# What does “percentage by mass” mean in the context of a compound dissolved in a solution?

Copper can be recovered from low-grade ores by 'leaching' the ore with dilute $\ce{H2SO4}$, which converts the copper compounds in the ore into $\ce{CuSO4(aq)}$. In a $\pu{50 cm^3}$ leach solution, the concentration of copper ions is $\pu{7.8 x 10^-3 mol/dm^3}$. Find the percentage by mass of copper in the leach solution.

Isn't it impossible to find the percentage by mass in this case? We don't know the total mass of the leach solution, and it could be anything. Is the question wrong, or is the definition of percentage by mass different than what I'm using?

Also, how does $\ce{H2SO4}$ react with copper compounds to form $\ce{CuSO4}$? I thought copper couldn't reduce hydrogen ions.

is the definition of percentage by mass different than what I'm using?

Basically, the questioner wants you to calculate mass by volume percentage contrary to your assumption that the question asks for a mass by mass percentage. The existence of a lot of terms referring to the same mass percentage can be a common confusion.

Nevertheless, as a rule of thumb, you'd say, I advise you to go for the calculation of mass by volume percentage unless the questioner explicitly mentions or you've been given enough data.

Find the percentage by mass of copper in the leach solution.

Now that you already have the concentration of $\ce{Cu^2+}$ ions with you, you just need to calculate the number of moles, then the mass and finally divide it with the volume of the mixture.

Also, how does $\ce{H2SO4}$ react with copper compounds to form $\ce{CuSO4}$? I thought copper couldn't reduce hydrogen ions.

Oxidic sources of copper are well known for their amenability to sulfuric acid leaching, for eg. azurite, malachite, tenorite, cuprite, chryosocolla, and brochantite dissolve easily in dilute sulfuric acid. Note that too dilute concentration is refrained from since we don't want hydrous ferric oxides to precipitate (they can adversely effect copper extraction).

And as a side-note, if you remove copper sulfides and iron sulfides, the $\ce{H2SO4}$ requirement would be about five times the weight of the dissolved copper (concentrated $\ce{H2SO4}$ is a good oxidizer).

Interestingly, this question was asked in University of Cambridge international examinations October/November 2010, your question is perhaps a simplification of this:

• Thank you! And yes, that is the question I intended to ask :) – John May 22 '17 at 6:11