of hydrated, MgSOu. xH20 was heated in a crucible until all of the crystallizati

Question

of hydrated, MgSOu. xH20 was heated in a crucible until all of the crystallization water is removed show calculation methods and take care of the significant figs). a iment 2: Introduction to gravimetric analysis). Based on the provided data fill in the empty spaces 40 pts) (Molar masses: MgSO-120.4 gmol, H2O-18.01 g/mol) tared mass mass of crucible, cover and hydrated salt mass of hydrated salt mass of crucible, cover and residue after heating mass of anhydrous salt moles of anhydrous salt mass of water lost percent by mass of water in hydrate moles of water lost mole ratio of water to anhydrous salt (the "x" in the MgS04. xH2O formula) probable formula for hydrate 20.123 g 20.818 g (show calculations here) 20.457 g

Explanation / Answer

Mass of hydrated salt = mass of crucible, cover and hydrated salt - tared mass

=( 20.818 - 20.123)

= 0.695 g

Mass of anhydrous salt = mass of crucible, cover and hydrated salt - tared mass

= 20.457 - 20.123

= 0.334 g

Moles of anhydrous salt = mass of anhydrous salt / molar mass

= 0.334 g / 120.4 gmol-1

= 2.77 * 10^-3 mol

Mass of water lost = mass of hydrated salt - mass of anhydrous salt

=( 0.695 - 0.334)g

= 0.361 g

Percent by mass of water in hydrate = (0.361 g / 0.691g) x 100

= 52.24%

Moles of water lost = 0.361 g / 18.01 gmol-1 = 0.020 mol

Mole ratio of water to anhydrous salt = 0.020 :0.0027

= 7.22 : 1

Probable formula for hydrate = MgSO4. 7 H2O

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