A.
%Co in Co3O4 = (3*atomic mass Co/molar mass Co3O4)*100 = ? You should get about 72% but get closer than that. How many grams Co is there in that sample? That is 0.7242 x 1.316g = about 0.95 g which means you must have started with 0.95 g Co in the 3.000 g sample of the hydrate.
The 2.000 g sample contained 0.9621g C2O4 which is 0.9621/2.000)*100 = about 48%. Again, you get closer than that.You want to know how much is in a 3.00 g sample. %C2O4 x 3.00 = about 1.44 g C2O4 in the initial 3.00 g sample.
So how much H2O is there? That is 3.00 g - 1.44g C2O4 - 0.95g Co = about 0.61g so %H2O = about (0.61/3.00)*100 = about 20%.
The problem doesn't ask for it but you can determine the empirical formula from these numbers. I won't do it here but the empirical formula is CoC2O4.2H2O. Post your work if you get stuck.
1.) A 3.000gram sample of a hydrate compound containing Co, C2O4, and H2O was combusted to give 1.316g of Co3O4. What is the percentage of Cobalt in the oxide?
3.) 2.000 grams of a sample containing Co, C2O4, and H2O showed there to be 0.9621g of oxalate in the sample. Was is the percentage of oxalate in 3.000grams of that sample?
What is the percentage of water in the sample for problems 1 to 3?
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