Something is haywire here.
mass sample = 1.664 g.
mass recovered Cu = 1.198 g.
percent Cu = (1.198/1.664)*100 = 71.995 which I would round to 72.00 to four s.f.
I don't know where you obtained 1.1664 nor 1.998. Check my work.
For percent error.
[(exp value - theoretical value)/theoretical value]*100 = ??
The 25.47% is the theoretical value; i.e., what it's supposED to be.
The mass of the CuSO4 x 5H2O sample is 1.664 g. The mass of the recovered copper is 1.198 g.
1. Compute the experimental percent by mass of copper in the sample:
1.1664/1.998x 100=71.81%
2. Calculate the percent copper in pure copper sulfate pentahydrate(this is the theoretical percent copper). One mole of copper has a mass of 63.55 g, and by addition, one mole of CuSO4 x 5H2O has a mass of 249.5 g:
63.55/249.5x100= 25.47%
3. In order to calculate the error of the % Cu and the percent error of the % Cu I know I am suppose to use 1.198 as the experimental value, but I am not sure what to use for the true value.
2 answers
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4 answers