Optimizing Gunpowder Composition - ChemPRIME

Optimizing Gunpowder Composition

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Stoichiometry of Black Powder Explosions

Throughout history, black powder has been made by blending 68-75% potassium nitrate (KNO3, known as saltpeter or saltpetre), 15-20% softwood charcoal , and 7-12% sulfur (elemental S). Charcoal is made by heating wood with limited air, and is mostly carbon (elemental C), but contains traces of other substances.

Which composition is the best?

As we will see in The Limiting Reagent in Forensics, there is no equation that accurately describes the combustion of black powder under all conditions, because the products, as well as the reactants, are numerous and varied.

A modern black powder substitute for muzzleloading rifles in FFG size[1]

But we can write a simplified equation[2]:

2 KNO3 + S + 3 C → K2S + N2 + 3 CO2     (1)


We can use the balanced chemical equation to calculate the optimal ratio of reactants. This will tell us how much of the solid and gaseous products will form. Solid products include like K2S in this case, but also K2CO3 under some conditions (which would require other equations). The solid products are detected in gunshot residue (GSR) test kits. The kits contain an adhesive coated surface that is dabbed on the suspect, and then viewed with an electron microscope.

Optimizing the amount of heat and gaseous substances produced (like N2 and CO2) makes the explosion more powerful.

Scanning Tunneling Microscopic (STM) Image of fine black gunpowder[3]
Scanning Tunneling Microscopic (STM) Image of modern Smokeless gunpowder[4]

The equation

2 KNO3 + S + 3 C → K2S + N2 + 3 CO2     (1)

not only tells how many atoms, molecules, or "formula units"[5] of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. Equation (1) says that 2 KNO3 formula units can react with 1 S atom and 3 C atoms to give 1 K2S formula unit, 1 N2 molecule, and 3 CO molecules. It also says that 2 mol KNO3 would react with 1 mol S and 3 mol C, yielding 1 mol K2S, 1 mol N2 and 3 mol CO.

The balanced equation does more than this, though. It also tells us that 2 × 2 = 4 mol KNO3 will react with 2 × 3 = 6 mol S, and that ½ × 2 = 1 mol KNO3 requires only ½ × 3 = 1.5 mol C. In other words, the equation indicates that exactly 1 C must react for every 2 mol KNO3 consumed. For the purpose of calculating how much C is required to react with a certain amount of KNO3 therefore, the significant information contained in Eq. (1) is the ratio


\frac{\text{3 mol C}}{\text{2 mol KNO}_{\text{3}}}


We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Eq. (1),


\text{S}\left( \frac{\text{C}}{\text{KNO}_{\text{3}}} \right)=\frac{\text{3 mol C}}{\text{2 mol KNO}_{\text{3}}}~~~(2)

The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.



Contents

EXAMPLE 1

Derive other possible stoichiometric ratios from Eq. (1)


Solution Any ratio of amounts of substance given by coefficients in the equation may be used:

\text{S}\left( \frac{\text{KNO}_{3}}{\text{S}} \right)=\frac{\text{2 mol KNO}_{3}}{\text{1 mol S}}~ ~ ~ ~ ~\text{S}\left( \frac{\text{KNO}_{3}}{\text{C}} \right)=\frac{\text{2 mol KNO}_{\text{3}}}{\text{3 mol C}}


\text{S}\left( \frac{\text{KNO}_{3}}{\text{N}_{2}} \right)=\frac{\text{2 mol KNO}_{3}}{\text{1 mol N}_{2}}~ ~ ~ ~ ~\text{S}\left( \frac{\text{KNO}_{\text{3}}}{\text{K}_{\text{2}}\text{S}} \right)=\frac{\text{2 mol KNO}_{\text{3}}}{\text{1 mol K}_{\text{2}}\text{O}}


\text{S}\left( \frac{\text{N}_{2}}{\text{K}_{\text{2}}\text{S}} \right)=\frac{\text{1 mol N}_{2}}{\text{1 mol K}_{\text{2}}\text{S}}~ ~ ~ ~ ~\text{S}\left( \frac{\text{C}}{\text{CO}_{\text{2}}} \right)=\frac{\text{3 mol C}}{\text{3 mol CO}_{\text{2}}}


There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Eq. (2) gives one of them.]



