Question

A 2.3995 g mixture containing malachite and an inert ingredient lost 0.2997 g of H₂O and CO₂ when decomposed by heating to constant mass.

A) how many moles of malachite in the mixture produced this mass of CO₂ and H₂O?

B) How many grams of malachite were present in the mixture?This is the equation:CuCO₃*Cu(OH)₂(s)--> 2 CuO (s)+H₂O (g) +CO₂

Answer 1

a) Moles of malachite in mixture :

Decomposition of malachite:

$$ \mathrm{Cu}_{2} \mathrm{CO}_{3}(\mathrm{OH})_{2} \longrightarrow 2 \mathrm{CuO}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} $$

That is, 1 mole of malachite decompose to produce 1 mole of water and 1 mole of \(\mathrm{CO}_{2}\).

That is \(221.12\) (MW of malachite) deocmpose to produce 18 (MW of water) \(\mathrm{g}\) of water and

44 ( MW of \(\mathrm{CO}_{2}\) ) g of \(\mathrm{CO}_{2}\). That is from \(221.12 \mathrm{~g}\) of malachite, total 62 g mass lost.

1) Percent of water lost \(=(18 \mathrm{~g} / 62 \mathrm{~g}) \times 100=29.03 \%\)

2) Percent of water \(\mathrm{CO}_{2}=(44 \mathrm{~g} / 62 \mathrm{~g}) \times 100=70.96 \%\)

All this for 1 mole of malachite.

Now, in from mixture, tolat mas lost is \(0.2997 \mathrm{~g}\).

1) Mass of water lost \(=(29.03 \times 0.2997 \mathrm{~g}) / 100=8.7002 / 100=0.0870 \mathrm{~g}\) water.

Moles of water lost = Mass / Molar mass \(=0.0870 \mathrm{~g} / 18 \mathrm{~g} / \mathrm{mol}=0.004833\) moles.

2) Mass of \(\mathrm{CO}_{2}\) lost \(=(70.96 \times 0.2997 \mathrm{~g}) / 100=21.266 / 100=0.2126 \mathrm{~g} \mathrm{CO}_{2}\)

Moles of \(\mathrm{CO}_{2}=\) Mass \(/\) Molar mass \(=0.2126 \mathrm{~g} / 44 \mathrm{~g} / \mathrm{mol}=0.004833\) moles.

That as per balanced equation, when malachite decomposes, it produces equal moles of water and \(\mathrm{CO}_{2}\). Therefore the stoichiometric ratio of water and \(\mathrm{CO}_{2}\) with malachite is \(1: 1\). That moles of water and \(\mathrm{CO}_{2}\) produces is equal to moles of malachite in mixture. So, moles of malachite in mixture \(=0.004833\) moles.

b) Mass of malachite in mixture \(=\) Moles \(\mathrm{x}\) molar mass \(=0.004833\) moles \(\times 221.12 \mathrm{~g} / \mathrm{mol}=1.0686 \mathrm{~g}\)

c) Mass percent of malachite \(=(\) Actual mass \(/\) Total mass ) \(\times 100\)

$$ \begin{array}{l} =(1.0686 \mathrm{~g} / 2.3995 \mathrm{~g}) \times 100 \\ =0.4453 \times 100 \end{array} $$

\(=44.53 \%\)

Summary:

a) Moles of malachite \(=0.004833\) moles. moles.

b) Mass of malachite \(=1.0686 \mathrm{~g}\)

c) Mass percent of malachite in mixture \(=44.53 \%\)