INQUIRY- What is the percentage by mass of water in Epson salt? DISCUSSION- When a crystal forms from water solutions, it takes a certain amount of water as part of the crystalline structures. The water is not covalently bonded to the host molecule or ion. Water is taken up in definite proportion in relation to the mass of the crystalline structure. This water that is trapped inside the hydrated solid structure is called water of crystallization or water of hydration. A salt with associated water of crystallization is known as a hydrate.
The formula of a hydrated salt is written as the number of moles of water present in one mole of crystalline structure. Deliquescent materials are substances that absorb water from the atmosphere and eventually become hydrated. These materials, usually salts, tend to have a high affinity for water. During this process, the actual structure of the crystalline changes to incorporate the water molecules. This causes the crystal to change in appearance.
In order to calculate the amount of water in the crystal, the weight by difference method was used.
The amount of water was calculated by measuring the differences in the mass of the hydrous and anhydrous salt. By using this mass, the percentage of water, the number of moles and the complete formula of the salt were derived. HYPOTHESIS- If the water is driven off of hydrated magnesium sulfate, then it would be colorless because it is made of Group 2 metals, which do not form colors. OBJECTIVES- a)To use the method of “weight by difference” to determine mass quantities. b)To determine the mass of an anhydrous salt by heating the sample to a constant mass. c)To convert grams to moles. )To determine the percent of water in a hydrate sample. e)To determine the formula of a hydrated compound. MATERIALS AND EQUIPMENT- 1)Balance 2)Clay triangle 3)Crucible 4)Crucible tongs 5)Burner 6)Iron ring 7)Ring stand 8)Scoopula 9)Wire gauze 10)Hydrated magnesium sulfate PROCEDURE- The crucible was cleaned with a tissue paper in order to wipe off particles in inside it. Then the clay triangle was placed on the iron ring. The Bunsen burner was fired up and a flame was created. The crucible was place on the clay triangle and was allowed to heat gently for two minutes.
The crucible was removed with a crucible tong, allowed to cool and its mass was then measured using the analytical balance. Then a certain amount of magnesium sulfate was put into the crucible and the mass was measured again. Both the measurements were recorded. Then the crucible containing the magnesium sulfate was heated for 15 minutes. After the time was up, the flame was turned off and the crucible was removed from the clay triangle using crucible tongs. After cooling down, the mass of the anhydrate and the crucible was measured using the analytical balance.
The final measurement was recorded. The anhydrate was disposed properly and the materials were cleaned and put back to their respective places. INDIVIDUAL DATA- Mass Empty Crucible16. 749g Crucible with Hydrate Magnesium Sulfate18. 293g Hydrated Magnesium sulfate1. 5440g Crucible with anhydrate magnesium sulfate17. 516g Anhydrate magnesium sulfate0. 77700g Amount of water left – Mass of hydrated MgSO4 – Mass of anyhdrate MgSO4 1. 544g – 0. 77700g = 0. 7670g of H2O INDIVIDUAL CALCULATIONS AND ACCURACY- 1. Calculate the percentage of water in the hydrate.
Explain why the size of the sample tested would not change this answer. Percent of water = (mass of H2O/ mass of hydrate) x 100 (0. 7670g of H2O/ 1. 544g of hydrate) x 100 = 49. 68% H2O 2. Determine the formula of the hydrated salted. a. Calculate the number of moles of water driven off from your sample. 0. 7670g H2O x 1 mole H2O = 0. 04261 moles H2O 18. 001g H2O b. Calculate the number of moles of anhydrous salt remaining in the crucible 0. 77700g anhydrate MgSO4 x 1 mole MgSO4 = 0. 006453 moles MgSO4 120. 4 g MgSO4 c.
Determine, to the nearest integer, the ration of the number of moles of water per mole of the anhydrous salt. (0. 04261 moles H2O : 0. 006453 moles MgSO4) 0. 006453 moles MgSO4 6. 6031 moles H2O : 1 mole MgSO4 ~ 7 moles H2O : 1 moles MgSO4 d. Write the complete formula for this salt. MgSO4 7 H2O e. Percent Error % error = (Measured Value- Accepted Value) / Accepted Value x 100 Measured Value = 49. 68% Accepted Value = 51. 17% (49. 68% – 51. 12%)/51. 12% X 100 % Error = -2. 817% CONCLUSION- CONFIDENCE REPORT- SUMMARY QUESTIONS- 1.
A student determines the percent of water hydration in a crystal to be 31%. The correct value is 15%. What might be a likely source of error? On likely source of error is that the 2. A solid hydrate weighing 2. 32 grams is heated and allowed to cool three times to drive off the water of hydration. The mass, after each of the three heating process are respectively 2. 21g, 1. 78g, and 1. 79g. If the anhydrous residue has a molar mass of 178, how many moles of water are present in one mole of salt? Mass of water in hydrate (2. 32g – 2. 21g) = 0. 11g H2O (2. 32g – 1. 78g) = 0. 54g H2O (2. 32g – 1. 9g) = 0. 53g H2O Avg mass of anhydrous solid= (2. 21g + 1. 78g + 1. 79g) / 3 = 1. 93g anhydrous solid Mole of anhydrous = (mass of anhydrous / molar mass of anhydrous) = (1. 93g) / (178 g/mol) =0. 0108 mol Avg mass of H2O = (0. 11g + 0. 54g + 0. 53g) / 3 = 0. 393 g H2O Mole of H2O = (mass of H2O driven off/Molar mass of H2O) = (0. 393g) / (18. 0 g/mol) = 0. 0219 mol Ratio of moles of H2O to anhydrous = (0. 0219 mol) : (0. 0108 mol) =2. 028 mol H2O : 1 mol anhydrous ~2. 000 mol H2O: 1. 000 mol anhydrous There are two moles of water present in one mole of salt. WHAT IF- What If the amount of
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