jadesource Posted April 9, 2006 Share Posted April 9, 2006 I need to know the oxygen persentage in potassium perchlorate .What Im really hoping to do is switch out some of the potassium perchlorate and add strontium nitrate in winoker #1 to get a more red glitter instead of just pink .I now from the skylighter article #71 about red rockets that strontium nitrate is 45.36% oxygen by moleculer weight so I need to know the percentage to keep the oxygen balanced . Or at least thats the angle Im shooting for to reduce the amount of experiments and wasted chemicals in my efforts thank you Bill Link to comment Share on other sites More sharing options...
BigBang Posted April 9, 2006 Share Posted April 9, 2006 I would like to just note that there isn't KClO4 in winokur #1. I am actaully testing this out, but I don't believe #1 is the best one to modify.... Link to comment Share on other sites More sharing options...
aquaman Posted April 9, 2006 Share Posted April 9, 2006 In a book that I have it states that Sr(NO3)2 has 18.9 (wt %) oxygen availiable where KClO4 has 46.2 (wt %) oxygen available. Link to comment Share on other sites More sharing options...
jadesource Posted April 9, 2006 Author Share Posted April 9, 2006 MY BAD potassium nitrate is the one I need not potassium perchlorate Im doing to many things at once . I have two comps drying one ready to mix on the table . And im mixing them up in my mind . Sucks getting old. Big bang I would love to now were you experiments have takin you .Also which glitter comp would you try and change thanks Bill Link to comment Share on other sites More sharing options...
Mumbles Posted April 9, 2006 Share Posted April 9, 2006 You guys need to learn chemistry. This shouldnt even be a question. I included potassium nitrate as it is in Win #1, and not perchlorate. The forumla, if you can even call it that, is also included so I don't have to see such dumbness in the future. Strontium Nitrate - 45.4% Potassium Perchlorate - 46.2% Potassium Nitrate - 47.5% Formula:(16 x number of oxygens) / (Molecular weight of the compound) Link to comment Share on other sites More sharing options...
Givat Posted April 9, 2006 Share Posted April 9, 2006 Just from curiosity, your talking about the mass of oxigen in the molecule of an oxidizer. But most of the oxidizers don't give all of there oxygen as "free oxygen" for combustion. So how this info helps you? Link to comment Share on other sites More sharing options...
Mumbles Posted April 9, 2006 Share Posted April 9, 2006 Thats true. He didn't ask about oxygen contribution, he asked about oxygen percentage. I'm not going to go out of my way for someone who can't figure something so simple out. Link to comment Share on other sites More sharing options...
Givat Posted April 9, 2006 Share Posted April 9, 2006 OK, so lets try to promote this topic. Is there any way to calculate the oxygen contribution of an oxidizer? I guess the answer will be no, and you can know this only from experiments. So if this the answer - Do you have some information about oxygen contribution of common used oxidizers in pyro? Link to comment Share on other sites More sharing options...
Rogue Chemist Posted April 9, 2006 Share Posted April 9, 2006 I could be wrong on this, but assuming that something like2KNO3--> K2O + N2 + 2.5O2 the ammount of usable oxygen could be calculated. KClO4 --> KCl+2O2 Of course, these may not be the exact reactions, pyrotechnic formulas produce many more products than you would think, and to ratio of products formed is more complex than simple stoich, and very variable dependign on conditions. Link to comment Share on other sites More sharing options...
Mumbles Posted April 10, 2006 Share Posted April 10, 2006 Another thing is what it is mixed with. In BP for instance, it primarily gives K2CO3, and K2S over K2O. For sake of calculation ease, just assume all the oxygen is given up. Link to comment Share on other sites More sharing options...
jadesource Posted April 10, 2006 Author Share Posted April 10, 2006 My apoligies mumbles im a newby and I made a mistake in my post Ive already googled for the answer didnt get squat . It was my first post ive been learning great things from you guys .Sorry I asked but you did still answer my question so I can start from there Thanks Bill PS. I will post in the newbie section next time Link to comment Share on other sites More sharing options...
Recommended Posts