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Homework answers / question archive / This is a classic chem lab! It involves massing a compound, then heating it to remove water and then measuring its mass a second time

This is a classic chem lab! It involves massing a compound, then heating it to remove water and then measuring its mass a second time

Chemistry

This is a classic chem lab! It involves massing a compound, then heating it to remove water and then measuring its mass a second time.

This can be approached in two different ways, calculating the percent of water in a hydrate from experimental data or analysis of a chemical formula.

Let's discuss experimental analysis first. Let's say that you are given a hydrated salt we'll call Y. Its general formula would be...

Y•X(##H_2O##).

If you heat the sample you will be able to dry it and remove all of the water leaving just Y. To calculate the percent of water you will divide the change in mass of your sample (mass of water removed) by the mass of the hydrated salt (original mass).

If you are given the formula you would do analysis of the information provided...

##CuSO_4## = 159.62g Mass of water = 18.02 x 5 = 90.1g ##CuSO_4##•5(##H_2O##) = 249.72g

% water = (90.1 / 249.72) x 100 = 36.08%

video from: Noel Pauller

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