Question

2. Based on your experimental data cal KAl(SO4)2.nH20. in your experimental data calculated the value of n in the chemical formula 3. Calculate the theoretical percentage of SO2- in pure KAl(SO4)2.nH2O using a periodic table for atomic weights and the correct literature value for n. 4. Is your sample alum? Use the results of the three tests to support your answer. Discuss the accuracy of your tests and possible sources of experimental error. Mass of crucible and cover (9) s of crucible, cover, and alum before heating (9) 28.5779 30.577g 29 Mass of alum before heating (g) of crucible, cover, and alum after 1st heating (9) qa of crucible, cover, and alum after 2nd heating (g) 29.6799 pf crucible, cover, and alum after 3rd heating (9)

Answer 1

1.Mass of crucible and cover=28.577g

mass of crucible and cover and alum before heating=30.577g

Therefore mass of alum= (30.577-28.577)g=2g

After 1st heating mass of crucible and cover and alum=29.719g

mass of alum after 1st heating (29.719-28.577)g=1.142g

mass lost (2-1.142)g=0.858g

After 2nd heating mass of crucible and cover and alum=29.679g

mass of alum after 2nd heating (29.679-28.577)g=1.102

mass lost (2-1.102)g=0.898g

After 3rd heating mass of crucible and cover and alum=29.649g

mass of alum after 3rd heating (29.649-28.577)g=1.072g

mass lost (2-1.072)g=0.928g

Molecular weight of alum =M=39.1+27+(32$\times$2)+2$\times$(16$\times$4)+n$\times$18=258.1+18n g/mol

total mass lost= 0.928 g

molecular weight of water=18 g

number of moles of water =(0.928/18)=0.05

mass of alum after third heating 1.072g

Molecular weight of dehydrated alum =258.1 g

number of moles of alum (1.072/258.1)=0.00415

0.00415 mol of alum contains 0.05 mol of water.

therefore 1 mol contains (0.05/0.00415)=12.03 mol of water.

therefore n=12.03 $\approx$12

3.molecular weight of alum =258.1+18n=258.1+12$\times$18=474.1

molecular weight of sulphate group =(32+16$\times$4)=96

theoretical percentage =$\frac{96 \times100}{474.1}$=20%