Abstract: In last week’s experiment we calculated that percent by mass of water in a hydrated salt by completing the gravimetric analysis. To complete this experiment, we calculated the difference in masses of the hydrated salt sample before and after it heated multiple times. This data was then averaged to find the %H 2 O within the ion for that unknown sample. The procedures to conduct this experiment were repeated multiple times for accuracy along with multiple trials of new samples of the same compound. After conducting the experiment, we discovered that our hydrated salt was MgSO4 x 7H2O because our average %H 2 O was 54.86%. In conclusion, we successfully completed the experiment since our results were very similar to what the %H2O should be.

You are watching: Percent water in a hydrated salt

Date: 14 September 2019

Objective: In this experiment, the percent by mass of water in a hydrated salt will be determined. Also, we will learn how to use lab apparatus.

Introduction: All forms of salt are usually hydrated which means that certain amounts of water molecules are connected to the salt ions. These water molecules are referred to as waters of crystallization. Typically, since the moles of water per moles of salt is constant, we add the amount of hydrates after the chemical formula. For example, iron(iii) chloride can be represented as Fecl 3 x 6H 2 O (hexahydrate). Sometimes, the water molecules are too weakly bound that heat can break them apart eventually forming anhydrous salts. If the salts lose water molecules without heat, they are known as efflorescent while salts that absorb water are known as deliquescent. To determine the percent by mass of H 2 O, you begin by measuring the mass of the salt before and after the waters of crystallization are heated and are driven off the salt. The difference between the masses is the amount of water that was in the salt. Therefore, if you divide that number by the total mass of the original sample and multiply it by 100, you calculated the percent of H 2 0 in a sample of hydrated salt. This process is called gravimetric analysis.

Procedures:1- heat the crucible for about 5 minutes with high temperatures. Allow them to slowly cools them measure the mass of the crucible and the lid separately. Record the data.2- Add about 0.5 g (±0.001) of the hydrated salt into the crucible then collect the data.3- Return the crucible with the salt inside to the holding tray (with tongs) and place the lid slightly off the crucible to allow the water molecules to escape. Gradually intensify the heat for about 5 minutes then measure the mass of the crucible with the dried salt.4- Reheat the sample for 2 minutes then measure the mass again. Repeat this step 2 more

times to make sure that all water molecules have escaped successfully.5- Repeat the experiment with 2 more samples of the same salt for a total of three trials thenclean up.

Hypothesis: If we heat a sample of hydrated salt, then the mass of the salt will decrease since water molecules escaped the ion.

Data: Trial 1 Trial 2Mass of fired crucible and lid (g) C: 38.L: 47.

### L: 47.

Mass of fired crucible, lid, and hydrated salt (g) 86.4415 86.Mass of crucible, lid, and anhydrous salt:1 st mass measurement (g) 86.1688 86.2 nd mass measurement (g) 86.1510 86.Final mass of crucible, lid, and anhydrous salt (g) 86.1510 86.

Mass of hydrated salt(g) 0.5096 g 0.Mass of anhydrous salt (g) 0.2191 g 0.2395 gMass of water lost (g) 0.2905 g 0.2669 gPercent by mass of volatile water in hydrated salt (%) 57.01% 52.71%

Average percent H 2 O in hydrated salt (%H 2 O) 54.86%Standard deviation of %H 2 O 3.Relative standard deviation of %H 2 O in hydrated salt (%RSD) 5.54%

Post lab:2- After the oils from the finger are burnt of in part B.1, the percent water in the hydrated saltwill be unaffected because the oils have been driven off and the mass of the oils on the cruciblewould not be accounted for.6- If the hydrated salt is overheated and the anhydrous salt thermally decomposes, then thereported percent water in the hydrated salt will be too high because there would be a largerdifference between the initial and final mass of the salt. The decomposed anhydrous salt that isreleased as a gas will be accounted for as the mass of the water molecules being escaped whichis inaccurate.8- Bill’s haste could result in his percent being too high because he did not let all the watermolecules escape the sample. Some of the water would’ve been left behind in the cruciblecausing the calculations to be higher than expected.

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Reference:Beran, J. A. Laboratory Manual for Principles of General Chemistry; 10th ed.; Wiley: Hoboken, 2014.