# How do I find the empirical formula for cobalt oxalate hydrate with mass of oxalate and oxide produced?

I have the formula $$\ce{Co(C2O4)*H2O}$$ in a reaction that forms cobalt oxide $$\ce{Co3O4}$$

My data here says that my oxalate hydrate weighed $$\pu{0.3283 g}$$ and my product oxide weighed $$\pu{0.1158 g}$$ which I have calculated as:

$$0.1158 \times \frac{\pu{176.79 g}\, \mathrm{(cobalt\,oxalate)}} {\pu{240.79 g}\, \mathrm{(cobalt\,oxide)}} =\pu{0.08502 g\,\mathrm{(cobalt)}}$$

Then $$\pu{0.3283 g (reactant)} − \pu{0.1158 g (product)}$$ gives me $$\pu{0.2125 g}$$ water removed divided by $$\pu{18.016 g/mol}$$ to get $$\pu{0.01179 mol}$$ $$\ce{H2O}$$ in the sample.

How do I use this data to determine the empirical formula? I am not sure how to determine the molar ratios from this point.

I'm not exactly sure - but I guess you are trying to find the amount of water in your sample and to determine the ratio to cobalt oxalate.

Some parts of your calculation are correct, you do however forget that $$\ce{CO2}$$ is released. This would be my approach:

$$n_\ce{Co3O4} = \frac{\pu{115.8 mg}}{\pu{240.79 g/mol}} = \pu{0.4809 mmol}$$

This means that there was three times as much $$\ce{CoC2O4}$$ in your sample.

$$n_\ce{CoC2O4} = \pu{1.443 mmol}$$

That equals a weight of $$\pu{212.1 mg}$$. Your original sample weighed $$\pu{328.3 mg}$$, so there is $$\pu{116.2 mg}$$ left that seem to be $$\ce{H2O}$$ ($$\pu{18.015 g/mol}$$), giving you.

$$n_\ce{H2O} = \pu{6.45 mmol}$$

The ratio of cobalt oxalate to water then is $$\left(\frac{6.450}{1.443} = 4.5\right)$$ which implies the empirical formula is $$\ce{2CoC2O4.9H2O}$$

I am doing another approach, to get another result.

Starting from $$\pu{0,1158 g}$$ $$\ce{Co3O4}$$, whose molar mass is $$\pu{241 g/mol}$$, this amount corresponds to $$\pu{\frac{0.1158 g}{241 g/ mol} = 0.4804 mol}$$ $$\ce{Co3O4}$$, and also to $$\pu{3*0.4804 = 1.4415 mmol }$$ $$\ce{Co}$$. This amount of cobalt is included in the original sample of oxalate, which weighs $$\pu{0.3283 g}$$ So the molar mass of this original oxalate is $$\pu{M = \frac{0.3283 g}{1.4415 mmol} = 227.7 g/mol}$$. As the molar mass of anhydrous $$\ce{CoC2O4}$$ is $$\pu{147 g|mol}$$, the difference $$\pu{227.7g - 147 g = 180.7 g}$$ is the mass of $$n$$ moles water. $$\pu{180.7 g}$$ is exactly the mass of $$10$$ moles water.

As a consequence the original oxalate has the formula $$\ce{CoC2O4·10 H2O}$$