# Concentrating H2O2

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Does anyone know how to concentrate hydrogen peroxide further?

I know of these methods:

1. Freezing: requires qutie low temperature, anyone want to explain fractional freezing any further, especially if it would be a good way for H2O2

2. Simple Distillation: BAD IDEA, doesn't it automatically combust (or at least the vapors)

4. Vacuum distillation: Quite good, but not easy to keep vacuum, anyone ever done this? (and wants to share their secret)

if i were to further distill it, I would like to expiriment with rocket propulsion, but first, I must get a higher concentration

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what concentration have you got so far?

hydrogen peroxide is fairly hazardous even at 30%

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hydrogen peroxide is fairly hazardous even at 30%
I know that you can buy it with a concentration 3-30%.
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Once it reaches 70% concentration, it becomes capable of instantly vaporizing all of its contents in a very large liquid-to-gas explosion. Careful there... You can get it to exploding concentrations (above 30%) just by evaporating off the water. (It helps if it's in solvent that evaporates faster. The stuff I had was 80% ethanol, 5% H2O2, and 15% mystery solvent.)

So, the easiest way is the simplest: evaporate off the water.

Tip: Do this outside. Or you will end up like my profile picture...

And if you want rocket propellant, try dripping it over silver mesh. That makes it decompose. Then you might need something to use all that oxygen. (By the way, under an oxygen-rich atmosphere, even a solid brick of iron will burn.)

Enjoy!

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Ditto above, H2O2 is pretty dangerous at high concentration. You can get easily get H2O2 at concentration of up to 50% (usually stabilized).

70% H2O2 is usually more troublesome to get as it doesnt have much laboratory use, unless you are making a rocket or smtg along that line.

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how exactly do you evaporate off the water??

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how exactly do you evaporate off the water??

If you are seriously incapable of evaporating water, you really shouldn't be dealing with things that explode. You know, when you blow on something to make it dry faster... evaporation's like that, except you don't do anything at all.

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sorry stupid moment !!!

is there a easy way to determine the H2O2 concentration ??

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Titration Or you could go decomp and measure the resultant gas in a syringe.

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decomposition with manganese oxide ?

what would the rotio of gas be to water at ,say 50% H2O2

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calculate it yourself!

work out what a mole of pure hydrogen peroxide weighs, and the ratio of Oxygen that will be liberated, then use the gas law to calculate how many moles of gas came off, and apply that.

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decomposition with manganese oxide ?

what would the rotio of gas be to water at ,say 50% H2O2

You will know when it reaches 70% because at that point, when it decomposes, all of the water in it will be vaporized, leaving little or no material left.

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ok i have never done this before (because im 15 and a lazy bum that thinks chemistry is awesome) so bare with me and pls correcrt me !!

1 mol H2O2 = (2x 1.0079 amu)hydrogen + (2x 15.999 amu)Oxygen=34.0138 grams

therefore for every 34.0138 g H2O2->18.015g H2O+ 15.9988g O2

So for a 50 % conc(by weight???)H2O2

(50%x 18.015)H2O + (50%x 34.0138)H2O2-> 18.015g H2O +(50%x15.9988) 7.9994g O2

therefore(50%) 26.014g H2O2-> 18.015g H2O + 7.9994g O2

I want to check the gas laws Quickly and see if i can work out the volume per mol O2 let me Know if the above is correct !!

thnx !!

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ok i have never done this before (because im 15 and a lazy bum that thinks chemistry is awesome) so bare with me and pls correcrt me !!

1 mol H2O2 = (2x 1.0079 amu)hydrogen + (2x 15.999 amu)Oxygen=34.0138 grams

therefore for every 34.0138 g H2O2->18.015g H2O+ 15.9988g O2

So for a 50 % conc(by weight???)H2O2

(50%x 18.015)H2O + (50%x 34.0138)H2O2-> 18.015g H2O +(50%x15.9988) 7.9994g O2

therefore(50%) 26.014g H2O2-> 18.015g H2O + 7.9994g O2

I want to check the gas laws Quickly and see if i can work out the volume per mol O2 let me Know if the above is correct !!

thnx !!

The mass of one mole of [ce]H2O2][/ce] is 34.016, so yes, you're correct on that. But this is 100% concentrated, and in 50% you have $34.016 /times 0.5 = 17.008g$

As you know [ce]H2O2[/ce] decomposes by the following reaction:

[ce]2H2O2 -> 2H2O + O2[/ce]

So if you want to know how much [ce]O2[/ce] is being released, you can easily calculate that since we have all the data we need. To calculate how much [ce]O2[/ce] is being released, I do it this way:

$\frac{2\times H_2O_2}{34.016g}=\frac{O_2}{x}$ from there x is:

$x=\frac{O_2 \times 34.016}{2 \times H_2O_2}$ and the result would be 16, or for 50% would be 8, so you're OK.

The same would apply to water.

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What formula can i use to determine the volume of gas per x weight ??

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What formula can i use to determine the volume of gas per x weight ??

You can use $V=\frac{m}{M} \times V_m$

where:

V=volume

m=mass

M= molar mass

Vm = 22.41

Suppose you want to know the volume of 10 g of [ce]O2[/ce]

Then you plug the data on the equation:

$V=\frac{m}{M} \times V_m$

$V=\frac{10}{32} \times 22.41$

$V=7.003124$

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so 7.999g O2;

V=m/M x Vm

V=7.999/31.998 x 22.41

V=5.602 of what unit ???(liter??)

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yes litres of gas.

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thnx guys i might consider working thing out by myself now

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