Final answer:
Copper(I) sulfide consists of two copper ions (Cu+) and one sulfide ion (S2-) for every formula unit of Cu2S.
Step-by-step explanation:
When we consider copper(I) sulfur, also known as copper(I) sulfide, we have to look at their chemical formula to determine the number of ions present. For copper(I) sulfide, the formula is Cu2S. This indicates that there are two copper ions (Cu+) for every sulfide ion (S2-).
Since copper(I), or cuprous ion, has a 1+ charge, and sulfur has a 2- charge, their charges balance out in a 2:1 ratio. Therefore, in copper(I) sulfide, for every one formula unit, there are two copper(I) ions present and one sulfide ion present.