How does the \" percent-water in the hydrated salt\" will be affected under the

Question

How does the " percent-water in the hydrated salt" will be affected under the following conditions? If the fired (heated) crucible is handled with (oily) fingers before its mass measurement. Explain If the original sample is unknowingly contaminated with a second anhydrous salt. Explain If the hydrated salt is overheated and the anhydrous salt thermally decomposes, one product being a gas. explain. Because of the lack of time, one student decided to skip the step of reheating the sample in the experimental procedure. explain What is the percentage mass of water present in a 4.63 g sample of MgSO_4-7H_2O?

Explanation / Answer

(1) Measuring the % water in a hydrated salt experimentally involves initially heating a known mass of the hydrated salt to eliminate the waters of hydration and then measuring the mass of the anhydrous salt. The difference between the two weights shall be equal to the mass of water lost. Dividing this value by the original weight of hydrated salt will give the fraction of water in the compound. Multiply this fraction by 100 to get the % water in the hydrated salt.

The % water in hydrated salt will be wrongly measured. The reason for the same would be due to oily fingers the weight of the crucible will increase leading to a lower calculation of the % water in hydrated salt. For example: assuming the weight of hydrated salt is 5g (w1). After heating the weight is 4.2g (w2). Thus the 0.8 g is the mass of water lost in ideal situation. But if oily fingers were used then w2 would be higher (say 4.3 g) and thus the mass of water lost in this situation will be 0.7 g leading to a lower calculation of the % water in hydrated salt.

Contamination with a second anhydrous salt will not affect the calculation of % water in hydrated salt of the original sample since the contaminating salt does not have water which will affect the weight after heating. So there shall be no error in calculation in this case.

If the hydrated salt is overheated resulting in the anhydrous salt decomposing and gas being released, in this case too, the final weight of the crucible shall be erroneously calculated to be less. This error will lead to a higher calculation of the % water in hydrated salt.

If the step of reheating is skipped, it is likely that the atmospheric moisture absorbed in the salt will not be lost leading to a lower calculation of the % water in hydrated salt; similar to situation (a).

(2)MgSO4.7H2O

% mass of water = mass of water/mass of entire compound

Now to calculate the molar mass of all the individual constituent in the above chemical

Mg =                      24.3

S =                          32

O = 16 x 4 =            64

H2O = 18 x 7 =    126 (the total mass of the entire compound will be 246.3)

Thus, using the above mentioned formula;

% mass of water = mass of water/mass of entire compound

                                = 126 / 246.3

                                = 0.511 x 100 (for % conversion)

                                = 51.15%

The amount of % water calculated in MgSO4.7H2O is considered for 100 gram and so for 4.63 gms the % water will be;

In 100 g of MgSO4.7H2O % mass of water = 51.15%

So, in 4.63 g of MgSO4.7H2O, the % mass of water will be = 2.36%

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