Pre-Laboratory Assignment For Hydrates And Anhydrates

Percentage of Water in HydratePre-Laboratory Assignment1.Reason why it is necessary to be extremely careful when working with Barium Chloride dihydrate;This is necessary because Barium is a very toxic element. It is soluble in water and, hence, can easily be absorbed in the body. It can be fatal if swallowed and harmful if inhaled. Careless handling of the compound can result in irritation to the skin, eyes, respiratory tract and nervous system complications.2.Difference between;I.CuSO4 (s) and CuSO4.5H2O (s)The main difference is that CuSO4 (s) is anhydrous while the CuSO4.5H2O (s)is a hydrous compound i.e. is hydrated with a 1.5 molar mix of CuSO4and H2Orespectively.CuSO4 (s) is also grey/white in colour while CuSO4.5H2O (s)is blue in colour. Additionally, the former is lighter in mass while the latter is heavier.II.CuSO4 (s)and CuSO4 (aq)The major difference between the two is that CuSO4 (s) is a solid while CuSO4 (aq)is a solution. CuSO4 (s)is also greyish in colour while the CuSO4 (aq) is blue. III.CuSO4.5H2O (s) and CuSO4 (aq)CuSO4.5H2O (s) is a crystalline solid compound and contains water of crystallization. It is a hydrated salt with water weakly bounded around the copper

Unformatted text preview: (53/3 Stoichiometry-Hydrates Prelaboratory Assignment Name and Drawer Number Q9 Q}: n I SCAN g‘fi ‘” £93 1. The data listed below was determined from the dehydration of the hydrate of magnesium sulfate. Mass of empty crucible = 35.794 g Mass of crucible and magnesium sulfate hydrate = 37.255 g Mass of crucible and anhydrous magnesium sulfate = 36.504 g 3. Write the chemical formulas for water and anhydrous magnesium sulfate. From the formulas calculate the molar masses of water and anhydrous magnesium sulfate. van-MU '23 \AarO ; Fiona 3imo‘ Our\\-“:§_\re'c$ \‘Ynsxnegwcm $‘0\\:O_5‘ Q— i\$ 1 at) o 5 l m0" b. Calculate the mass of water removed from the hydrate during the drying process from the difference between the mass of the crucible and hydrate and the mass of the crucible and the anhydrous salt. 0 a'\ ‘53\ E) be 0 3%“; guitalri‘wj / c. Use the mass of water (part lb) and the molar mass of water to calculate the moles of water in the sample. CncfiM-r W‘hu\ use (:1. Determine the mass of anhydrous magnesium sulfate by subtracting the mass of the empty crucible from the mass of the crucible containing the anhydrous compound. 0.110 5 /( e. Use the molar mass of the anhydrous compound and the mass of the anhydrous compound (part 1d) to determine the number of moles of anhydrous salt. ODOD%%$ mo\ M5504 f. Determine the ratio of the number of moles water in the hydrate to the number of moles of anhydrous salt in the hydrate by dividing the answer from part 1c by the answer from part 1c. This value is the number of moles of water in the hydrate for each mole of the anhydrous salt. Use this value to write the chemical formula for the hydrate. O I“ N W : 7.0a me" “BO/MM “as” 090°5%% chmvixgg} Qou W -41- "‘ Prelaborator)‘ Assignment - continued 2. A11 unknonn hydrated salt has the general formula Salt~nH20 where “n” is the number of moles of water for each mole of the hydrated compound. Although the chemical formula for the anhydrous salt is not known, it is known that the molar mass of the anhydrous salt is 151.92. The data listed below were obtained for a portion of the salt. (3) Determine the number of moles n of water in the hydrate for each mole of the anhydrous salt. (b) Write the formula for the salt, i.e., substitute the value of n into the general formula written above. Hint: It might be useful to follow the general solution outline used in question 1 of the Prelaboratory Assignment. Show the calculations. Mass of empty crucible = 35.107 g Mass of crucible and hydrated salt = 36.274 g Mass “crucible and anhydrous salt = 35.745 g _ “is “ (“Mafia Sn“ Ema. MA as.“ I c F - 3‘5. \0‘1 5 ml vaxix‘ow. ’ 0 .t; 35% so. \ - \\.o "t :5 \\\5qu¥: tux 93CL\¥- 3) mm of: \—\a0 \fi‘b\ N30: 3.x\.,00%-\\ + unmet: -(3 \q .Q:\\o % \XQD \\ (mask -»a\ A a o k 0 , muse 0'53 9&0 Mngb — O, oacvkwxcfiQsaG Ff, ‘ _ H \a \w'] S\\%d\ro}st¢3~ bc‘\¥ x \% ic\\o$ 9‘30 -W o .594; g) ‘R'axo f— so‘\1r -— O. QQQ‘AO ’mo\ 3Q\* O‘KPE’Q $30“ \:3\_O\';%SCLU¢ ® V&k\0 \e\ 0 0ean «WA 9‘30 A: "\ r“‘°_\______.—9‘ _—————r*""" mw-\ ww Ado - DO‘VBD “‘5‘ aflk , —42- 930A": ‘_l \'\ 9‘0 ...
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