Question

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

Answer 1

Approx. `20` `g`.

Explanation:

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

`CuCO_3(s) + Delta rarrCuO(s) + CO_2(g)uarr`

We start with `(31*g)/(123.57*g*mol^-1)` `=` `0.25` `mol`.

Copper oxide, `CuO`, has a formula mass of `79.6` `g*mol^-1`, of which `0.25` `mol` 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, `Cu_2(OH)_2CO_3`. Given the context of the question, an identity of `CuCO_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 `CO_2` would be the same molar quantity, but of course would have a different mass and volume.