Question

on heating 12.4 grams of copper (II) carbonate in a crucible only 7.0g of copper(II) oxide was produced. what was the percentage purity in copper carbonate ?

Answer 1

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,

$CuCO_3\rightarrow CuO+CO_2$

First we have to calculate the moles of CuO.

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

Now we have to calculate the moles of $CuCO_3$

From the balanced reaction we conclude that

As, 1 mole of CuO produced from 1 mole of $CuCO_3$

So, 0.088 mole of CuO produced from 0.088 mole of $CuCO_3$

Now we have to calculate the mass of $CuCO_3$.

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

The pure mass of $CuCO_3$ = 10.868 g

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

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

Therefore, the percentage purity in copper carbonate is, 87.6 %

Answer 2

Answer:

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