaction is important as working fluid for causing the explosion and associated report. The exact dynamics are probably quite complicated, but one could consider the Aluminium as a high energy fuel that propagates the decomposition of the oxidiser and heating of the intermediate reaction products. The practical salute devices probably disassemble before the reaction is complete. Part of the reaction occurs in the liquid phase, the oxidiser melts prior to decomposition and liberation of oxygen. Does oxygen nucleate out as bubbles? Just what happens to the alumina produced? Does the Aluminium melt and change physical form before/during it's oxidation? Does the potassium chloride flux the Aluminium and help prevent passivation, speeding the reaction? Are the concepts of liquid and gas even valid at the nanoscale local environment of the reaction? How important is optical transport of energy through the composition? How does DDT occur? Actual adiabatic shock formation in flash, does it really happen?

I think it is highly simplistic to assume the deflagration dynamics are not highly dependent on device geometry and physical form of the composition reactants.

I only wish I had several lifetimes and an unlimited budget... ### #12 Potassium chlorate

Potassium chlorate

Posted 01 September 2010 - 05:37 PM

I disagree that stoichiometric flash would necessarily produce louder reports. It may, or may not? I am unsure how much study has been performed on optimal flash report compositions. In fact I am curious just how much we really know about how something as simple as binary flash actually works.

Reports are produced by motion of air, the flash reaction is ultimately "gasless" if truly stoichiometric, so I strongly suspect air inside the composition and unreacted oxygen from the reaction is important as working fluid for causing the explosion and associated report. The exact dynamics are probably quite complicated, but one could consider the Aluminium as a high energy fuel that propagates the decomposition of the oxidiser and heating of the intermediate reaction products. The practical salute devices probably disassemble before the reaction is complete. Part of the reaction occurs in the liquid phase, the oxidiser melts prior to decomposition and liberation of oxygen. Does oxygen nucleate out as bubbles? Just what happens to the alumina produced? Does the Aluminium melt and change physical form before/during it's oxidation? Does the potassium chloride flux the Aluminium and help prevent passivation, speeding the reaction? Are the concepts of liquid and gas even valid at the nanoscale local environment of the reaction? How important is optical transport of energy through the composition? How does DDT occur? Actual adiabatic shock formation in flash, does it really happen?

I think it is highly simplistic to assume the deflagration dynamics are not highly dependent on device geometry and physical form of the composition reactants.

I only wish I had several lifetimes and an unlimited budget... Flash as a subject of its own is a bit of a no-no on this board. However, when I was younger I was pretty into flash, so I have some experience with it.

The most powerful binary flash is probably chlorate/magnalium. Both potassium chlorate and barium chlorate make tremendous bangs in a 3:2 or 1:1 ratio with magnalium, though both are probably very dangerous, especially the barium chlorate one.

Next to that comes chlorate and German Black/Dark aluminium or very fine magnesium. These are also dangerous but not as dangerous as the magnalium variety. Here the ratio should be 70:30 also stochiometrically.

With perchlorate and German Black/Dark the best is the already mentioned 66:34 or 2:1. This one can be very much improved with sulfur, in which case the ratio would be:

64% potassium perchlorate
23% dark aluminium
13% sulfur.

Though you could probably make some kind of "superflash" by adding a catalyst like potassium dichromate or manganese dioxide to the binary mixture. A little touch of zirkonium might also help.Not more than +5% though, I think.

Edited by Pyroswede, 01 September 2010 - 05:37 PM.

"This salt, formerly called hyperoxymuriate of potassa, is
used for sundry preparations, and especially for experimental
fire-works."

Dr. James Cutbush

### #13 rick87

rick87

Posted 22 September 2010 - 09:03 PM

just like to say the pinned threads across the whole of this forum are immense.

### #14 LanaSanchez

LanaSanchez

Posted 17 May 2014 - 02:46 PM

Very well! i found this post helpful. yeah! that was a good decision to pin this post.

### #15 Crazy Cat

Crazy Cat

Posted 10 June 2014 - 02:37 AM

This site, BALANCE CHEMICAL EQUATION - ONLINE BALANCER. http://www.webqc.org/balance.php

Example: KNO3 + C + S gives The equation is not complete (products are missing). Please either complete your equation or click on one of the suggestions below: KNO3 + C + S = CO2 + K2S + N2

Clicking on http://www.webqc.org...+C+S=CO2+K2S+N2

KNO3 + C + S = CO2 + K2S + N2 gives Balanced equation: 2 KNO3 + 3 C + S = 3 CO2 + K2S + N2

Two things are infinite: the universe and human stupidity; and I'm not sure about the universe. ― Albert Einstein ― Insanity is doing the same thing, over and over again, but expecting different results.

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