For an ordered groupoid  $G$, we denote by $G^0$ the ordered

groupoid arising from $G$ by the adjunction of a zero element. 

We first prove the following: If $S$ is an ordered groupoid, 

then there exists a proper prime ideal of $S$ if and only if 

there exists and ordered group $G$ and a mapping $f$ of $S$ 

onto $G^0$ which is a homomorphism. The following question 

arises: Given an ordered groupoid $S$ under what conditions 

there exists and ordered group $G$ such that $S$ is isomorphic 

to $G^0$ ? We prove the following: If $S$ is an ordered 

groupoid with a zero element $\theta$ such that 

$S\backslash \{\theta \}$ is a subgroup of $S$, then there 

exists an ordered group $G$ such that  $S$ is isomorphic to 

$G^0$. The converse statement also holds: If $S$ is an 

ordered groupoid and $G$ an ordered group such that $S$ is 

isomorphic to $G^0$, then there exists a zero element $\theta$ 

of $S$ such that $S\backslash \{\theta \}$ is a subgroup of $S$.