Chemistry question- equivalent hexahydrate to anhydrous CoCl2

[Darklight]

I know some of you are slacking off work reading this, so maybe you can put your brains to this ;)

It's been years ( about 15 ) since I did this manually and I can't seem to work it out now

I need to make up some stock solution Murashige and Skoog basal medium at 10x concentration

One of the micronutrients I need is cobalt chloride ( anhydrous ) at 0.25mg for the stock solution. MW=129.84

I only have cobalt chloride hexahydrate, MW= 237.93

How much CoCl2.6H2O is needed to give the same number of CoCl2 molecules as 0.25mg of CoCl2

For the life of me I can't get the same answer twice.

[Khala]

n=m/M

n=0.00025/129.84

.:. n(CoCl2) = 1.92544 x 10^-6

.:. n (CoCL2.6H2O) = 1.92544 x 10^-6

.:. n x M = 4.58x10^-4 g

So you need 4.58 x 10^-4 g of CoCl2.6H2O

In milligrams, 0.458mg

Hopefully I did that right, I'm pretty rusty. Hope I helped! Cheers

Edited November 21, 2012 by Khala

[Alice]

Covert everything to moles.

moles = mass/MW

so moles in 25 mg of anhydrous (remember to convert mg to grams first) = 0.025g / 129.84 = 1.925 x 10^-4

So what mass of the hydrated compound do we need to weigh out to have the same number of moles as the anhydrous?

mass = mol x MW = 1.925 x 10^-4 x 237.93 = 0.0458 g

which when converted back to mg is 45.8 mg

And I always do a quick check on whether my calculation seems reasonable:

The MW of the hydrated form is almost twice that of the MW of the anhydrous. Therefore my answer should logically be almost twice 25mg, which it is.

45.8 mg

Edit: Ooh beat me to it Khala.

Edited November 21, 2012 by Alice

[Khala]

Heheh Alice, trying to stay sharp for next year...I was off by a significant figure though! Big mistake! Sigh.

Edit: No I wasn't actually, in the given figures he said 0.25mg rather than 25mg. HSC proofreading skills for the win.

Edited November 21, 2012 by Khala

[Darklight]

Thanks you two Khala's answer was pretty close to one of mine

OK for my next question:

I need to substitute Manganese chloride for Manganese sulphate ( because for some reason I don't have any of the sulphate )

Stock solution is 169mg of MnSo4.H2O at MW 169.02

I need the equivalent number of manganese atoms for this amount using Manganese chloride tetrahydrate MW 197.91

And an opinion on whether the substitution of chloride for sulphate is going to create any significant changes in response from plants? Taking away a few chloride ions and adding more sulphur ... is the outcome likely to be trivial?

Is there software to help me with this?

When doing mg/uM calculations for hormone solution use I use an old program called Chemical Calculator ( I like it so much I paid for the shareware, it's super awesome for those moments where I used to misplace the decimal point.) But I can't seem to parse the questions above in a way that might make equivalence questions clear

Edited to remove stupid reference which wasn't relevant

Edited November 21, 2012 by Darklight

[Darklight]

Hey where'd it go?

I had the workings out to the last question but a page refresh ate it all

Something something the amount of manganese in the sulphate is 32.5%

In the chloride form it's 30.6%

You all do know I count on my fingers... right?

[Alice]

Heheh Alice, trying to stay sharp for next year...I was off by a significant figure though! Big mistake! Sigh.

Edit: No I wasn't actually, in the given figures he said 0.25mg rather than 25mg. HSC proofreading skills for the win.

 

Aah crap. I read it as 25mg, stupid brain lol.

[Alice]

Thanks you two Khala's answer was pretty close to one of mine

OK for my next question:

I need to substitute Manganese chloride for Manganese sulphate ( because for some reason I don't have any of the sulphate )

Stock solution is 169mg of MnSo4.H2O at MW 169.02

I need the equivalent number of manganese atoms for this amount using Manganese chloride tetrahydrate MW 197.91

 

work out moles of MnSO4.2H2O

mol = mass / MW = 0.169 / 169 = 0.001 mol

you want to weigh out that exact same number of mol of the MnCl2.4H2O, since both 1 mol of MnSO4.2H2O and 1 mol of MnCl2.4H2O each contain the same amount (i.e. 1 mol) of Mn.

