OK i am setting up a new tank once again and usually when i am doing a fishless cycle i just keep adding and testing ammonia levels until i reach my magic number of around 4ppm. So this time i decided I would try and figure out exactly how much ammonia it would take to get my 130g system up to 4ppm. I have checked and estimated the 130g to be total system volume including tank and sump minus all decorations and media's. -----------So this is what i have come up with. To figure parts per million wiki says you take the substance to be calculated (ammonia) and divide that into our total system volume. Then multiply that by 1million. Here is there example for calculating salt in water in gallons of each http://wiki.answers.com/Q/What_is_parts_per_million. Now we can not use gallons cause a gallon of ammonia in a 130g system is about 7692ppm am the fumes from that might kill you over time. lol-------So we need milliliters which is what is used in all the test kits. So i found that 1 gallon is 128.52 fluid ounces and 1 ml is .03 fluid ounce. And that means a tank volume of 130g is 16707.6 fluid ounces. And if you do our simple math .03 (1ml in fl oz) divided by 16707.6 (130g in fl oz) = .000001796 then multiply that by 1,000,000 (as in PPM) you get 1.7955 ppm. That means 1ml of ammonia will equal 1.8ppm in a 130g system. ------------------------------------------------------------------------------------------------------------------I know that was long winded and I hope no one fell a sleep while reading it. But please check over my math and or if you know a easier way to calculate PPM figures that are useful for us post em up. And if this is right maybe a extremely cut down version can be added to our cycling sticky. Well i just checked all the threads listed on the cycle sticky say no matches found so maybe we need a new one?
POP QUIZ-- how to figure PPM
- Thread starter hybridtheoryd16
- Start date
Parts per million calculations work by masses. You need to figure out total masses of each item present. Since you're considering ammonia parts per million in water, you need the total mass of water present and the total mass of ammonia present. Divide the total mass of ammonia present by the total mass of water present, multiply by 10^6 and you should get ppm.
ppm = [ {mass of solute} / {mass of solvent} ] * 10^6
Water = 18g/mole
Ammonia = 17 g/mole
6.022*10^23 molecules/mole (a mole is the required number of molecules of any element to have a weight equal in magnitude to its atomic weight with units of grams. Example: 1 mole of Oxygen = 6.022*10^23 molecules and weighs 15.99g)
1 g = 0.0022046 lb
Mass of water:
Mass = [Water volume] * [Water density]
Water density = 9807 N/m^3 or 62.4 lb/ft^3 or 8.345 lb/gal.
Ammonia required calculation:
Volume = {Ammonia density} / {Ammonia mass required}
Ammonia density = 0.00073g/mL or 0.0456 lb/ft^3
You can take the desired value of ammonia in ppm and put it on the left (of the first equation) and solve for the mass of ammonia on the right.
Keep in mind that your ammonia solution in the bottle is diluted. You'll need to measure out the appropriate volume of solution to acquire the desired amount of ammonia.
Helpful tip: make sure you write out EVERY unit with your calculations. Units will be your friends in these kinds of calculations. If you don't write out the units, these numbers will positively kick your butt (similar to mine in chemistry class and environmental engineering systems class)
I hope this helps more than it hurts!
By the way, I looked at your numbers. Your math works well, however, I think 1gal = 128 fl. oz. If you rework with your method (which works very well), I find:
Tank volume = 16640 fl. oz. (minus your decor and substrate, maybe 12 gal), 15104 fl. oz.
4 ppm ammonia yields 1.787 mL of ammonia
The only error I see is in your conclusion. If I use a tank volume of 16640 fl. oz and a volume of 1 ml of ammonia, I find your concentration is 2.03 ppm. If I use my tank volume, I find the concentration to be 2.238 ppm. Using the reduced tank volume, I figure you need 1.787 mL of ammonia to bring you up to 4 ppm ammonia.
ppm = [ {mass of solute} / {mass of solvent} ] * 10^6
Water = 18g/mole
Ammonia = 17 g/mole
6.022*10^23 molecules/mole (a mole is the required number of molecules of any element to have a weight equal in magnitude to its atomic weight with units of grams. Example: 1 mole of Oxygen = 6.022*10^23 molecules and weighs 15.99g)
1 g = 0.0022046 lb
Mass of water:
Mass = [Water volume] * [Water density]
Water density = 9807 N/m^3 or 62.4 lb/ft^3 or 8.345 lb/gal.
