Data Sheet Compound selected for analysis:CS Trial First Second Part A: Determin

Question

Data Sheet Compound selected for analysis:CS Trial First Second Part A: Determining the mass of water evolved Mass of crucible and cover Mass of hydrated sample 2 A 1st weighing Mass of crucible, cover, and dehydrated sample 2nd weighing: Mass of crucible, cover, and dehydrated sample 3rd weighing: Mass of crucible, cover, and dehydrated sample 4th weighing: Mass of crucible, cover, and dehydrated sample Mass of dehydrated sample 3 Mass ofwater evolved -- Part B: Calculating n Moles of H20 (show calculations) 1.1 23. o67 23.607 delbydrated sample 13 21,387 272993 2 2. g 0.6566222 0.656 g I- -6.222 360 g mol mol C.020 mol mol Moles of dehydrated sample (show calculations) 2x36322 Ino) Ise.6 Value of n (show calculations) 3 5.13 s 2.21 dur 2 of Percent Water vl.2 Determination

Explanation / Answer

Trial-1

Mass of Crucible +Cover = 20.729 g

Mass of Crucible + cover + hydrated sample= 21.747 g

Mass of hydrated sample = 21.747 g -20.729 g = 1.018 g

Mass of Crucible + cover + dehydrated sample = 21.387 g

Mass of dehydrated sample = 21.387 g - 20.729 g = 0.658 g

Mass of water evolved = 1.018 g - 0.658 g = 0.360g

From mass of H2O, calculate moles H2O

0.360 g H2O x 1 mol /18.02 g = 0.01997 moles

Moles of dehydrated sample = 0.658 g/159.609 g/mol = 0.00412 moles

Value of n= moles H2O / Moles of dehydrated sample

n= 0.01997 moles/0.00412 moles = 4.85 approximately 5

Actual formula of hydrate = CuSO4.5H2O

PART C % error

Theoretical mass % of water in the hydrate -

Cu = 63.55 g/mol

S = 32.07 g/mol

O x 4 = 16.00 g/mol x 4 = 64.00 g/mol

H2O x 5 = 18.02 g/mol x 5 = 90.10 g/mol

Now add them together 63.55g/mol + 32.07 g/mol + 64.00 g/mol + 90.10 g/mol = 249.72 g/mol

In 249.72 grams of hydrate, 90.10 grams is water,

So, the percentage of water = (90.10 g / 249.72 g) x 100

= 36.08% water

Experimental mass % of water in the hydrate -

Mass % H2O in sample = (Mass of water evolved/ Mass of hydrated sample) x100

= (0.360 g/1.018 g) x 100%

= 35.36%

Percent error = [experimental value - theoretical value] / theoretical value x 100%

                        = [35.36 – 36.08]/ 36.08 x 100%

                       = -1.99% ~ 2%

Discard any negative sign. Take the absolute value so percent error is 2%

Trial-2

Experimental mass % of water in the hydrate -

Mass % H2O in sample = (Mass of water evolved/ Mass of hydrated sample) x100

= (0.476 g/1.298 g) x 100%

= 36.67 %

Percent error = [experimental value - theoretical value] / theoretical value x 100%

                        = [36.67 – 36.08]/ 36.08 x 100%

                       = 1.6 %

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