2e3ef
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5.00g of a mixture of MgSO4 . 7H2O and CuSO4 . 5H2O was heated at 120 degrees Celsius until a mixture of the anhydrous salts was formed, which weighed 3.00g. Calculate the percentage by mass of MgSO4 . 7H2O in the mixture.

I do this in different methods, if I say n1 = mol mgso4.7h2o and mgso4 and n2= mol cuso4.5h2o and cuso4 and then use 246.4n1 + 249.6n2 = 5
then create a 2nd equation 120.4n1 + 159.6n2 = 3 and solve for n1 and then find the mass i get it as 14%

But if I simply do 5-3=2g h2o, then as 7:1 ratio 1/9 x 7/13 /7 (Mr of MgSO4.7H2O) = 2.28g so 45.6%

I'm confused as to what is correct as I believe the answer is actually 74%!
thanks so much
Last edited by 2e3ef; 3 weeks ago
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charco
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(Original post by 2e3ef)
5.00g of a mixture of MgSO4 . 7H2O and CuSO4 . 5H2O was heated at 120 degrees Celsius until a mixture of the anhydrous salts was formed, which weighed 3.00g. Calculate the percentage by mass of MgSO4 . 7H2O in the mixture.

I do this in different methods, if I say n1 = mol mgso4.7h2o and mgso4 and n2= mol cuso4.5h2o and cuso4 and then use 246.4n1 + 249.6n2 = 5
then create a 2nd equation 120.4n1 + 159.6n2 = 3 and solve for n1 and then find the mass i get it as 14%

But if I simply do 5-3=2g h2o, then as 7:1 ratio 1/9 x 7/13 /7 (Mr of MgSO4.7H2O) = 2.28g so 45.6%

I'm confused as to what is correct as I believe the answer is actually 74%!
thanks so much
Fun question:

Let the mass of magnesium sulfate heptahydrate = xg
The mass of copper sulfate pentahydrate = 5 - x

Hence the mol of magnesium sulfate heptahydrate = x/246
Hence the mol of copper sulfate pentahydrate = (5 - x)/249.5

After heating the moles of water produced from magnesium sulfate = 7x/246
After heating the moles of water produced from magnesium sulfate = 5(5-x)/246

After heating the mass of water produced from magnesium sulfate = 18*7x/246
After heating the mass of water produced from magnesium sulfate = 18*5(5-x)/246

and the sum of masses of water = mass loss = 2.0g

After heating the moles of each anhydrous salt is the same as the hydrated salt
the mol of magnesium sulfate = x/246
the mol of copper sulfate = (5 - x)/249.5

So you can work out the masses of each in terms of x
the mass of magnesium sulfate = 120x/246
the mass of copper sulfate = 159.5(5 - x)/249.5

And the sum of these masses = 3.0g

120x/246 + 159.5(5 - x)/249.5 = 3.0

0.4878x + 3.196 - 0.6393x = 3

0.1515x = 0.196

x = 1.294g

Hence mass of CuSO4.5H2O = 1.294g
Hence mass of MgSO4.7H2O = 3.706g

Percentage by mass magnesium sulfate heptahydrate = 100*3.706/5 = 74.12%
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heptanitrocubane
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(Original post by 2e3ef)
5.00g of a mixture of MgSO4 . 7H2O and CuSO4 . 5H2O was heated at 120 degrees Celsius until a mixture of the anhydrous salts was formed, which weighed 3.00g. Calculate the percentage by mass of MgSO4 . 7H2O in the mixture.

I do this in different methods, if I say n1 = mol mgso4.7h2o and mgso4 and n2= mol cuso4.5h2o and cuso4 and then use 246.4n1 + 249.6n2 = 5
then create a 2nd equation 120.4n1 + 159.6n2 = 3 and solve for n1 and then find the mass i get it as 14%

But if I simply do 5-3=2g h2o, then as 7:1 ratio 1/9 x 7/13 /7 (Mr of MgSO4.7H2O) = 2.28g so 45.6%

I'm confused as to what is correct as I believe the answer is actually 74%!
thanks so much
Yeah I think your method of using h2o moles is wrong because you're assuming the amount of MgSO4.7H2O and CuSO4.5H2O is the same because you're combining the molar ratio from both reactions which is invalid
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