help figuring this answer out!?

Question

For 8.7 mmole of the compund MgSO4*7 H2O:

What is the percent water in this amount of the compound?
What weight of sulfur is there in this amount of compound?

Answer 1

% water = weight of water/weight of MgSO4*7 H2O

for % you don't need a specific weight (just use 1 mole):

MgSO4 wt/mol = [24.3 + 32 + (4 x 16)] = 120.3g/mol

Water wt/mol = (2 x 1) + 16 = 18g/mol

(7 x 18)/[120.3 + (7 x 18)] = 126/[120.3 + 126) =

126/246.3 x 100% = 51.2% water

For weight of sulfur in 8.7mmole of MgSO4*7 H2O

Easy, each mole of this material contains one mole of sulfur, so 8.7 mmoles will contain 8.7 mmoles of sulfur; just convert to grams:

32g/mol x 0.0087 mol = 0.278g