# Tetraamminecopper(II) sulphate hydrate Write-up

Categories: Copper

Function

The purpose of this experiment is to form tetraamminecopper( II) sulphate hydrate and figure out the yield.

Products

CuSo4ï ¿ 1/2 5H2O.

NH3 (concentrated).

Ethanol.

50 cm3 measuring cylinder.

250 cm3 beaker.

Spatula.

Devices for vacuum purification.

Procedure.

Weigh out roughly 5.0 g of CuSo4ï ¿ 1/2 5H2O.

Dissolve it in 30 cm3 water in the beaker.

Add 10 cm3 concentrated ammonia (NH3) and stir the option.

Include 40 cm3 ethanol and stir carefully for a couple of minutes.

Filter the solution through devices for vacuum purification.

Transfer the item to a clean weighing boat and leave to dry.

Procedure and observations in class.

First 5.01 g of CuSo4ï ¿ 1/2 5H2O was weighed out. After it was liquified in 30 cm3 water, in the beaker, the solution got the colour blue. Next was 10 cm3 concentrated ammonia (NH3), which was included into the option and the colour dark blue was observed. Then 40 cm3 ethanol was included and the service got the colour intense blue. Then the service was filtered through a Buchner flask and the final item was weighed in a plastic weighing boat.

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The total mass was 5.98 g, from which the weight of the boat, 1.16 g, needs to be deducted. So the mass of the end product was 5.98 – 1.16 = 4.82 g.

Data processing.

1. “Compute the variety of moles CuSo4ï ¿ 1/2 5H2O utilized.”.

To find out the variety of moles the formula n = m/ Mr needs to be utilized.

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

m = 5.01.

n = 5.01/ 250 = 0.02004 ï ¿ 1/2 0.0200 moles (3 s.f.).

2. “Focused ammonia consists of 25% NH3 by mass. The density of focused ammonia is 0.91 g/cm3. Calculate the number of moles of NH3.

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”.

Density of con. ammonia = 0.91 g/cm3 and in the procedure there was utilized 10 cm3, so for that reason mass of ammonia utilized: 0.91 x 10 = 9.1 g.

Considering that just 25% of ammonia is NH3, mass of NH3: 9.1 x 0.25 = 2.275 g.

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?”

CuSo4ï¿½5H2O

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 theoretical yield of Cu(NH3)4SO4 . H2O”

From the equation above it can be seen that the ratio between CuSO4 . 5H2O and Cu(NH3)4SO4 . H2O is 1 : 1. Therefore 0.02 moles of CuSO4 . 5H2O will give 0.02 moles of Cu(NH3)4SO4 . H2O. By using the formula m = Mr x n the theoretical yield can be calculated:

n = 0.02

Mr = 246

m = 0.02 x 246 = 4.92 g

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

The yield in percentage can be calculated by the “formula”: actual mass / expected mass. 4.82 / 4.92 ï¿½ 97.9% (3 s.f)

Because the difference is so small (2.1%) the experiment can be considered successful. The difference could have been caused by different things like: a small measurement mistake, a little bit was spilt or not transferred when the solution was held in the Buchner flask.