Electrolysis of copper sulphate using copper electrodes: Mass lost from anode is greater than mass deposited on cathode
$\pu{3 V}$ is too high. Kinetics of deposition of $\ce{Cu}$ is saturated, so another electroactive system kicks on, creation of hydrogen. Similarly for anode. Try to evaluate volt-ampere characteristic of the system and stay on the current plateau before gases start to be created.
Generally, is is more advantageous to use a power source keeping constant current rather than constant voltage.
If gases are being evolved, you cannot expect the dissolved and deposited mass of copper would match, as various portions of the current are spent on gases, instead of copper. The current density that Maurice suggests is a good one.
For low enough voltage, copper deposition + dissolution can absorb the current. At particular current density, one of these reactions is saturated and cannot take more current any more, being diffusion controlled. Then either hydrogen, either oxygen starts to be created, depending on the electrode where it occurs first. When it happens for the other reaction as well, the other gas gets created as well.

Each electrode has a voltampere characteristic given by I=f(E) function, that has a shape of irregular rounded step-wise slope. When a cathodic potential gets low enough value for a given compound/ion, current starts to raise. When reaction gets saturated, i becomes being controlled by ion diffusion and there is zero concentration at the electrode. Then the current reaches a plateou. The plateou ends at a potential where another redox system starts to interact. Similarly for anode.
When an external voltage is applied, electrodes gain such potentials to meet 2 conditions:

The potential difference equals to the external voltage.
Currents of both electrodes, according to the above function, are equal but the sign.

Lower current density means more positive cathode potential, where $\ce{Cu^2+}$ gets selectively reduced. Higher density means lower cathode potential where both $\ce{Cu^2+}$ and $\ce{H3O+}$ may get reduced. It is like if you used mildly reducing reagent, reducing just copper ions versus strongly reducing reagent, reducing also hydronium ions. Similarly for the anode.
The only necessery thing is the cathodic current matches the anodic current. Processes on both electrodes are independent. There is no &quot;quantum entanglement&quot; between these processes. If $\ce{Cu^2+}$ diffusion and $\ce{Cu}$ deposition can absorb all the cathodic current, there is no need of hydrogen creation.
Try to measure the steady current as a function of the applied voltage. You can use a small resistor is a serie with a cell, measuring voltage on it, which is equivalent to the passing current as $I = U_\mathrm{R} / R.$ Find the first current plateau. Or the &quot;like to be a plateau, just if the 2 voltage steps were not so close together.&quot;
Choose the voltage from this plateau, but be aware it may change with the change of the cell geometry or electrolyte composition.

Even in cased hydronium ions do not participate on electrode reactions directly, they can participate indirectly by influencing I=f(U) characteristics of the cell via:

Keeping or affecting chemical state of involved ions. ( E.g. pushing equilibrium $\ce{[Cu(H2O)4]^2+ + H2O &lt;=&gt; [Cu(H2O)3(OH)]+ + H3O+}$ to the left.
Balancing the local gradient due charge disbalance
Increasing conductivity and therefore the solution potential difference, that subtracts from the applied voltage.