Mass of aluminum a) Mass of empty 125-mL Erlenmeyer flask (b) Mass of 125-mL Erl

Question

Mass of aluminum a) Mass of empty 125-mL Erlenmeyer flask (b) Mass of 125-mL Erlenmeyer flask + aluminum foil (c) Temperature of cooled alum crystals Mass of alum (a) Mass of empty 50-mL beaker (b) Mass of 150-mL beaker + alum Calculations Mass of aluminum metal used Mass of alum obtained Calculate the number of moles of aluminum that you used. Show your calculations: moles Al Calculate the molar mass of alum: g/mole Noting that one aluminum atom is in each formula unit of alum we know that each mole of aluminum will give rise to one mole alum. Based on the number of moles of aluminum you used, calculate the maximum number of grams of alum that you could produce (theoretical yield), and assuming that aluminum is the limiting reagent: (show your work) g alum Using the number of grams that you actually obtained, calculate the percent yield of alum you synthesized: (show your work)

Explanation / Answer

1. No. of moles of Aluminium used = Mass of Aluminium / Molar mass of Aluminium = 0.561 g / 26.982 g/mol

No. of moles of Aluminium used= 0.021 mol

2. Formula of Alum is  KAl(SO4)2·12H2O

Molar mass of Alum = Molar mass of K + Molar mass of Al + 2 X [(Molar mass of S) + 4 (Molar mass of O)] + 12  X [2 X (Molar mass of H) + (Molar mass of O)]

Molar mass of Alum = 39.098 g/mol + 26.982 g/mol + 2 X [(32.065) + 4 (15.999)] + 12  X [2 X (1.008) + (15.999)]

= 39.098 g/mol + 26.982 g/mol + 192.122 g/mol + 216.180 g/mol = 474.382 g/mol

Molar mass of Alum = 474.382 g/mol

3. Maximum no. of grams of alum = No. of moles of Aluminium used X Molar mass of Alum

= 0.021 mol X 474.382 g/mol = 9.962 g

Maximum no. of grams of alum that can be produced (theoretical yield) = 9.962 g

4. Percent yield of alum = (Grams of alum obtained / theoretical yield of alum) X 100%

= (8.458 g / 9.962 g) X 100% = 84.9%

Percent yield of alum = 84.9%

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