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Chemistry Essay

1712 words - 7 pages

Chemistry
Aim: I am trying to find out whether the equation 2CuCO3 ® Cu2O + 2CO2
+ ½O2 or the equation CuCO3 ® CuO + CO2 is correct for the
decomposition of copper carbonate.

Introduction: This is my background research, this information may
help me when I am trying to see which equation is correct. I have used
the book Chemistry 1 by OCR as my reference.

Relative Atomic Mass: This is the mass of an atom of the element
relative to the mass of an atom of carbon-12 which has a mass of
exactly 12.

Relative Molecular Mass: This is the mass of a molecule of a compound
relative to an atom of carbon-12.

Mole: This is the unit of an atom of substance. One mole of a
substance is the mass that has the same number of particles
(Molecules, Ions or Electrons) as there are atoms in exactly 12g of
carbon-12

Avagadro’s Constant: This is the number of atoms or molecules in one
mole of a substance. It is calculated using (L = 6.01 x 1023)

I will also include the equation for number of moles and volume of gas
formed. I shall use my class notes as my reference.

Number of Moles of gas = Volume of Gas (cm3) = Volume of Gas
(cm3)

Molar Volume (cm3) 24000cm3

Volume of Gas = Number of Moles x Molar Volume (cm3)

= Number of Moles x 24000 cm3

Planning:

Copper has two oxides; Cu2O and CuO

Copper carbonate, CuCO3 decomposes on heating to form one of these
oxides and an equation can be written for each possible reaction:

Equation 1:

2CuCO3 ® Cu2O + 2CO2 + ½O2

Mol Ratio 1 : 0.5 : 1 : 0.25

Total number of mols of gas in equation one: 1.25

Equation 2:

CuCO3 ® CuO + CO2

Mol Ratio: 1 : 1 : 1

Total number of mols of gas in equation two: 1

The total number of mols of gas in equation one is 1.25 and in
equation two is 1

\ I need to choose a mass of CuCO3 that produces a suitable amount of
gas to fit a 100cm3 gas syringe.

Equation 1 produces the most gas.

Number of moles = Volume of Gas (cm3) = 80cm3

Molar Volume (cm3) 24000cm3

= 0.0033 moles of gas

1.25 moles of gas is produced from 1 mole of copper carbonate (CuCO3)

» 0.0033 moles of gas is produced from (0.0033)

1.25

= 0.00264 moles of Copper Carbonate

m = n x M \ M[CuCO3] = [63.5 + 12 + (3 x 16)] = 123.5g mol 1

\ 0.00264 x 123.5g mol 1

= 0.33g of CuCO3

I can now work out how much gas 0.33g of CuCO3 will give in equation
2:

n = m = 0.33g = 0.00267 moles of Copper
Carbonate

M 123.5g mol 1

1 mole of CuCO3 gives 1 mole of gas

\ 0.00267 moles of CuCO3 gives 0.00267 moles of gas.

Volume of gas (cm3) = number of moles x molar volume (cm3)

...

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