= Nucleoside monophosphate
A \b[nucleoside monophosphate] is a <nucleotide> with one phosphate group attached to a <nucleoside>. The <nucleoside> supplies a <nucleobase> linked to a <sugar>; the monophosphate specifies the phosphate count. Attachment position and sugar identity still matter. A common example is <adenosine monophosphate>, with a phosphate ester at the 5′ position of <ribose>; ChEBI identifies it specifically as adenosine 5′-monophosphate.<Adenosine 5'-monophosphate in ChEBI>[CHEBI:16027, Definition and structure]{ref}
The corresponding acid structure has <molecular formula> $\mathrm{C_{10}H_{14}N_5O_7P}$. Its single phosphorus atom provides a quick count, but the <chemical structure> is what identifies the 5′ attachment. The neutral acid formula is a stated representation; proton loss changes charge and hydrogen count without adding a second phosphate group.<Adenosine 5'-monophosphate in ChEBI>[Formula, Net Charge and structure]{ref} A <nucleoside diphosphate> instead contains a linked pair of phosphate groups.
A useful reaction is the <adenylate kinase> interconversion
$$
\mathrm{ATP}+\mathrm{AMP}\rightleftharpoons2\mathrm{ADP},
$$
where the abbreviations denote <adenosine triphosphate>, <adenosine monophosphate> and <adenosine diphosphate>. The phosphate-unit balance is $3+1=2+2$: a phosphate group is transferred rather than created.<Enzyme Nomenclature by the IUBMB>[EC 2.7.4.3, Reaction: ATP + AMP = 2 ADP; https://iubmb.qmul.ac.uk/enzyme/EC2/7/4/3.html]{ref}
For an original reaction-extent example, start with $0.5\,\mathrm{mmol}$ of ATP, $0.3\,\mathrm{mmol}$ of AMP and no ADP. A forward extent of $0.2\,\mathrm{mmol}$ leaves $0.3\,\mathrm{mmol}$ ATP, $0.1\,\mathrm{mmol}$ AMP and produces $0.4\,\mathrm{mmol}$ ADP. Both sides of this bookkeeping contain $1.8\,\mathrm{mmol}$ of phosphate units. This is a <stoichiometry> calculation for a specified extent, not a prediction of <chemical equilibrium>.
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