When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Eq. (1) as an example, this means that the ratio of the amount of KNO3 consumed to the amount of C consumed must be the stoichiometric ratio S(KNO3/C):


\frac{n_{\text{KNO}_{3}}\text{ consumed}}{n_{\text{C}}\text{ consumed}}~ ~=~ ~\text{S}\left( \frac{\text{KNO}_{3}}{\text{C}} \right)=\frac{\text{2 mol KNO}_{3}}{\text{3 mol C}}


Similarly, the ratio of the amount of K2S produced to the amount of KNO3 consumed must be

S(K2S/KNO3):


\frac{n_{\text{K}_{\text{2}}\text{S produced}}}{n_{\text{KNO}_{\text{3}}\text{ consumed}}}=\text{S}\left( \frac{\text{K}_{\text{2}}\text{S}}{\text{KNO}_{3}} \right)=\frac{\text{1 mol K}_{\text{2}}\text{S}}{\text{2 mol KNO}_{3}}


In general we can say that


\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}~ ~ ~ ~ ~(3a)


or, in symbols,


\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}~ ~ ~ ~ ~ ~ (3b)


Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios.



EXAMPLE 2

Find the amount of CO2 that will be produced when 3.68 mol KNO3 is consumed according to Eq. (1).


Solution The amount of CO2 produced must be in the stoichiometric ratio S(CO2/KNO3) to the amount of potassium nitrate consumed:


\text{S}\left( \frac{\text{CO}_{2}}{\text{KNO}_{\text{3}}} \right)=\frac{n_{\text{CO}_{2}}\text{ produced}}{n_{\text{KNO}_{\text{3}}}\text{ consumed}}


Multiplying both sides by nKNO3 consumed, by we have


n_{\text{KNO}_{\text{3}}\text{ consumed}}\times \text{S}\left( \frac{\text{CO}_{2}}{\text{KNO}_{\text{3}}} \right)~=~n_{\text{CO}_{2}}\text{ produced}~=~\text{3}\text{.68 mol KNO}_{\text{3}}\times \frac{\text{3 mol CO}_{2}}{\text{2 mol KNO}_{\text{3}}}=\text{5.52 mol CO}_{2}



This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form


\text{amount of X consumed or produced}\overset{\begin{smallmatrix} 
 \text{stoichiometric} \\ 
 \text{   ratio X/Y} 
\end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}


or symbolically.


                                      n_{\text{X   consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y   consumed or produced}}

When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol KNO3 cancels 1 mol KNO3 but does not cancel 1 mol CO2.

The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.



EXAMPLE 3

Calculate the masses of C and S that will react with 75 g of KNO3 according to equation (1), and the masses of all products that will form. This will solve the problem of which recipe for gunpowder is optimal, assuming that equation (1) is an accurate representation of the reaction under the conditions of interest.


Solution The problem asks that we calculate the masses of reactants and products. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of any substance to its mass. So we can calculate the amount of KNO3 available:

n_{{KNO}_{3}}= \frac{m_{\text{KNO}_{3}}}{M_{\text{KNO}_{3}}}~=~\frac{75 g}{\frac{101.1 g}{mol ~KNO_{3}}} =~0.742~ mol~ KNO_{3}

Now we need to calculate the amounts of reactants and products from the amount of KNO3 consumed. This is similar to Example 2. It requires stoichiometric ratios like


\text{S}\left( \frac{\text{C}}{\text{KNO}_{\text{3}}} \right)=\frac{\text{3 mol C}}{\text{2 mol KNO}_{\text{3}}}


