I’m confused as I am unable to find a completely trusted way of finding coordinately bonded compounds from a bunch of other compounds.
Answer
Coordinate bonds can be distinguished from (other or ‘normal’) covalent ones experimentally.
Let’s look at the bond creation first as Newbie suggested. Typically, one would use a Lewis base and a Lewis acid to create a coordinate bond between them.
$$\ce{NH3 + BH3 -> H3N\bond{->}BH3}\tag{1}$$
In contrast, one would usually consider a normal covalent bond to be created from two radicals:
$$\ce{H3C^{.} + {}^{.}CH3 -> H3C-CH3}\tag{2}$$
But we can also create covalent $\ce{C-C}$ bonds by using a carbanion source and a carbocation source (slightly different example, because both ions are actually known and react with each other):
$$\ce{(H3C)3C+ + {}^{-}CN -> (H3C)3C-CN}\tag{3}$$
Thus, we can’t really tell a difference here. The difference in fact lies in the way a bond dissociates if the bond dissociation enthalpy $D = \Delta_\mathrm{D} H^0$ is provided. Standard covalent bonds will dissociate homolytically by the following equation:
$$\ce{H3C-CH3 -> H3C^. + ^.CH3}\tag{4}$$
A coordinate bond will, however, dissociate heterolytically thus liberation the original Lewis acid and base:
$$\ce{H3N\bond{->}BH3 -> NH3 + BH3}\tag{5}$$
In cases such as the ammonium ion ($\ce{NH4+}$) or say $\ce{[AlCl4]-}$ this means we have a certain number of normal bonds and another certain number of coordinate bonds (three and one, respectively in both examples). Although they are all between the same atoms, we can distinguish if we remember that neither ion can exist by itself, there will always be a corresponding counterion. Thus, if we consider $\ce{NH4Cl}$ or $\ce{Na[AlCl4]}$, we realise how the first bond is broken differently:
$$\begin{align}\ce{NH4Cl &-> NH3 + HCl}\tag{6}\\[0.6em] \ce{Na[AlCl4] &-> NaCl + AlCl3}\tag{7}\end{align}$$
And all further bonds are cleaved homolytically. We gain two different bond dissociation energies, one corresponding to the coordinate bond and one corresponding to the remaining bonds. (Note that salts such as $\ce{AlCl3}$ or $\ce{NaCl}$ are cleaved homolytically by heating in the gas phase — flame-colouring would not happen if we had $\ce{Na+}$ in the flame.)
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