Useful Information

                                              General Solubility Information

The following is a listing of available materials that are (among other things) used as a binder in pyrotechnic type formulations. Those so marked with " * " preceding the item indicate that the material so used in solution, that is to say, it is dissolved in the appropriate solvent before mixing into the formulae. It should be noted that all of these materials may be used in solution but those so noted must be dissolved first because of their more difficult to solvise properties. Do not allow complete solvation before evaporation and drying the mixture resulting in adhesive bond failure. Some of these materials may take as long as 24 hours at room temperature to completely dissolve.
If maximum strength is required, binder solutions should be used instead of moistening the dry mixture containing the binder. A solution allows a more thorough, homogenous mixing with adequate wet-out and a cured strength many times greater than the same binder when used dry.
Those so indicated with "+" also have good moistureproofing properties (some better than others) because they are impervious to moisture themselves and seal the ingredients in the formula that are hygroscopic from the air and moisture they may otherwise absorb from it.

PREPARING PERCENTAGE STRENGTH SOLUTIONS

Whenever a formulation calls for a chemical solution of a particular strength, the below listed table exemplifies an easy way to accomplish it. As depicted, the table shows how much of a dry substance to mix with the liquid vehicle to obtain a solution of the desired or required percentage strength. These percentages, of course, depend upon the dry chemical agent dissolving completely in the liquid. If by chance a percentage of the dry chemical does not dissolve into the liquid, the strength of the solution would be reduced accordingly. There is an upper limit to be considered and that limit is the actual amount of any chemical or substance that will dissolve into a pre-determined amount of the appropriate liquid. Therefore, solutions are usually of low percentage strength. Concentrated solutions can be obtained only when the solutions’ upper limit has been met but not exceeded and are often called "slurries". Exceeding the upper solvation limit can be expressed as "solids loadings" at one end and "putties" at the other. Paints and epoxies can be included in this example.

A good source of information about the dry chemical in question can be obtained from the spec sheet or “MSDS” supplied with the chemical. Yours truly uses the Merck Index, an encyclopedia of chemicals, drugs and biologicals. In it can be found all the technical information about a particular chemical including its solvents, rate of solvation and at what temperature(s).

“Solutions” are generally referenced as using water as the liquid agent (solvent) unless otherwise specified according to the procedure listed. The table to follow lists the amount of dry chemical that is required to produce a particular solution strength in one pint (16 oz) of liquid. Solution strengths not listed may be obtained by adding various (dry ingredient) strengths listed together.

Note: to obtain true solids loading percentages, subtract the chemical/substance gram weight from the liquid quantity weight. Example: for the 1% solution, the weight of liquid to solid would be 900.11 grams (31.75 oz) to 9.07 grams.

CONVERTING PARTS BY WEIGHT TO PERCENTAGE (100%)

Whenever formulations are listed in “parts”, they can be easily converted into percentage by using the mathematics to follow. No matter what the total “parts” add up to, the formulation can be converted into a total 100% figure to simplify the exact percent each ingredient carries in the total formulation. Expressed in this manner, the formulation becomes easier to understand and change if need be.

The procedure is a simple one. First add up the “parts” to obtain their total, divide 100 by this total to obtain a decimal figure (factor) and multiply this factor by the quantity of parts per each chemical in the formulation. Example;

29 parts        ingredient A
38 parts        ingredient B
72 parts        ingredient C
  2 parts        ingredient D
  3 parts        ingredient E
144 parts total

100 (%) ÷ 144 = factor of .694444444(may be expressed as .694 or .6944)

29 x .694444 = 20.1388%  A
38 x .694444 = 26.3888%  B
72 x .694444 = 49.9999%  C
2 x .694444 =   1.3888%  D
3 x .694444 =   2.0833%  E
Total = 99.9996%

To simplify, the percentages may be rounded off to the nearest quarter / whole number and still be within operational limits in the formulation. For this formulation the percentages will be;
A                  =        20.00%
B                  =        26.50%
C                  =        50.00%
D                  =          1.50%
E                  =          2.00%
Total = 100%

MAKING SOLUTIONS LISTED AS “SOLIDS LOADINGS”

Let’s try the math on an equation when we wish to obtain a particular “solids loading” solution strength. Since many lacquers and resins are offered for sale as a particular strength commonly referred to as “solids loadings”, we can use the math we have previously studied to obtain the exact amount of the appropriate dry substance it will take to produce the desired (solids) strength. For our test, we will list the basic agents for making nitrocellulose lacquer;

MAKING NITROCELLULOSE LACQUER

The following equation uses a factor of .78125 (120 + 8 = 128 / 100(%) ÷ 128 = .78125)
To make one gallon:  

   120 oz acetone = 120 x .78125 = 93.75% solvent
    8 oz dry NC = 8 x .78125 = 6.25 % NC solids

128 oz total = 93.75% acetone, 6.25% NC = 100%

This is the strength most used as a binder when making slurries for primes, electric igniter pyrogens, Thermalite fusing, etc... For use as a binder for pressing smoke compositions, stars, etc., a 3 or 4% solution may be used.

