And we don't even have to know that the element is gold. If $Z$ $=$ $79$, there are 79 positively charged nuclear particles, 79 protons, in the nucleus. If 79 nuclear charges, the NEUTRAL atom contains 79 electrons. You specified an ion, and gold forms both $Au^+$ and $Au^(3+)$ ions so $78$ and $76$ electrons respectively.
Now, it is a fact that the of gold is $196.97$ $g*mol^-1$, roughly close to $197$. Because the atomic mass is $197$ $"amu"$, there must be $197-79= 118 " neutrons"$, massive, neutrally charged "nucular" particles. Atomic mass is to a first (and second) approximation, the sum of the protons and neutrons (there will be a mixture of because of different numbers of neutrons), because electrons have negligible mass.
Now if $Z$ $=$ $78$, can you tell me the element, the number of protons, the number of neutrons, and the number of electrons?
