How Much Anhydrous Magnesium Sulfate Remains After Dehydration?

  • Thread starter Thread starter Raza
  • Start date Start date
  • Tags Tags
    Chemistry
AI Thread Summary
To determine how much anhydrous magnesium sulfate remains after dehydrating 125g of magnesium sulfate heptahydrate, one must calculate the molar masses of both compounds. The molar mass of magnesium sulfate heptahydrate is 246.3 g/mol, while the anhydrous form is 120.3 g/mol. Using the stoichiometric ratio of 7 moles of water for every mole of the hydrate, the calculation shows that approximately 61 grams of anhydrous magnesium sulfate will remain after dehydration. Understanding the definitions of dehydration and anhydrous is crucial for solving the problem. The discussion concludes with a participant expressing clarity on the solution process.
Raza
Messages
203
Reaction score
0
Hello, I got this really hard question on my chemistry assignment and I don't know where to start; I am completely lost.

If 125g of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?

Any help would be greatly appreciated!
 
Physics news on Phys.org
Raza said:
Hello, I got this really hard question on my chemistry assignment and I don't know where to start; I am completely lost.

If 125g of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?

Any help would be greatly appreciated!
Do you know the formula for the above mixture ? What does the formula tell you about how much each part contributes to the mass ?
 
is it Mg2SO3 . 7H2O?

I could probably solve this if I knew what anhydrous meant.
 
That's not SULPHATE,that's a wrong version of SULPHYTE.You need another formula...

Anhydruous=without water...

Daniel.
 
I am sorry, I do not get this; could you please rephrase it for me?What's sulfyte?
and what is a difference between dehyration and anhydrate?
 
Last edited:
Raza said:
I am sorry, I do not get this; could you please rephrase it for me?What's sulfyte?
and what is a difference between dehyration and anhydrate?

MgSO_{3} is MAGNESIUM SULPHYTE and MgSO_{4} is MAGNESIUM SULPHATE.One of them comes from the sulphurous acid,the other from sulphuric acid...

DEHYDRATION=losing water.
ANHYDROUS=without water...

Daniel.
 
I finally get the question but how do I solve it?
I done this before but with knowing the weight of each elements.
 
Last edited:
Well,1 mole of MgSO_{4}\cdot 7 H_{2}O has a certain mass...Can u compute it...?How about the mass of 1 mole of MgSO_{4} ??

Daniel.
 
You can observe the formula and from this find the final mole value of magnesium sulfate heptahydrate; find the molar mass of the compound and then find the realistic mole value using the 125 grams.

You know that for every one mole of the compound, there is 7 moles of H20. This should give you a stoichiometric ratio 7 moles of H20/1 mole of the compound. Convert using this ratio to moles of water. From this mole value, convert to grams of water. Subtract this value from 125, and that should be your answer.
 
  • #10
I'd like to help:

mol. wgt of MgS04*7H20 = 246.3
mol.wgt of MgS04 = 120.3

Thus, the ratio of anhydrous/hydrous (without water/with water) = 0.488

Thus, when you remove the water from 125 grams of the hydrate, the remaining is:

125*0.488=61 grams MgSO4 left.

I believe that's right. It's been a while and I don't want to make it worst for you.

Salty
 
Last edited:
  • #11
Thank you all, I got how you solve it now.
It seems so easy after you get what is doing on
 

Similar threads

Back
Top