# What mass of copper oxide will result after a 31*g mass of copper carbonate is subjected to fierce and prolonged heating?

Feb 4, 2016

Approx. $20$ $g$.

#### Explanation:

Metal carbonates undergo decomposition under heating to give the metal oxide and carbon dioxide, i.e.:

$C u C {O}_{3} \left(s\right) + \Delta \rightarrow C u O \left(s\right) + C {O}_{2} \left(g\right) \uparrow$

We start with $\frac{31 \cdot g}{123.57 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.25$ $m o l$.

Copper oxide, $C u O$, has a formula mass of $79.6$ $g \cdot m o {l}^{-} 1$, of which $0.25$ $m o l$ represents a $20$ $g$ mass. Here we use the reaction stoichiometry above: one equiv of copper salt gives one equiv of copper(II) oxide.

Note that copper carbonate is as far as anyone knows, $C {u}_{2} {\left(O H\right)}_{2} C {O}_{3}$. Given the context of the question, an identity of $C u C {O}_{3}$ is assumed. Typically, at A level, a student would be asked to estimate the volume of carbon dioxide gas evolved under standard conditions. This $C {O}_{2}$ would be the same molar quantity, but of course would have a different mass and volume.