ANALYSIS OF COPPER OXALATE HYDRATE 1. (a) What happens to copper oxalate hydrate

Question

ANALYSIS OF COPPER OXALATE HYDRATE 1. (a) What happens to copper oxalate hydrate when it is heated in the test tube for this experiment? (b) What is the chemical reaction and the mole ratio between oxalate and permanganate in the titration reaction? Chemical reaction: Mole ratio: C2042: MnO4 (c) How is the percent water in copper oxalate hydrate determined in this experiment? 2. Describe the procedure for this experiment. This can be in an outline format or a flow chart or paragraph format. It should not be copied directly from the manual. (This should be between ½ and one page, you may use the back of the page.) 3. Briefly discuss the safety precautions related to this experiment. (chemical and equipment related)

Explanation / Answer

1.(a)copper oxalate usually decompose to metalloc copper under high temperature. But since its given in a test tube the water of hydration get escaped and it becomes dehydrated copper oxalate.

b. Titration would b a redox titration hence Reaction is a redox reaction. There will be a change in oxidation state of Mn

and C. The reaction would be

MnO4(1-) + C2O4(2-)------>Mn(+2) +CO2 here the oxidation state of Mn changes from +7 to +2 and that of C changes from +3 to +4

Mole ratio would be 2:5 for Mno4(1-):C2O4 (2-)

(c) since the water of hydration escapes from the oxalate. One can trap it using previously weighed anhy.cacl2 crystals. By taking the weight after it is hydrated. The difference between the weight can give the amount of water liberated.

2. Heat the copper oxalate in excess of air in closed tube which is connected to a u- tube having previously weighed anhy cacl2 which adsorbs hydrated water liberated. Weight once agin and take their difference. The difference gives amount of water liberated. Divide this amount of water liberated by molar mass of water and multiply by 100 which gives you the percentage of water liberated.

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