Tetraamminecopper ii sulphate hydrate write up

Purpose

The objective of this experiment is to contact form tetraamminecopper(II) sulphate hydrate and determine the yield.

Components

CuSo4�5H2O

NH3 (concentrated)

Ethanol

50 cm3 measuring cylinder

250 cm3 beaker

Spatula

Equipment intended for vacuum filtration

Procedure

Consider approximately five. 0g of CuSo4�5H2O

Break down it in 30 cm3 water in the beaker

Put 10 cm3 concentrated ammonia (NH3) and stir the perfect solution is

Add forty five cm3 ethanol and stir carefully for a few minutes.

Filtering the solution through equipment pertaining to vacuum filtration.

Transfer the merchandise to a clean weighing vessel and keep to dry.

Procedure and observations in class

First 5.01g of CuSo4�5H2O was weighed out. After it was dissolved in 30 cm3 water, in the beaker, the solution got the colour blue. Next was 10 cm3 concentrated ammonia (NH3), which was added into the solution and the colour dark blue was observed. Then 40 cm3 ethanol was added and the solution got the colour bright blue. Then the solution was filtered through a Buchner flask and the final product was weighed in a plastic weighing boat.

The total mass was 5.98g, from which the weight of the boat, 1.16g, has to be subtracted. So the mass of the final product was 5.98 – 1.16 = 4.82g.

Data processing

1. “Calculate the number of moles CuSo4�5H2O used.”

To find out the number of moles the formula n = m / Mr has to be used.

Mr = 64 + 32 + (16 x 4) + (5 x 16) = 250

m = 5.01

n = 5.01 / 250 = 0.02004 � 0.0200 moles (3 s.f.)

2. “Concentrated ammonia contains 25% NH3 by mass. The density of concentrated ammonia is 0.91g/cm3 . Calculate the number of moles of NH3 .”

Density of con. ammonia = 0.91g/cm3 and in the procedure there was used 10 cm3, so therefore mass of ammonia used: 0.91 x 10 = 9.1g

Since only 25% of ammonia is NH3 , mass of NH3 : 9.1 x 0.25 = 2.275g

From here the amount of moles can be calculated by the formula n = m / Mr.

Mr = 14 + (1 x 3) = 17

m = 2.275g

n = 2.275 / 17 = 0.134 moles (3 s.f.)

3. “Which of the reactants is in excess? Which is the limiting reagent?”

CuSo4�5H2O

NH3

Number of moles (n)

0.02

0.134

Divide by smallest ratio

0.02 / 0.02 = 1

0.134 / 0.02 = 6.7

Divide by stoichiometric co-efficient from equation

(Equation below this table)

1 / 1 = 1

6.7 / 4 = 1.675

Reactant in excess or limiting reagent

Limiting reagent

Reactant in excess

(1)CuSO4 . 5H2O + 4NH3 –>Cu(NH3)4SO4. H2O + 4H2O

4. “Calculate the assumptive yield of Cu(NH3)4SO4. H2O”

From the formula above it is usually seen that the ratio between CuSO4. 5H2O and Cu(NH3)4SO4. H2O is 1: 1 ) Therefore zero. 02 moles of CuSO4. 5H2O will give 0. 02 moles of Cu(NH3)4SO4. INGESTING WATER. By using the formulation m = Mr back button n the theoretical produce can be determined:

n = 0. 02

Mr = 246

meters = zero. 02 by 246 sama dengan 4. 80 g

“Calculate the yield in percentage of the theoretical and comment on any difference. “

The yield in percentage could be calculated by “formula”: actual mass as well as expected mass. 4. 82 / four. 92 � 97. 9% (3 t. f)

For the reason that difference is so small (2. 1%) the experiment may very well be successful. The difference could have been due to different things just like: a small measurement mistake, somewhat was spilt or not really transferred if the solution occured in the Buchner flask.

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