With these figures in mind, if you wish to make one gallon of a concentrated solution of 15.63% solids NC lacquer, use the equation to follow:

The following equation uses a factor of .78125 (108 + 20 = 128 / 100(%) ÷ 128 = .78125
To make one gallon:

108 oz acetone = 108 x .78125 = 84.375% solvent
20 oz dry NC4000 = 20 x .78125 = 15.625% solids
128 oz total = 84.375% acetone, 15.625% NC4000 = 100%

To make one quart, divide these figures by 4:

27 oz acetone
       5 oz dry NC4000

32 oz total =  84.375% acetone, 15.625% NC4000 = 100%

FIGURING PERCENTAGES WHEN A “COMPOSITION”
IS LISTED WITHIN THE FORMULATION

In many of the older fireworks literature, it is not uncommon to find a composition in itself listed with the other ingredients of the formulation. One that comes to mind most often is the use of “meal powder” (fine black powder) included in a formulation (gerbs, drivers, glitter stars, comets, prime compositions, fountains, etc.) for various reasons such as combustion ease, increasing the burn rate or lowering the ignition point of the formulation to achieve various end results. In most of these cases, you are simply adding other ingredients to alter the characteristics of a functioning formulation to achieve different yet specific end results.

This is normally not a problem but with black powder becoming a very difficult and a dear commodity to obtain, the formulation may be broken down chemically with these ingredients included in the formulation individually as a percentage. In most cases, the substitution of the meal powder with its ingredients is acceptable with slight variations in result from the original formulation including meal powder. Commercial made meal powder generally burns more vigorously than the simple mixture of its three ingredients and therefore the alteration may have different but acceptable burn rate and energy end results.

The basic components of black powder may be expressed as follows;

Potassium nitrate    75%
Charcoal                 15%
Sulfur                     10%

The first step is to convert the formulation into percentages. For example, let’s try an improved first fire/prime formulation based on black powder as shown;

Meal powder                     60 parts x .925925925  =  55.55555555%             (55.56%)
Potassium perchlorate      18 parts x .925925925  =  16.66666665%              (16.67%)
Iron oxide, red                  10 parts x .925925925  =    9.259259250%           (  9.30%)
Aluminum, spherical          10 parts x .925925925  =    9.259259250%           (  9.30%)
Red gum                             8 parts x .925925925  =    7.407407407400%      (  7.41%)
Boric acid                           2 parts x .925925925  =     1.851851850%           (  1.85%)

108 parts = a multiplication factor of .925925925

This formulation shows the meal powder as 55.5555555 percent. Since the formula for meal powder is potassium nitrate 75%, charcoal 15% and sulfur 10%, we need to take 55.5555555% of each of the three values then add them back to the formulation expressed as percentages;

55.5555555 x 75 = 41.66666662%    = (round to 41.67)
55.5555555 x 15 =   8.333333325%  = (round to   8.33)
55.5555555 x 10 =   5.55555555%    = (round to   5.56)

 Potassium nitrate         41.67%
Charcoal                        8.33%
 Sulfur                             5.56%
Potassium perchlorate 16.67%
Iron oxide, red, ferric     9.30%
Aluminum, spherical       9.30%
Red gum                       7.41%
Boric acid                      1.85%
100.09% (.09 is from rounding up)

This same procedure may be used to convert other similar “compositions” into active ingredients in a formulation when the composition ingredients and their percentages are known.

