Question

i need help with questions A-E please

The Mole Concept: Chemical Formula of a Hydrate- Lab Report Assistant Exercise 1: Water of Hydration Data Table 1. Alum Data. Object Mass (g) Aluminum Cup (Empty) 3 Aluminum Cup +2.00 grams of Alum5 Aluminum Cup + Alum After 1 Heating 4.44 Aluminum Cup + Alum After 2 Heating 3.52 Mass of Released HO 1.48 Molecular Mass of H,0 18.015 Moles of Released H,0 OU54 Questions: A) Calculate the moles of anhydrous (dry) KAI(SO), that were present in the sample. Show all work including units. B) Calculate the ratio of moles of H.O to moles of anhydrous KAl(SO), Show all work including units. Note: Report the ratio to the closest whole number. C) Write the empirical formula for the hydrated KAl(SO.), based on your experimental results wer to Question 2. Show all work including units. Hint: if the ratio of moles of H.O to moles of anhydrous KAl(SO), was 4, then the empirical formula would be: KAI(SO),H,O. 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, will these errors cause the calculated number of moles of water in the hydrate to be artificially high or low? a. The student ran out of time and did not do the second heating. Explain how this error will affect the calculation for the number of moles of water in the hydrate? Will the final 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 of hydrated compound but used 2.00 g in the calculations. Explain how this error will affect the calculation for the number of moles of water in the hydrate? Will the final answer be artificially high or low? How do you know?

Answer 1

Mass of alum taken = 2.0 g

Mass of water released = 1.48 g( the value calculated is correct)

Mass of KAl(SO4)2 dry = mass of alum - mass of water released = 2.0 g - 1.48 g = 0.52 g

A). Molar mass of dry KAl(SO4)2 = 258.21 g/ mol

Mass of KAl(SO4)2 = 0.52 g

Moles of KAl(SO4)2 = mass / molar mass = 0.52g/ 258.21g/mol = 0.002 mol

B). Moles of water = 0.082 mol

Moles of KAl(SO4)2 = 0.002 mol

Ratio of moles of water to moles of dry KAl(SO4)2 = 0.082 mol/ 0.002 mol = 41.07 rounding off to nearest whole number we get the ratio of 41:1

C). We have already calculated the ratio of moles of water to the moles of dry KAl(SO4)2. So the empirical formula is KAl(SO4)2 . 41 H2O. You have 41 moles of water for eachole of KAl(SO4)2

For calculating empirical formula you need to divide the moles of all the components of the formula by the component that has least moles.

KAl(SO4)2 has least moles 0.002 mol

So for KAl(SO4)2 0.002 mol/ 0.002 mol = 1

For H2O 0.082 mol/ 0.002 mol = 41 ( nearest whole number)

So you have 1 KAl(SO4)2 and 41 H2O

D) the answer to this will be based on your observation in the lab

E).

Part a. If the second heating is not done, then the student will calculate the mass of water lost based on the value obtained after first heating (4.44 g). This will give the mass of water lost to be much lower than the actual mass of water lost. So the calculated moles of water will be much lower than the actual moles.

Part b

If the student starts with 2.2 g of initial sample but uses 2.0g for the calculations, then the mass used for calculations will be lower than the actual mass. When thebstudent weighs the final sample after drying, the difference in mass is going to be higher ( 0.2 g error will be seen). The mass recorded will show 0.2 g less mass. Student will add this less mass as the mass of water lost. So the calculated mass of water will be much higher than the actual mass. This means that the calculated moles of water lost will be much higher than the actual moles of water lost.