The amount of C consumed is then

n_{\text{C}}\text{ consumed} ~=~ n_{\text{KNO}_{\text{3}}}\text{ consumed}~ ~\times~\text{  conversion factor}

 ~=~ \text{0.742 mol KNO}_{\text{3}}\times \frac{\text{3 mol C}}{\text{2 mol KNO}_{\text{3}}}=\text{1.11 mol C}


The mass of C is


\text{m}_{\text{C}}=\text{1.11 mol C}\times \frac{\text{12}\text{.01 g C}}{\text{1 mol C}}~=~\text{13.4 g C}


With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O2 to moles of SO2 and the molar mass will convert moles of SO2 to grams of SO2. A schematic road map for the one-step calculation can be written as


n_{\text{KNO}_{\text{3}}} ~\xrightarrow{S\text{(C}\text{/KNO}_{\text{3}}\text{)}}  ~ n_{\text{C}} ~ \xrightarrow{M_{\text{C}}} ~ m_{\text{C}}


Thus


\text{m}_C ~=~ \text{0.742 mol KNO}_{\text{3}} ~ \times ~\frac{\text{3 mol C}}{\text{2 mol KNO}_{\text{3}}} ~\times ~\frac{\text{12.01 g}}{\text{1 mol C}}=\text{13.4 g}

These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation.

2 KNO3 + S +3 C → K2S + N2 + 3 CO2
m (g) 75.0 11.9 13.4 40.9 10.4 48.9
M (g/mol) 101.1 32.1 12.01 110.3 28.01 44.01
n (mol) 0.742 0.371 1.11 0.371 0.371 1.11

The masses of products are calculated in a similar way, for example:

\text{m}_{CO_{2}} ~=~ \text{0.742 mol KNO}_{\text{3}} ~ \times ~\frac{\text{3 mol CO}_{2}}{\text{2 mol KNO}_{\text{3}}} ~\times ~\frac{\text{44.01 g}}{\text{1 mol CO}_{2}}=\text{48.9 g}




EXAMPLE 4

What mass of oxygen would be consumed when 3.3 × 1015 g, 3.3 Pg (petagrams), of octane (C8H18) is burned to produce CO2 and H2O?


Solution First, write a balanced equation


2C8H18 + 25O2 → 16CO2 + 18H2O


The problem gives the mass of C8H18 burned and asks for the mass of O2 required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O2 consumed. Finally, the molar mass of O2 permits calculation of the mass of O2. Symbolically


m_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{M_{\text{C}_{\text{8}}\text{H}_{\text{18}}}}\text{ }n_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/C}_{\text{8}}\text{H}_{\text{18}}\text{)}}\text{ }n_{\text{O}_{\text{2}}}\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}}


m_{\text{O}_{\text{2}}}=\text{3}\text{.3 }\times \text{ 10}^{\text{15}}\text{ g }\times \text{ }\frac{\text{1 mol C}_{\text{8}}\text{H}_{\text{18}}}{\text{114 g}}\text{ }\times \text{ }\frac{\text{25 mol O}_{\text{2}}}{\text{2 mol C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\times \text{ }\frac{\text{32}\text{.00 g}}{\text{1 mol O}_{\text{2}}}=\text{1}\text{.2 }\times \text{ 10}^{\text{16}}\text{ g }


Thus 12 Pg (petagrams) of O2 would be needed.



The large mass of oxygen obtained in this example is an estimate of how much O2 is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 1021 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O2. Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen.



References

  1. http://en.wikipedia.org/wiki/Gunpowder
  2. http://en.wikipedia.org/wiki/Gunpowder
  3. http://www.state.nj.us/njsp/divorg/invest/criminalistics.html
  4. http://www.state.nj.us/njsp/divorg/invest/criminalistics.html
  5. "formula units" is the term used to describe the compostion of ionic compounds which are not made up of molecules.
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