FIGURING PERCENTAGES ACCORDING TO
A “SET” QUANTITY OF MATERIAL

Those of us in the business are very familiar with “binary mixtures” (such as A & B) requiring a certain number of material(s)contained in two or more individual packages for safety’s sake. The final mixture of application does not become useable or ”volatile” until all the packages are mixed together for use. Naturally, for ease of use in the field, the binary ingredients in each package are pre-weighed for precise measurement to avoid doing so in the field. Other examples would include applications where a certain number of material(s) are required to complete a set number of devices, such as in a “kit” form. Since the quantity of devices will determine the amount of mixture required to complete them, how do you know how much of each ingredient in the mixture will be required?
                                                                                                                              100 % or parts =    “X”
                                                                                                                              *17       “X”      = ounce
                                                                                                                               oz   x    30      = grams

* The number 17 here represents the total amount of material required to complete the application. This figure will change according to the amount required in each case. We use figure 17 here to demonstrate the equation as listed below.

  Let’s assume a mixture totaling 108 parts;

        Ingredient A                       8.0 parts
Ingredient B                      20.0
Ingredient C                      25.0
Ingredient D                      20.0
Ingredient E                      10 0
Ingredient F                      15.0
Ingredient G                     10.0
       108 parts

Let’s also assume the application will require a total of 17 oz to complete the devices in question. The equation takes form as follows;

17 x (  8 ÷ 108) = 1.259 oz Ingredient A (rounded to 1.26)
17 x (20 ÷ 108) = 3.148 oz Ingredient B (rounded to 3.15)
17 x (25 ÷ 108) = 3.935 oz Ingredient C (rounded to 3.94)
 17 x (20 ÷ 108) = 3.148 oz Ingredient D (rounded to 3.15)
17 x (10 ÷ 108) =  1.574 oz Ingredient E (rounded to 1.57)
17 x (15 ÷ 108) = 2.361 oz Ingredient F (rounded to 2.36)
17 x (10 ÷ 108) = 1.574 oz Ingredient G (rounded to 1.57)

Total 17 oz

Therefore, if the “binary mixture” is to be contained in two separate packages and all the candidates except ingredient A (for example) are to be placed in one package and ingredients B in a separate package for safety’s sake then ingredients B through G are first weighed then mixed and placed in one package and ingredient A is placed in the second package. If the formulation contained more than one oxidizer (ingredients A and B for example) then ingredients A and B are weighed and mixed, and ingredients C through G are weighed and mixed in a separate package. This keeps the oxidizers from the flammables, which in most cases will keep the mixture from reacting as intended until the two packages are properly mixed. This also applies to any ingredients that may react with one another when combined in a mixture which does not necessarily only include the separation of oxidizers and flammables. There are many chemicals that will react violently when they encounter a non-compatible material. Be aware of this when designing formulations. Consult MSDS on each ingredient prior to their use. The MSDS usually list incompatible materials that may or may not affect its use in formulation or react immediately but may require a catalyst such as moisture or heat to set them in motion. Each mixture will react differently depending on the candidates. The cognizant designer will be aware of applicable material(s) (such as stabilizers) that may be able to be included to counter any unwanted reaction.

The math will work for any total percentage or parts, any number of ingredients or any quantity of total required material for the application. Simply make the changes in the equation as needed.

FIGURING PERCENTAGE / WEIGHTS FOR TWO PART
BINARY “A & B” MIXTURES

An example application for this type of math would be mixtures that are sold and purchased without one of the necessary ingredients. An example here would be smoke mixtures and other formulations that are usually supplied without the necessary oxidizer for shipping purposes.

Firefox supplies smoke mixtures without the necessary oxidizer. The ingredients for these mixtures are weighed out and blended without the percentage of oxidizer which must be added by the customer before he/she uses it. Since we supply these mixtures by the pound (16 oz) of dye mix only, the instruction sheet must reflect how much oxidizer to add to the 16 oz of dye mix. Here is one way to accomplish this. First let’s figure the percentages of dye mix and oxidizer;

                            “A” Dye mix          68% (all ingredients except oxidizer)
   “B” Oxidizer          32% (oxidizer total)

Since the dye mixture is supplied in 16 oz containers, we first move the decimal point over one place on each of the two percentages and apply the math as follows;

16 oz (qty of “A”) ÷ 6.8 (% A) = 2.3529 x 3.2 (% B) = 7.529 oz oxidizer

The math shows that 7.529 oz of oxidizer (rounded off to 7.5 oz) is required for 16 oz of smoke dye mix thus making a total of 23.50 oz of finished smoke mixture. Let’s try another one. How about 18 oz of a 50/50 mixture?

“A” Dye mix             50%
“B” Oxidizer             50%

18 oz ÷ 5.0 = 3.6 x 5.0 = 18.0 oz

Legend: a “+” indicates retained on the screen, a “-“ indicates passing through the screen, a “<” indicates less than