Question
Needing Help!!! Did I calculate the moles released H2O correctly???
Also unsure how to solve B and C.
The Mole Concept: Chemical Formula of a Hydrate Lab Report Assistant Exercise 1: Water of Hydration Data Table 1. Alum Data Mass (g) Object Aluminum Cup (Empty) Aluminum Cup + 2.00 grams of Alum Aluminum Cup+Alum After 1 Heating Aluminum Cup + Alum After 2nd Heating Mass of Released H2O Molecular Mass of H20 Moles of Released H2O mol Questions A) Calculate the moles of anhydrous (dry) KAI(SO4)2 that were present in the sample. Show all work including units. 00115 mo EA(S)2 BCalculate the ratio of moles of H20 to moles of anhydrous KAl(SO4)h. Show all work including units. Note: Report the ratio to the closest whole number. gt20 mil H70 XC) Write the empirical formula for the hydrated KAl(SO4)2, based on your experimental results and answer to Question 2. Show all work including units. Hint: if the ratio of moles of H20 to moles of anhydrous KAI(SO4)2 was 4, then the empirical formula would be: KAI(SO4)2.4H0 D) Describe any visual differences between the hydrated sample and the dried, anhydrous form E) How would the following errors affect the empirical formula for the compound? That is, wil these errors cause the calculated number of moles of water in the hydrate to be artificiall high or low? The student ran out of time and did not do the second heating. Explain how this error wi affect the calculation for the number of moles of water in the hydrate? Will the firn a. answer be artificially high or low? How do you know? b. The student recorded the mass of the cup + sample incorrectly and started with 2.20 g hydrated compound but used 2.00 g in the calculations. Explain how this error will affe the calculation for the number of moles of water in the hydrate? Will the final answert artificially high or low? How do you know?
Explanation / Answer
Mass of anhydrous alum = mass of Aluminum cup + alum after 2nd heating - mass of aluiminum cup = 0.8 gms
Moles of alum = mass /molar mass = 0.8 / 258 = 0.0031
Moles of H2O = 1.1/18 = 0.0611
Moles of H2O / Moles of anhydrous alum = 0.0611 / 0.0031 = 20
Formula is KAl(SO4)2 . 20 H20
THIS is the way for calculating alum formula
if u find anything difficult please comment
