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2008-01-29 12:34:15
HurricaИe
Banned
+877|6446|Washington DC
So I kinda goofed off in class and didn't take down all the notes on formulas for hydrates. I have two homework problems that, sure enough, require knowledge of the hydrate formula.

I wrote this much down:

-Mass the hydrate
-Heat it up
-Mass the anhydrous salt
-Subtract the mass of the anhydrous salt from the mass of the hydrate to get the mass of the water
-Use mole ratio to find the number of waters (what?!)

Since I don't have access to a Bunsen burner, naturally we already have the mass of the hydrate and the mass of the anhydrous salt. I did the whole subtraction thing but IDK what to do next.

TL;DR: Tell me the rest of the hydrate formula
2008-01-29 12:35:42
Surgeons
U shud proabbly f off u fat prik
+3,097|6975|Gogledd Cymru

if you take the mass of the anhydrate away from the hydrate you will get the mass of the water, yes?

Then divide by 18 to get no. moles of water.
2008-01-29 12:37:24
HurricaИe
Banned
+877|6446|Washington DC

The Sheriff wrote:

if you take the mass of the anhydrate away from the hydrate you will get the mass of the water, yes?

Then divide by 18 to get no. moles of water.
Yeah, I did that but end up with .042 mol. There's gotta be something else missing. The compound is hydrated MgSO3, so:

MgSO3 * X H2O where X is the number of H2O molecules

I can't have .42 H2O molecules =/
2008-01-29 12:46:54
SgtSlutter
Banned
+550|7123|Amsterdam, NY
Coulda sworn I posted something in this thread
2008-01-29 12:50:23
avman633
Member
+116|6849

HurricaИe wrote:

The Sheriff wrote:

if you take the mass of the anhydrate away from the hydrate you will get the mass of the water, yes?

Then divide by 18 to get no. moles of water.
Yeah, I did that but end up with .042 mol. There's gotta be something else missing. The compound is hydrated MgSO3, so:

MgSO3 * X H2O where X is the number of H2O molecules

I can't have .42 H2O molecules =/
.042 is the number of moles, not molecules for x. To get that you would have to multiply that and the coefficient infront of MgSO3 by a number that would make .42 whole or close to it. I did a lab similar to this except we got Copper (II) Sulfate hydrate.
2008-01-29 12:51:41
Aries_37
arrivederci frog
+368|7060|London

HurricaИe wrote:

The Sheriff wrote:

if you take the mass of the anhydrate away from the hydrate you will get the mass of the water, yes?

Then divide by 18 to get no. moles of water.
Yeah, I did that but end up with .042 mol. There's gotta be something else missing. The compound is hydrated MgSO3, so:

MgSO3 * X H2O where X is the number of H2O molecules

I can't have .42 H2O molecules =/
0.42 mol =/= 0.42 molecules. 0.42 moles should be right if you haven't got your subtraction and division wrong.

EDIT: my bad read the question wrong. You don't want the number of moles, you want the ratio in the formula.

In this case you know the number of moles of water (0.42). Now you work out the number of moles of anyhydrous salt that you had. The ratio should be pretty obvious from this.

If you give us your figures I'll get an answer for you to check against.

Last edited by Aries_37 (2008-01-29 13:10:00)

2008-01-29 12:55:12
avman633
Member
+116|6849
To add on to mine, you need the number of moles of Magnesium Sulfite.
2008-01-29 13:28:11
HurricaИe
Banned
+877|6446|Washington DC

Aries_37 wrote:

HurricaИe wrote:

The Sheriff wrote:

if you take the mass of the anhydrate away from the hydrate you will get the mass of the water, yes?

Then divide by 18 to get no. moles of water.
Yeah, I did that but end up with .042 mol. There's gotta be something else missing. The compound is hydrated MgSO3, so:

MgSO3 * X H2O where X is the number of H2O molecules

I can't have .42 H2O molecules =/
0.42 mol =/= 0.42 molecules. 0.42 moles should be right if you haven't got your subtraction and division wrong.

EDIT: my bad read the question wrong. You don't want the number of moles, you want the ratio in the formula.

In this case you know the number of moles of water (0.42). Now you work out the number of moles of anyhydrous salt that you had. The ratio should be pretty obvious from this.

If you give us your figures I'll get an answer for you to check against.
Let's see if I did this right then

Hydrate: 1.500g
Anhydrous salt: .737g
Water: .763g
Mol H2O: .042mol
Mol MgSO3: .007

Then I did the thing for empirical formula, meaning you divide both amount of mol by the smaller amount

So I got: 1 mol of MgSO3 (which is right; there's only one molecule of MgSO3) and 6 mol (.042/.007) of H2O, so it'd be Magnesium Sulfite Hexahydrate.
2008-01-29 13:34:13
S.Lythberg
Mastermind
+429|6932|Chicago, IL
the weight lost will be the weight of the water, which divided by 18g/mol, gives you your moles of water.  mol(water)/mol(reactant) gives you your ratio.


eg. if you lose 36g after heating, that equates to two moles of water. if you had .5 moles of reactant, that gives you a ratio of 4:1, four H20 for every base molecule.
2008-01-29 13:35:45
HurricaИe
Banned
+877|6446|Washington DC

S.Lythberg wrote:

the weight lost will be the weight of the water, which divided by 18g/mol, gives you your moles of water.  mol(water)/mol(reactant) gives you your ratio.


eg. if you lose 36g after heating, that equates to two moles of water. if you had .5 moles of reactant, that gives you a ratio of 4:1, four H20 for every base molecule.
Sounds like I got it then thanks BF2s!

 

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