Question

2.On heating 12.4 g of copper (II) carbonate in a crucible only 7.0g of copper (II) oxide was produced. What was the % yield of copper (II) oxide ? [Cu=64,C=12,O=16]

Answer 1

Answer:

Answer: The percentage purity in copper carbonate is, 87.6 %

Explanation : Given,

Mass crude copper carbonate = 12.4 g

Mass of cooper(II) oxide = 7 g

Molar mass of copper carbonate = 123.5 g/mole

Molar mass of copper(II) oxide = 79.5 g/mole

The balanced chemical reaction will be,

CuCO_3\rightarrow CuO+CO_2CuCO

3

→CuO+CO

2

First we have to calculate the moles of CuO.

\text{Moles of }CuO=\frac{\text{Mass of }CuO}{\text{Molar mass of }CuO}=\frac{7g}{79.5g/mole}=0.088molesMoles of CuO=

Molar mass of CuO

Mass of CuO

=

79.5g/mole

7g

=0.088moles

Now we have to calculate the moles of CuCO_3CuCO

3

From the balanced reaction we conclude that

As, 1 mole of CuO produced from 1 mole of CuCO_3CuCO

3

So, 0.088 mole of CuO produced from 0.088 mole of CuCO_3CuCO

3

Now we have to calculate the mass of CuCO_3CuCO

3

.

\text{Mass of }CuCO_3=\text{Moles of }CuCO_3\times \text{Molar mass of }CuCO_3=(0.088mole)\times (123.5g/mole)=10.868gMass of CuCO

3

=Moles of CuCO

3

×Molar mass of CuCO

3

=(0.088mole)×(123.5g/mole)=10.868g

The pure mass of CuCO_3CuCO

3

= 10.868 g

Now we have to calculate the percentage purity in copper carbonate.

\%\text{ Purity in }CuCO_3=\frac{\text{Mass of pure }CuCO_3}{\text{Mass of crude }CuCO_3}\times 100=\frac{10.868g}{12.4g}\times 100=87.6\%% Purity in CuCO

3

=

Mass of crude CuCO

3

Mass of pure CuCO

3

×100=

12.4g

10.868g

×100=87.6%

Therefore, the percentage purity in copper carbonate is, 87.6