Calculations for Hydrated Crystals Lab
You need to remember that we are making the water attached to the crystal leave. The
difference in the weight before you heated the crucible and after is the mass of the water
attached to the compound. This lab shows calculations for MgSO4. When using BaCl2 or
CuSO4 all observations are the same except for 9, 12, 13.
Obviously in observ #9 we would divide by the molecular wt of barium chloride. (137.33 +
35.45+35.45) or copper(II) sulfate.
In #12 the correct formula is BaCl2 ·2 H2O, CuSO4 ·5 H2O
#13 then 36.00 is the water and 208.23 is the whole for BaCl2 90 is water and 159.5 is the
whole for CuSO4
Remember, when you calculate a formula you need a mole to mole ratio of the parts. In this
lab the MgSO4 is one part and H2O is the other part. Moles are found by grams and the
Periodic Table
You will have a data table that looks something like this
(the numbers you will see correspond to the data in the table. 1 = mass of crucible; 3 =
mass after 1st heating)
1.mass of crucible
your lab data goes here
2.mass of crucible+hydrate
your lab data goes here
3.mass after 1st heating
your lab data goes here
4.mass afer 2nd heating
your lab data goes here
Now let's calculate what we know base upon this data
5. mass of hydrate
#2 - #1(cru+hyd-cru)
6. mass of water
2-4
7. mass of anyhdrous MgSO4
5-6
The lab asks for % change(%= part/whole) so:
8. % change
6/5 x 100 (water is part; crystal is whole)
Cal moles of anyhydrous MgSO4and moles of water (Mole=g/PT)
9. Mole of Any MgSO4
7/120.3
10. Moles of Water
6/18.0
Determine the formula. (mole:mole ratio)
11.Experimental formula
9/9 MgSO4 ·10/9 H2O
(remember our problem solving for empirical
formulas. change g into moles; divide each by the
smallest to get a 1: ratio. MgSO4 is the smaller of the
two moles so divide both by MgSO4)
Find the actual formula and cal the % of water in it
12. Actual formula
MgSO4 · 7 H2O
13. Theoretical % water
(126.0)/276.3 x 100
Calculate % error = (exp-acc)/acc x 100
14. % error
(8 - 13) / 13 x 100