Answer
Chalcopyrite ($CuFeS_{2}$) has the lowest amount of copper.
Work Step by Step
The mass of a single atom is equal to its molar mass multiplied by the atomic mass unit.
The percent composition by mass is defined as the mass of a component divided by the mass of the entire system.
We also have to take into account all the atoms of a single element that are in a compound.
In order to find which compound has the lowest percent by mass of copper we have to calculate the percent by mass of copper in every compound.
Azurite, $Cu_{3}(CO_{3})_{2}(OH)_{2}$
% $Cu =\frac{m(Cu)}{m(Cu_{3}(CO_{3})_{2}(OH)_{2})}=\frac{3\times M(Cu)amu}{M(Cu_{3}(CO_{3})_{2}(OH)_{2})amu}=\frac{3 \times 63.55 amu}{344.686 amu} = 55.31$ %
Chalcocite, $Cu_{2}S$
% $Cu =\frac{m(Cu)}{m(Cu_{2}S)}=\frac{2\times M(Cu)amu}{M(Cu_{2}S)amu}=\frac{2 \times 63.55 amu}{159.16 amu} = 79.86$ %
Chalcopyrite, $CuFeS_{2}$
% $Cu =\frac{m(Cu)}{m(CuFeS_{2})}=\frac{M(Cu)amu}{M(CuFeS_{2}) amu}=\frac{63.55 amu}{183.52 amu} = 34.63$ %
Covelite, $CuS$
% $Cu =\frac{m(Cu)}{m(CuS)}=\frac{M(Cu)amu}{M(CuS) amu}=\frac{63.55 amu}{95.61 amu} = 66.47$ %
Cuprite, $Cu_{2}O$
% $Cu =\frac{m(Cu)}{m(Cu_{2}O)}=\frac{2\times M(Cu)amu}{M(Cu_{2}O) amu}=\frac{2\times 63.55 amu}{143.1 amu} = 88.82$ %
Malachite, $Cu_{2}CO_{3}(OH)_{2}$
% $Cu =\frac{m(Cu)}{m(Cu_{2}CO_{3}(OH)_{2})}=\frac{2\times M(Cu)amu}{M(Cu_{2}CO_{3}(OH)_{2}) amu}=\frac{2\times 63.55 amu}{221.126 amu} = 57.48$ %
Therefore, chalcopyrite has the lowest amount of copper.
