Final answer:
The reactant is copper(II) sulfate heptahydrate (CuSO4.7H2O), which loses water to form anhydrous copper sulfate when heated.
Step-by-step explanation:
The name of the reactant in the reaction where a wet 43.2 g sample of copper sulfate heptahydrate is heated until only copper sulfate remains, and water is lost, is copper(II) sulfate heptahydrate. This compound is generally shown as CuSO4.7H2O. Upon heating, the water of crystallization in the heptahydrate is evaporated, leaving anhydrous CuSO4.
The mass of water lost during the heating process is the difference between the mass of the hydrate before heating and the mass of the remaining anhydrous copper sulfate. Since 34.1 g of water was lost, it can be inferred that the compound started as a heptahydrate, which means it had seven water molecules associated with each formula unit of copper(II) sulfate.