Ammonia required calculation:
Volume = {Ammonia density} / {Ammonia mass required}
Ammonia density = 0.00073g/mL or 0.0456 lb/ft^3
You can take the desired value of ammonia in ppm and put it on the left (of the first equation) and solve for the mass of ammonia on the right.
Keep in mind that your ammonia solution in the bottle is diluted. You'll need to measure out the appropriate volume of solution to acquire the desired amount of ammonia.
Helpful tip: make sure you write out EVERY unit with your calculations. Units will be your friends in these kinds of calculations. If you don't write out the units, these numbers will positively kick your butt (similar to mine in chemistry class and environmental engineering systems class)
I hope this helps more than it hurts!
By the way, I looked at your numbers. Your math works well, however, I think 1gal = 128 fl. oz. If you rework with your method (which works very well), I find:
Tank volume = 16640 fl. oz. (minus your decor and substrate, maybe 12 gal), 15104 fl. oz.
4 ppm ammonia yields 1.787 mL of ammonia
The only error I see is in your conclusion. If I use a tank volume of 16640 fl. oz and a volume of 1 ml of ammonia, I find your concentration is 2.03 ppm. If I use my tank volume, I find the concentration to be 2.238 ppm. Using the reduced tank volume, I figure you need 1.787 mL of ammonia to bring you up to 4 ppm ammonia.
Ok, like I said, please disregard the garbage I posted above. I have consciously reworked the problem this morning and have come up with reasonable numbers.
The error I spotted that was in both your work and mine last night is as follows: Fluid ounces are a measurement of volume, not mass. So, as simple as it was to do that calculation, it doesn't quite work. Plus, I made a few other sloppy errors in the calculations. Here's my corrected version:
Water volume = 130 gal.
Water density = 8.345 lb/gal.
Water weight = 1084.85 lb = 492084.7 g
Ammonia density (liquid) = 681.9 kg/m^3 = 0.6819g/mL
With
{ [1g NH3] / [492084.7g H20] } *10^6 = 2.032ppm Ammonia
Now, 1g of ammonia:
[1g] / [0.6819 g/mL] = 1.47mL ammonia ~ 1.50mL ammonia.
4 ppm Ammonia (within 130 gal) yields:
4 = { [Xg NH3] / [492084.7g H20] } *10^6
x = 1.9683g NH3, which is 2.887mL ~ 3mL ammonia.
Recheck my work, this kind of chemistry type stuff has always been my weakest area.
I hope THIS helps, unlike my gibberish above.
The error I spotted that was in both your work and mine last night is as follows: Fluid ounces are a measurement of volume, not mass. So, as simple as it was to do that calculation, it doesn't quite work. Plus, I made a few other sloppy errors in the calculations. Here's my corrected version:
Water volume = 130 gal.
Water density = 8.345 lb/gal.
Water weight = 1084.85 lb = 492084.7 g
Ammonia density (liquid) = 681.9 kg/m^3 = 0.6819g/mL
With
{ [1g NH3] / [492084.7g H20] } *10^6 = 2.032ppm Ammonia
Now, 1g of ammonia:
[1g] / [0.6819 g/mL] = 1.47mL ammonia ~ 1.50mL ammonia.
4 ppm Ammonia (within 130 gal) yields:
4 = { [Xg NH3] / [492084.7g H20] } *10^6
x = 1.9683g NH3, which is 2.887mL ~ 3mL ammonia.
Recheck my work, this kind of chemistry type stuff has always been my weakest area.
I hope THIS helps, unlike my gibberish above.
There is only one problem with the whole idea that we both missed. And unlike all the math i threw out there this is simple. The ammonia we can get ahold of is diluted and not 100%. So you would have to now the percentage of pure ammonia to even start a correct calculation. Its like the teachers allways said, you have to now where you are to now how to get to a destination. I did not---------sorry for making you go thru all that.
Not a problem. I actually looked at my ammonia bottle when I got through with all of that to see what the ammonia concentration was.... and I came up empty handed like you! Just get a little syringe from the pharmacy (like is used for insulin injections) that has graduations of 0.1mL, going up to 1.0 mL. Put it in a little at a time and just check it periodically!