Sunday, November 6, 2011

Determination of the Percentage of Water in Hydrated Copper (II) Sulfate Lab

Observations:

Copper (II) Sulfate Crystals:
                                            Before: small (sand grain like) blue pieces with some white pieces
                                           
                                            During: - a light blue rim forms around the edges of the crystals
                                                         - light blue rim spreads closer towards center
                                                         - blue crystals completely light blue
                                                         - white rim forms around edges, moves closer towards center
                                                         - light blue crystals completely white

                                             After:  - broken up into solid-like pieces
                                                         - light green on powder/crystals on bottom

Calculations:


Evaporating dish: 50.04g
Watch glass: 57.67g
Dish with Copper (II) Sulfate Crystals: 59.38g
Dish with CuSOcrystals after cooling: 56.03g

1) 59.38g - 50.04g = 9.34g CuSO
56.03g - 50.04g = 5.99g CuSO
(9.34g - 5.99g = 3.35g)
3.35g/9.34g = 35.87% water lost

2) 35.87 - 36.08  =  0.0058 =  0.58% error
        36.08

3)


=  0.06  mol CuSO4
    0.06  = 1

                                                                                                                       

=  0.19  mol H2
    0.06  = 3 (X=3)

   CuSO4 ∙ 3H2O

1 comment:

  1. Excellent job except that 0.19/0.06 = 6. You were thinking 18/6 which would be 3. That extra decimal place changes the answer. 18/20

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