= Atomic number
{title2=$Z$}
{wiki}
The <atomic number> $Z$ is the number of <protons> in an <atomic nucleus>, and fixes the <chemical element>. It is a count, with no units. An <atomic nucleus> with six <protons> is always <carbon>, regardless of how many <neutrons> accompany them or how many <Electrons> surround it. See the https://goldbook.iupac.org/terms/view/A00499[IUPAC Gold Book, atomic number].
Three independent bookkeeping questions become easy once their symbols are kept separate. Let $N$ count <neutrons>, $n_e$ count <Electrons>, $A$ be the <mass number>, and $Q$ be the net <electric charge>. A <proton> carries <electric charge> $+e$ and an <Electron> carries $-e$, where $e$ is the positive <elementary charge>; a <neutron> carries zero net <electric charge>. Hence
$$
A=Z+N,\qquad N=A-Z,\qquad Q=(Z-n_e)e,\qquad n_e=Z-Q/e.
$$
The reasoning is just counting: add <protons> and <neutrons> for $A$, and subtract negative <Electron> contributions from positive <proton> contributions for $Q$. A neutral <chemical atom> has $Q=0$, so $n_e=Z$. For a positively charged <ion>, fewer <Electrons> remain. Notation ${}^{A}_{Z}\mathrm X$ gives the <mass number> at upper left and <atomic number> at lower left, with $\mathrm X$ standing for the <chemical element>; a right superscript such as $+$ specifies the <electric charge> in units of $e$. See https://openstax.org/books/chemistry-2e/pages/2-3-atomic-structure-and-symbolism[Flowers et al., Chemistry 2e, atomic structure and symbolism].
For neutral <carbon-12>, ${}^{12}_{6}\mathrm C$, the <atomic number> gives six <protons>, the subtraction $N=12-6=6$ gives six <neutrons>, and neutrality gives six <Electrons>. Neutral <carbon-13>, ${}^{13}_{6}\mathrm C$, instead has $N=13-6=7$ <neutrons>, with the same six <protons> and six <Electrons>. Thus these two <isotopes> occupy the same <carbon> box in the <periodic table>.
For neutral <sodium> of <mass number> $23$, ${}^{23}_{11}\mathrm{Na}$, the counts are eleven <protons>, $23-11=12$ <neutrons>, and eleven <Electrons>. Removing one <Electron> gives the <ion> ${}^{23}_{11}\mathrm{Na}^{+}$: $Q/e=1$, so $n_e=11-1=10$. Both its <atomic number> and its twelve <neutrons> stay unchanged. This is <ionization>, not conversion into the <chemical element> <neon>, even though a neutral <neon> <chemical atom> also has ten <Electrons>.
\Image[/atomic-number-isotopes-and-ions.png]
{title=<Atomic number> fixes <chemical element> identity: changing <isotopes> changes <neutrons>, while forming a <sodium> <ion> changes <Electrons>}
{description=The dots are counting tokens, with <protons> and <neutrons> in an <atomic nucleus> inventory and <Electrons> in a separate inventory. Their layout does not depict physical positions, sizes, or orbits. Neutral <carbon-12> and <carbon-13> differ by one <neutron>; neutral <sodium> of <mass number> 23 and its singly positive <ion> differ by one <Electron>.}
This bookkeeping is useful whenever a symbol must be translated into a physical inventory. <Mass number> answers how many <protons> plus <neutrons> there are; <electric charge> answers the imbalance between <protons> and <Electrons>; <atomic number> alone answers which <chemical element> it is. Keeping those questions distinct prevents an <isotope> or an <ion> from being mistaken for a different <chemical element>.
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