Determining a molecular formula of the catalyst for polymerisation of butadiene
Despite the absurd lack of data, this problem attracted my attention, probably because it closer resembles a real-life challenge rather than a textbook problem.
It looks like it's an adaptation of the problem from Schaum’s Outline of Theory and Problems of College Chemistry [1, p. 39]:

3.37 What is the empirical formula of a catalyst that can be used in the polymerization of butadiene if its composition is $23.3\%$ $\ce{Co},$ $25.3\%$ $\ce{Mo},$ and $51.4\%$ $\ce{Cl}.$
Ans. $\ce{Co3Mo2Cl11}$

Here we are dealing with zero knowledge as to what the remaining element(s) are, but with a couple of assumptions we actually can come up with an answer from the textbook, and an answer on top of that with a justification.
Let's dive in and assume we are dealing with an unknown compound with the following formula:
$$\ce{Co_xCl_y}\sum_{i=1}^N \ce{El}_{i,z_i}$$
where we account for $i$ other elements $\ce{El_$i$}$ with the respective coefficients $z_i.$
The first assumption required to advance with the solution is to treat the compound as stoichimetric. e.g. $x, y, z_i \in\mathbb{N}.$
This would immediately allow us to find the exact values for both $x$ and $y$:
$$y : x = \frac{m(\ce{Cl})}{M(\ce{Cl})} : \frac{m(\ce{Co})}{M(\ce{Co})} = \frac{\pu{25.7 g}}{\pu{35.45 g mol-1}} : \frac{\pu{11.65 g}}{\pu{58.93 g mol-1}} = 3.67\tag{1}$$
To satisfy $x, y \in\mathbb{N},$ and taking into account the &quot;hint&quot; decimal part $.67$ (which implies a triple factor to an integer):
$$x = 3, 6, 9, \ldots, 3n~(n\in\mathbb{N})\tag{2}$$
To pinpoint the exact allowed value of $x$ (and $y,$ as $y = 3.67x),$ we can use the total molar mass:
$$M = x × M(\ce{Co}) + 3.67x  × M(\ce{Co}) + \sum_{i=1}^N z_iM_i\tag{3}$$
$$\pu{759 g mol-1} = x × \pu{58.93 g mol-1} + 3.67x × \pu{35.45 g mol-1} + \sum_{i=1}^N z_iM_i\tag{3a}$$
$$\sum_{i=1}^N z_iM_i = (759 - 189x)~\pu{g mol-1}\tag{3b}$$
Since molar mass is a positive number, $759 - 189x &gt; 0$ and the only $x$ to satisfy this criteria would be $x = 3.$
Accordingly, $y = 3.67 × 3 = 11,$ and at this point we are left with the following formula:
$$\ce{Co3Cl11}\sum_{i=1}^N \ce{El}_{i,z_i}$$
and the remaining sum
$$\sum_{i=1}^N z_iM_i = (759 - 189 × 3)~\pu{g mol-1} = \pu{192 g mol-1} \tag{3c}$$
The second assumption is that we are dealing with the common oxidation numbers of the elements, no exotic stuff.
With this in mind, and knowing that cobalt catalyst used in polymerization are vastly only $\ce{Co^{II}}$ and $\ce{Co^{III}}$ species, we can deduce the possible charge of the remaining sum:
$$(\ce{Co3Cl11})^{q-}\left(\sum_{i=1}^N \ce{El}_{i,z_i}\right)^{q+}$$
Since it's likely only $\ce{Cl-},$ the $q$ can adopt only the values defined by $\ce{Co^{II}}:\ce{Co^{III}}$ ratio.
For the total of three cobalt(II,III) atoms $q = 2,3,4,5.$
This will assist us in proposing missing element(s), and one can start to iterate over numbers.
The third assumption is that $i = 1$ and there is only one extra element with single coefficient $z_1$.
The table below summarizes possible outcomes:
$$
\begin{array}{ccccc}
\hline
z_1 &amp; M_1/\pu{g mol-1} &amp; \ce{El_1} &amp; M(\ce{El_1})/\pu{g mol-1} &amp; \text{Formula Example} \\
\hline
1 &amp; 192 &amp; \ce{Ir} &amp; 192.22 &amp; \ce{Co^{II}2Co^{III}Ir^{IV}Cl11} \\
  &amp; &amp; &amp; &amp; \ce{Co^{II}Co^{III}2Ir^{III}Cl11} \\
2 &amp; 96 &amp; \ce{Mo} &amp; 95.95 &amp; \ce{Co^{II}3Mo^{II}Mo^{III}Cl11} \\
  &amp; &amp; &amp; &amp; \ce{Co^{II}2Co^{III}Mo^{II}2Cl11} \\
3 &amp; 64 &amp; \ce{Cu} &amp; 63.55 &amp; \ce{Co^{II}Co^{III}2Cu^I3Cl11} \\
  &amp; &amp; &amp; &amp; \ce{Co^{II}3Cu^{I}Cu^{II}2Cl11} \\
4 &amp; 48 &amp; \ce{Ti} &amp; 47.87 &amp; \ce{Co^{?}3Ti^{?}4Cl11} \\
\hline
\end{array}
$$
Starting with $z = 4,$ there doesn't seem to be an appropriate set of oxidation numbers for the element to comply with the charge balance, so I'd say the candidates for the third elements are only iridium, molybdenum and copper.
Judging from the application (catalysis) as well as from the typical oxidation states, I'd actually propose iridium cobalt chloride $\ce{Co3IrCl11}$ as the answer, yet $\ce{Co3Mo2Cl11}$ would also be possible (AFAIK, molybdenum(II,III) isn't a common composition for halide salts), and $\ce{Co3Cu3Cl11}$ would be a stretch both due to higher deviation from the declared molar mass and presence of copper(I).
References

Rosenberg, J. L.; Epstein, L. M. Schaum’s Outline of Theory and Problems of College Chemistry, 8th ed.; Schaum’s outline series; McGraw-Hill: New York, 1997. ISBN 978-0-07-053709-5.