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5.00g of a mixture of magnesium sulfate crystals, MgSO4.7H2O, and copper sulfate crystals, CuSO4.5H2O, was heated until both salts has lost all water of crystallisation. There was a decrease in mass of 2.00g. Find the percentage of each hydrated salt in the mixture.
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(Original post by CtrlAltDelirious)
5.00g of a mixture of magnesium sulfate crystals, MgSO4.7H2O, and copper sulfate crystals, CuSO4.5H2O, was heated until both salts has lost all water of crystallisation. There was a decrease in mass of 2.00g. Find the percentage of each hydrated salt in the mixture.
5.00g of a mixture of magnesium sulfate crystals, MgSO4.7H2O, and copper sulfate crystals, CuSO4.5H2O, was heated until both salts has lost all water of crystallisation. There was a decrease in mass of 2.00g. Find the percentage of each hydrated salt in the mixture.
If you let the mass of magnesium sulfate = m
Then the mass of copper sulfate = (5 - m)
Now work out the moles of each
mol MgSO4.7H2O = m/246
mol CuSO4.5H2O = (5-m)/249.5
Now calculate the moles of water lost by each when heated
mol water from MgSO4.7H2O = 7m/246
mol water from CuSO4.5H2O = 5(5-m)/249.5
Now calculate the mass of water lost by each
mass water from MgSO4.7H2O = 18 x 7m/246
mass water from CuSO4.5H2O = 18 x 5(5-m)/249.5
and you know that the sum of the water masses = 2.00g
Therefore
[18 x 7m/246] + [18 x 5(5-m)/249.5] = 2
Now solve for m
126m/246 + 450/249.5 - 90m/249.5 = 2
0.512m + 1.804 - 0.361m = 2
0.151m = 0.196
mass of magnesium sulfate heptahydrate = 1.3g
mass of copper(II) sulfate pentahydrate = 3.7g
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#3
(Original post by charco)
Yes. Not an easy question, but it's more maths than chemistry.
If you let the mass of magnesium sulfate = m
Then the mass of copper sulfate = (5 - m)
Now work out the moles of each
mol MgSO4.7H2O = m/246
mol CuSO4.5H2O = (5-m)/249.5
Now calculate the moles of water lost by each when heated
mol water from MgSO4.7H2O = 7m/246
mol water from CuSO4.5H2O = 5(5-m)/249.5
Now calculate the mass of water lost by each
mass water from MgSO4.7H2O = 18 x 7m/246
mass water from CuSO4.5H2O = 18 x 5(5-m)/249.5
and you know that the sum of the water masses = 2.00g
Therefore
[18 x 7m/246] + [18 x 5(5-m)/249.5] = 2
Now solve for m
126m/246 + 450/249.5 - 90m/249.5 = 2
0.512m + 1.804 - 0.361m = 2
0.151m = 0.196
mass of magnesium sulfate heptahydrate = 1.3g
mass of copper(II) sulfate pentahydrate = 3.7g
Yes. Not an easy question, but it's more maths than chemistry.
If you let the mass of magnesium sulfate = m
Then the mass of copper sulfate = (5 - m)
Now work out the moles of each
mol MgSO4.7H2O = m/246
mol CuSO4.5H2O = (5-m)/249.5
Now calculate the moles of water lost by each when heated
mol water from MgSO4.7H2O = 7m/246
mol water from CuSO4.5H2O = 5(5-m)/249.5
Now calculate the mass of water lost by each
mass water from MgSO4.7H2O = 18 x 7m/246
mass water from CuSO4.5H2O = 18 x 5(5-m)/249.5
and you know that the sum of the water masses = 2.00g
Therefore
[18 x 7m/246] + [18 x 5(5-m)/249.5] = 2
Now solve for m
126m/246 + 450/249.5 - 90m/249.5 = 2
0.512m + 1.804 - 0.361m = 2
0.151m = 0.196
mass of magnesium sulfate heptahydrate = 1.3g
mass of copper(II) sulfate pentahydrate = 3.7g
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(Original post by charco)
Yes. Not an easy question, but it's more maths than chemistry.
If you let the mass of magnesium sulfate = m
Then the mass of copper sulfate = (5 - m)
Now work out the moles of each
mol MgSO4.7H2O = m/246
mol CuSO4.5H2O = (5-m)/249.5
Now calculate the moles of water lost by each when heated
mol water from MgSO4.7H2O = 7m/246
mol water from CuSO4.5H2O = 5(5-m)/249.5
Now calculate the mass of water lost by each
mass water from MgSO4.7H2O = 18 x 7m/246
mass water from CuSO4.5H2O = 18 x 5(5-m)/249.5
and you know that the sum of the water masses = 2.00g
Therefore
[18 x 7m/246] + [18 x 5(5-m)/249.5] = 2
Now solve for m
126m/246 + 450/249.5 - 90m/249.5 = 2
0.512m + 1.804 - 0.361m = 2
0.151m = 0.196
mass of magnesium sulfate heptahydrate = 1.3g
mass of copper(II) sulfate pentahydrate = 3.7g
Yes. Not an easy question, but it's more maths than chemistry.
If you let the mass of magnesium sulfate = m
Then the mass of copper sulfate = (5 - m)
Now work out the moles of each
mol MgSO4.7H2O = m/246
mol CuSO4.5H2O = (5-m)/249.5
Now calculate the moles of water lost by each when heated
mol water from MgSO4.7H2O = 7m/246
mol water from CuSO4.5H2O = 5(5-m)/249.5
Now calculate the mass of water lost by each
mass water from MgSO4.7H2O = 18 x 7m/246
mass water from CuSO4.5H2O = 18 x 5(5-m)/249.5
and you know that the sum of the water masses = 2.00g
Therefore
[18 x 7m/246] + [18 x 5(5-m)/249.5] = 2
Now solve for m
126m/246 + 450/249.5 - 90m/249.5 = 2
0.512m + 1.804 - 0.361m = 2
0.151m = 0.196
mass of magnesium sulfate heptahydrate = 1.3g
mass of copper(II) sulfate pentahydrate = 3.7g
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