a). It's easier to explain with an example. Suppose we start with MgSO4.xH2O and we want to determine x.
Suppose we took a 10.000 g sample of MgSO4.xH2O and it lost 5.116 g on heating. So the water content is 5.116 g and the MgSO4 content is 10.000 - 5.116 = 4.884 g.
Here is what we do.
5.116/18.015 = 0.284 = moles H2O.
4.884/120.37 = 0.0406 = moles MgSO4.
Then 0.284/0.0406 = 7 so the formula is
MgSO4.7H2O.
NOW, if we heated too fast and some of the sample spattered, what happens. When you reweigh the sample too much mass has been lost, so the mass H2O is too high and when you subtract from 10.000, the mass of MgSO4 too low. So the percent H2O is too high (70% using the numbers from b part) and the MgSO4 is too low (30% using the numbers from b part).[Percent MgSO4 should be 48.84% and percent H2O should be 51.16%.]
b) Suppose the loss in spattering is 7.0 g just to make things simple. Then mass H2O = 7 g and 10-7 = 3 g MgSO4.
Now 7/18=0.389
and 3/120 = 0.025, then
0.389/0.025 = 15.6 which we might round to 16 and the formula would be MgSO4.16H2O.
I hope this helps.
this is for a lab - determining the chemical formula of a hydrate.
suppose the hydrate was heated too quickly and some of it was lost of spattered out of the container.
explain how hthis would effect
a) the calculated percent by mass of water in the compound
b) the molecular formula u determined
THX
1 answer
1 answer