Question

The "alum" used in cooking is potassium aluminum sulfate hydrate, KA1(SO4)2 x H2O. To find the value of x, you can heat a sample of the compound to drive
off all of the water and leave only KA1(SO4)2. Assume you heat 4.74 g of the hydrated compound and that the sample loses 2.16 g of water. What is the value of x?

Answer 1

@aaronq

Answer 2

mass of \(\sf KAl(SO_4)_2*xH_2O\) = 4.74 g

Find moles of the anhydrate and of water, separately, using: \(\sf moles=\dfrac{mass}{Molar~mass}\)

\(\sf moles~of ~ KAl(SO_4)_2=\dfrac{4.74g-2.16g}{258.21~g/mol }=0.01 ~moles\)

\(\sf moles~of ~H_2O=\dfrac{2.16g}{18~g/mol}=0.12~moles\)

Normalize by dividing both by the lowest number

\(KAl(SO_4)_2=\dfrac{0.01}{0.01}=1\)

\(H_2O=\dfrac{0.12}{0.01}=12\)

the number of x is 12, the formula of the hydrate is \(\sf KAl(SO_4)_2*12~H_2O\)