1- Alum used in cooking is potassium aluminum sulfate hydrate, KAl(SO4)2. XH2O. To find the value of X, you can heat the sample of compound. Assume you heat 4.74 g of the hydrated compound and that sample loses 2.16 g of water. What is the value of X

THE VALUE OF X IS 7 AND THE FORMULA OF THE HYDRATED SALT IS KAl(SO4)2.7H20

Explanation:

1. write out the varibales given in thequestion:

Mass of the hydrated salt = 4.74 g

Mass of water lost = 2.16 g

Formula of the hydrated salt = KAl(SO4)2. XH20

2. calculate the molar mass of the salt and that of water of crystallization:

Molar mass of anhydrous salt = ( K = 39, Al = 27, S = 32, 0=16)

= ( 39 + 27 + 32*2 + 16 * 8

= (39 + 27 + 64 + 128)

= 258 g/mol

Molar mass of water = 18 g/mol

3. Use this expression to calculate X:

The expression,

XH20 / molar mass of anhydrous salt = Mass of water lost / Mass of hydrated salt.

X = molar mass of anhydrous salt * mass of water lost / mass of anhydrous salt * H20

where XH20 is the molar mass of water of crystallization, is used to calculate the value of X.

4. Solve for X:

So therefore:

X = 258 * 2.16 / 4.74 * 18

X = 557.28/ 85.32

X = 6.53

X is approximately 7.

The value of X is 7 and the formula pf the hydrated salt is KAl(SO4)2.7H20

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o