Ineed with calculating the percent of hydrate and mass
info:
a dry, empty crucible was obtained weighing 10.00 grams. the hydrate (beso4 ● 4 h2o) was placed into this crucible and the total mass was recorded at 15.00 grams. after heating the hydrate to remove the water, the mass of the anhydrous compound and crucible was 12.94 grams.
questions:
a) determine the mass of the original hydrate.
b) determine the mass of just the anhydrous salt.
c) calculate the number of moles of anhydrous salt using the molar mass of just beso4.
d) determine the mass of water lost.
e) calculate the number of moles of water lost using the molar mass of water.
f) determine the smallest whole number ratio of moles of water : moles of anhydrous salt.
g) did the experiment results corroborate with the formula beso4 ● 4 h2o?

a) 5.0 g.

b) 2.94 g.

c) 0.02798 mol.

d) 2.06 g.

e) 0.1144 mol.

f) 4.0 mol/mol.

g) Yes, the experiment results corroborate with the formula BeSO₄.4H₂O.

Explanation:

a) Determine the mass of the original hydrate.

The mass of the original hydrated salt = The mass of the crucible with the hydrated salt - the mass of the empty crucible = 15.0 g - 10.0 g = 5.0 g.

b) Determine the mass of just the anhydrous salt.

The mass of the anhydrous salt = the mass of the crucible with the anhydrous salt (after heating) - the mass of the empty crucible = 12.94 g - 10.0 g = 2.94 g.

c) Calculate the number of moles of anhydrous salt using the molar mass of just BeSO₄.

The no. of moles of the anhydrous salt = mass of anhydrous salt/molar mass of BeSO₄ = (2.94 g)/(105.075 g/mol) = 0.02798 mol.

d) Determine the mass of water lost.

The mass of water lost = the mass of the crucible with the hydrated salt (before heating) - the mass of the crucible with the anhydrous salt (after heating) = 15.0 g - 12.94 g = 2.06 g.

e) Calculate the number of moles of water lost using the molar mass of water.

The no. of moles of water lost = the mass of water lost/molar mass of water = (2.06 g)/(18.0 g/mol) = 0.1144 mol.

f) Determine the smallest whole number ratio of moles of water : moles of anhydrous salt.

The smallest whole number ratio of moles of water : moles of anhydrous salt = The no. of moles of water lost/The no. of moles of the anhydrous salt = (0.1144 mol)/(0.02798 mol) = 4.089 mol/mol ≅ 4.0 mol/mol.

g) Did the experiment results corroborate with the formula BeSO₄.4H₂O?

Yes, the experiment results corroborate with the formula BeSO₄.4H₂O.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.