# How do I find the empirical formula for Co(C2O4)*H2O with mass of Oxalate and Oxide Product?

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 0.3283 g and my product oxide weighed 0.1158 g which I have calculated as 0.1158 × (176.79 g cobalt / 240.79 g cobalt oxide) to equal 0.08502 g of cobalt in the original sample. 0.3283 g of reactant − 0.1158 product gives me 0.2125 g water removed divided by 18.016 g/mol to get 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. Any help would be greatly appreciated as I’ve been at this all day.

Some parts of your calculation are correct, you do however forget that $\ce{CO2}$ is released. This would be my approach:
115.8 mg of $\ce{Co3O4}$ (240.79 g/mol) equal 0.4809 mmol. This means there was three times as much $\ce{CoC2O4}$ (146.9522 g/mol) in your sample: 1.443 mmol. That equals a weight of 212.1 mg. Your original sample weighed 328.3 mg, so there are 116.2 mg left that seem to be $\ce{H2O}$ (18.01528 g/mol), giving you 6.450 mmol. The ratio of cobalt oxalate to water then is 6.450/1.443 ~ $\ce{CoC2O4}$ ⋅ 4.5 $\ce{H2O}$.
• This definitely helps but I think you got the molar mass of $\ce{Co3O4}$ wrong. Cobalt x3 is 176.79 and $\ce{Co3O4}$ is 240.79 – Alec Crudele Jun 27 '15 at 2:52
• Sorry for that, I edited the answer so that the molar mass of $\ce{Co3O4}$ is now correct. Everything else should be fine... I did convert to grams when doing the calculation (note that mmol is used here, which effectively is a division by 1000: mg/(g/mol) = mmol) – snurden Jun 27 '15 at 6:22