Question

A hydrate of MgSO4 is placed in a test tube with a total mass of 49.10 g. After heating, the mass is 33.65 g. If the test tube weighs 18.83 g, what is the formula of the hydrate?

A. MgSO4 · 5Η2Ο

B. MgSO4 · 7Η2Ο

C. MgSO4 · 8Η2Ο

D. MgSO4 · 6Η2Ο

Answer #1

Given weight of test tube=18.83 g

Mass of MgSO4.xH2O+test tube=49.10 g

Therefore mass of MgSO4xH2O=49.10-18.83=30.27 g

After heating, anhydrous MgSO4+test tube mass=33.65 g

Therefore anhydrous MgSO4=33.65-18.83=14.82 g.

We know that molar mass of MgSO4=120.4 g/mol, and water=18 g.

Mass of water= mass of MgSO4xH2O-mass of MgSO4

=30.27g-14.82 g=15.45 g

Moles of MgSO4=mass of MgSO4/molar mass=14.82 g/120.4 g/mol

Moles of MgSO4=0.123 moles.

Moles of water=15.45 g/18 g/mol=0.858 moles

Moles water/moles of anhydrous MgSO4=0.858/0.123=6.975~7.

So the formula is MgSO4.7H2O.

Option B is correct.

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Please show the work. Thank you
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Data:
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cover
51.729g
Mass of crucible, cover and
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______ ?
Percentage H2O in the
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______ ?
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