Using stoichiometry to split a complex mineral
Is it possible to separate out $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ into simply the $\ce{(Mg_{x}Cu_{1-x})(CO3)}$ part that reacts with the dilute sulfuric acid to produce the carbon dioxide, the moles of which I know?



$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{CO2}$ implies $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$.
$4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2(CO3)(OH)2}$ is 0.8167 g.
Then, considering the atomic weights of C, O and H, subtract the contribution of these elements from 0.8167 g to determine the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{(Mg_{x}Cu_{1-x})2}$.
Then divide by 2 to get the mass of $4.492 * 10^{-3} \,\mathrm{mol}$ of $\ce{Mg_{x}Cu_{1-x}}$;
Then use the atomic weights of Mg and Cu to solve for $x$.