Which Equation Is Correct?

Aim: To design and plan an experiment to measure a volume of gas that will prove which of the following two equations is correct.

Equation 1: 2CuCO3(s) → Cu2O(s) + 2CO2(g) + ½O2(g)

Ratio of 4:5

Equation 2: CuCO3(s) → CuO(s) + CO2(g)

Ratio of 1:1

A known mass of copper carbonate will be heated, and the volume of gas given off will be measured. From this, the number of moles of gas can be calculated. This will allow you to find out which equation is correct, because it is stated that each reaction produces a different number of moles of gas when heated.

Safety: CuCO3 is harmful and irritant. Goggles and a lab coat must be worn.

Apparatus/Chemical List:

Conical Flask – 250cm3
100cm3 Gas Syringe.
Clamp Stand
Clamp and Boss
Delivery Tube with bung
Boiling Tube
Bunsen Burner
Top-Pan Balance
Spatula
Copper Carbonate, CuCO3

Introduction: As the gas syringe can hold a maximum volume of 100cm3 I am going to work out how much CuCO3 to use so that the maximum volume isn’t reached and error is reduced. The aim is to collect about 90cm3 of gas.

In standard conditions, the volume of one mole of gas is 24dm3. ...

This is a preview of the whole essay

Safety: CuCO3 is harmful and irritant. Goggles and a lab coat must be worn.

Apparatus/Chemical List:

Conical Flask – 250cm3
100cm3 Gas Syringe.
Clamp Stand
Clamp and Boss
Delivery Tube with bung
Boiling Tube
Bunsen Burner
Top-Pan Balance
Spatula
Copper Carbonate, CuCO3

In standard conditions, the volume of one mole of gas is 24dm3. From this the amount gas each reaction will produce can be worked out, which allows you to work out how much CuCO3 to use.

90/24000 = 0.00375 moles of gas.

As the reactant and products in equation 2 are in the same ratio, you can see that equation 1 will produce more gas as the ratio of that equation is 4:5.

Ratio of 4:5 – 0.00300 : 0.00375 moles of gas.

Therefore, to work out the mass of the CuCO3 the following equation should be used:

Number of moles = Mass/Molecular Mass

Mass = Number of moles x Molecular Mass

0.00300 x 123.5 = 0.3705g.

This means that 0.3705g of CuCO3 will give me 90cm3 of gas, so the amount of CuCO3 to be used is 0.37g, which is 0.00296 moles.

Method:

Collect apparatus and chemicals and set up equipment.
Measure 0.37g of CuCo3 on the balance and put into boiling tube.
Attach gas syringe to the delivery tube and connect the boiling tube
Light the Bunsen burner.
Heat the CuCO3 in the boiling tube until no more gas is released and the powder has become black.
Repeat twice for reliable and accurate results.

Using the Results:

The volumes of gas produced can be used to work out which equation is correct. To find the number of moles of gas given off, the volume needs to be divided by 24dm3.

Number of moles = Volume (dm3) / 24 (dm3)

Equation 1 – Ratio 4:5 – 0.00296 : 0.00375 x 24000 = 90cm3

Equation 2 – Ratio 1:1 – 0.00296 : 0.00296 x 24000 = 71.04cm3

This means that if 90cm3 of gas is produced, equation 1 is correct. However, if 71.04cm3 of gas is produced, equation 2 is correct.

Sources Used/Bibliography:

Chemistry 1, Cambridge University Press, pages 25-29

Essential AS Chemistry, Lister and Renshaw, page 21

http://en.wikipedia.org/wiki/Copper%28II%29_carbonate