# Determining a molecular formula of the catalyst for polymerisation of butadiene

I've got the following question in a workbook which I need to solve:

A $$\pu{50 g}$$ sample of an experimental catalyst used in the polymerisation of butadiene is made up of $$\pu{11.65 g}$$ of $$\ce{Co}$$ and $$\pu{25.7 g}$$ of $$\ce{Cl}$$. If the molar mass of the compound is $$\pu{759 g/mol},$$ find the molecular formula of the catalyst.

I attempted to first find the percentage by mass of each given element by dividing their masses ($$\pu{11.65g}$$ and $$\pu{25.7 g},$$ respectively) by the mass of the entire catalyst $$(\pu{50 g}).$$ From there, I divided both percentages ($$23.3$$ and $$51.4)$$ by the smallest one, and multiplied the results $$(1$$ and $$2.2)$$ by $$5$$ to get two whole numbers. The formula I ended up with was $$\ce{Co5Cl11}.$$

Now that made sense to me, but somehow the correct answer is $$\ce{Co3Mo2Cl11}$$. I don't understand how I'm supposed to figure out that $$\ce{Mo}$$ is required? Could somebody please shed some light on this for me?

• At least you should know there is a third element in the catalyst other than $\ce{Co}$ and $\ce{Cl}$. You should include that in your calculations. Nevertheless, this question is very poorly written. It is missing one important information. Commented Dec 24, 2019 at 1:58
• I agree that the question is poorly written. Does this mean there is no way to determine that the third element is Mo? Or are you inferring that this question impossible because there is too little data? Commented Dec 24, 2019 at 2:38
• It is still possible to solve, but cannot pinpoint $\ce{Mo}$ is the third element. If you show your work, we may able to help you. Otherwise, this may be closed as a homework question. Commented Dec 24, 2019 at 4:43
• You say "$\pu{11.65g}$ of $\ce{Co}$ and $\pu{25.7g}$ of $\ce{Co}$." -- It seems that one of the masses should be for $\ce{Co}$ and the other for $\ce{Mo}$.
– MaxW
Commented Dec 24, 2019 at 7:09
• @MaxW This was a typo introduced by copy-pasting elements during third-party editing. OP included 11.65 g of Co and 25.7 g of Cl (fixed now). Commented Dec 24, 2019 at 11:14

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 "hint" 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, $$759 - 189x > 0$$ 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:

$$\begin{array}{ccccc} \hline z_1 & M_1/\pu{g mol-1} & \ce{El_1} & M(\ce{El_1})/\pu{g mol-1} & \text{Formula Example} \\ \hline 1 & 192 & \ce{Ir} & 192.22 & \ce{Co^{II}2Co^{III}Ir^{IV}Cl11} \\ & & & & \ce{Co^{II}Co^{III}2Ir^{III}Cl11} \\ 2 & 96 & \ce{Mo} & 95.95 & \ce{Co^{II}3Mo^{II}Mo^{III}Cl11} \\ & & & & \ce{Co^{II}2Co^{III}Mo^{II}2Cl11} \\ 3 & 64 & \ce{Cu} & 63.55 & \ce{Co^{II}Co^{III}2Cu^I3Cl11} \\ & & & & \ce{Co^{II}3Cu^{I}Cu^{II}2Cl11} \\ 4 & 48 & \ce{Ti} & 47.87 & \ce{Co^{?}3Ti^{?}4Cl11} \\ \hline \end{array}$$

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

1. 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.
• Thanks andselisk for the detailed answer. Even though this is a little above my level, I can make sense of most of it, so I'll mark it as correct. I think that perhaps this question was just worded incorrectly and simply doesn't contain enough information, however. Commented Dec 25, 2019 at 2:14
• Fun answer, but what about $\ce{Co^{II}2Co^{III}Mo^{II}2Cl11}$? Commented Dec 26, 2019 at 2:32
• @user5713492 Thank you, good point, I appended this one too. Commented Dec 26, 2019 at 6:50

It is no use to calculate the mass ratio of the atoms in the catalyst, as Doofilator did. What is important is the number of moles. The trouble is that the data are wrong. As MaxW said it previously, the following data is wrong : "11.65 g of Co and 25.7 g of Co". Cobalt cannot be mixed with Cobalt. And worse ! There is nowhere any information about a third element.

I have tried to rewrite the whole problem thinking that the 25.7 g are Chlorine atoms. With this assumption, the given masses correspond to 0.197 mol Co and 0.725 mole Cl. The ratio Cl/Co would be 3.68, which could be explained with 0.197 mol of CoCl2, but it remains 0.725 - 2·0.197 = 0.331 mol of Chlorine. And this Chlorine must be combined to some other element.

Of course the third element weighs 50 g - 25.7 g - 11.65 g = 12.65 g. If the third element is Molybdenum, the sample contains 12.65/95.94 = 0.132 mol Mo. The ratio Cl/Mo would be 0.331/0.132 = 2.5. This may correspond to Mo2Cl5.

So the sample may be made of CoCl2 and Mo2Cl5, both in a ratio that must be calculated. I let it to Doofilator. This ratio can be obtained from the molar mass of the compound.