So mass = mol x MW = 0.001 x 197.91 = 0.198g = 198 mg of MnCl2.4H2O is what you need.

Logic check: Both compound have only 1 Mn in their formula, so we want the same mol of each, i.e. equimolar amounts. Check.

The MW of the MnCl2.4H2O is a bit larger, (maybe 10-15%?) than the MnSO4.2H2O, so considering we want equimolar amounts, our answer should be a bit (maybe 10-15%) larger than the starting compound. Check.

[Alice]

And no idea on the plant stuff sorry.

[Darklight]

Thanks sooo much for your replies, both here and via PM/ The amount and types of knowledge available and shared here are still inspirational to me after all this time

Alice, I'm trying to work through your solution in my head so I don't need to ask these questions again. It's frustrating for me to not be able to do this. I'm still stuck on the notion that 1M of MnSO4.2H2O would have half the volume of H2O by weight when compared to the MnCl2.4H2O, but maybe the H2O isn't significant overall in terms of mass and by comparison. Is it compensated for by the increased mass of the SO4 when compared to the Cl2?

I was trying to break it down in terms of individual elements, figuring that all I needed was to derive the mass of Mn molecules needed for the solution, but maybe that was the wrong starting point

Obviously there is some faulty logic going on in my head, I'm putting it out there so I can try to work out what's wrong with my method

[Alice]

I need the equivalent number of manganese atoms for this amount using Manganese chloride tetrahydrate MW 197.91

^ this is the question you are asking

I was trying to break it down in terms of individual elements, figuring that all I needed was to derive the mass of Mn molecules needed for the solution, but maybe that was the wrong starting point

Because you want the equiv number of atoms, all you need to derive is the moles of Mn atoms needed for the solution, the other atoms and molecules (H2O, SO4^2-, Cl- etc) are just taking up space, you don't care what they are, but you do need to take them into account when turning your moles answer back to a mass answer so you can actually weigh the stuff out. Like would be much easier if someone invented digi scales where you could plug in the MW and the display read in moles lol.

When try to work out these sorts of questions, the first thing you do, always, is convert everything into moles. Always always always. The once you've done your calculation, the last thing you do is convert back to mass using the MW of the compound.

I'm still stuck on the notion that 1M of MnSO4.2H2O would have half the volume of H2O by weight when compared to the MnCl2.4H2O, but maybe the H2O isn't significant overall in terms of mass and by comparison. Is it compensated for by the increased mass of the SO4 when compared to the Cl2?

 

I'm not exactly sure what you are asking.

1 mol of MnSO4.2H2O contains 1 mol of Mn2+, 1 mol of SO4^2-, and 2 mol of H2O

1 mol of MnCl2.4H2O contains 1 mol of Mn2+, 2 mol of Cl-, and 4 mol of H2O

If all you want is a solution of know Mn2+ molarity, then all you really need to recognise is that both those compound each contain only 1 Mn2+ in their formula, and then we just need to figure out how many moles are in the manganese sulfate, then we know how many moles are in the manganese chloride, ie the same amount. It doesn't really matter that one compound has 4 waters of hydration compared with 2. Sure you need to know they are there, for when you convert back to mass in the final step, but when doing the initial moles calculation, all you need to know is that each compound has only 1 Mn2+.

First step: Everything gets converted to moles.

Middle step: Do the calculation. In this case, both compounds have only 1 Mn2+, so it's 1:1. If you only had say Mn2O3 at hand, then you'd only need half as many moles of it compared with MnSO4, as each Mn2O3 contains 2 Mn2+ atoms, etc.

Last step: Then you convert back to mass.

Edited November 21, 2012 by Alice

[Darklight]

Thanks heaps Alice, I got this. No need for me to convert at all, I wanted the same M.

Up til then I was overthinking it. Let's hope the info stays in there

[Alice]

Not a problem, feel free to post any more Q's or shoot me a pm. Those sort of questions used to confuse the hell out of me, until had to do them a gazillion times demonstrating to undergrads. It just takes practice, lots and lots of practice lol. No easy way around that.

One can never have too much chemistry in their lives

Edited November 22, 2012 by Alice