HELP with Chemistry Please!?

Question

1. Given a 10.00 g sample of Na2SO4 ∙ 10H2O, what weight of anhydrous sodium sulfate can be obtained after driving off the water?

2. Calculate the percentage, by weight of each of the elements (C,H) in cyclohexame. [---- we were not given the weights of these]

3. The molecular formula of cyclohexane, an organic solvent, is C6H12. Calculate the mass of a mole of material in grams.


Please show work. Any help is appreciated -- Thank you very much in advance!!

Answer 1

1.
Calculate the Mr (Na2SO4.10H2O)
Na x 2 = 23 x 2 = 46
S x 1 = 32 x 1 = 32
O x 4 = 16 x 4 = 64
H x 20 = 1 x 20 = 20
O x 10 = 16 x 10 = 160

46 + 32 + 64 + 20 + 160 = 322 (Mr (a2SO4.10H2O)
46 + 32 + 64 = 142 (Mr(Na2SO4 - anhydrous))
moles (Na2SO4.10H2O) = 10. / 322 = 0.03106
NB the molar ratios are 1 mole Na2SO4 to 10 moles of water in the crystal.
Overall there are 1 + 10 = 11 molar ratios present.
So taking the actual moles and dividing by 11 to find the value of one molar ratio.
0.03106/11 = 0.002823 (moles Na2SO4)
Value of 10 molar ratios
0.03106 x 10/11 = 0.02823 (moles H2O)

Mass of anhydrous Na2SO4 = 0.002823 x (Mr(Na2SO4))
Mass of anhydrous Na2SO4 = 0.002823 x 142

Mass of anhydrous Na2SO4 = 4.008 grams.

2.
??Careful with chemical names, they can be different substances. I think you mean cyclohexane - not cyclohexame.
Ar(C) = 12
Ar(H) = 1
Formula for cyclohexane is C6H12
Moles(C) = 12 x 6 = 72
Moles(H) = 1 x 12 = 12
72 + 12 = 84
% mass of C is 72/84 x 100 = 85.7%
% mass of H is 12/84 x 100 = 14.3%

3.
Remember the equation:-
moles = mass/Mr
mass = moles x Mr
mass = 1 mole x 84 Mr = 84 grams.

Answer 2

1. Let Na2SO4*10H2O be called H
Atomic weights: Na=23 S=32 O=16 H=1 O=16 Na2SO4*10H2O=299 Na2SO4=119

10.00gH x 1molH/299gH x 1molNa2SO4/1molH x 119gNa2SO4/1molNa2SO4 = 3.980g Na2SO4

2. Atomic weights: C=12 H=1 C6H12=84

C6=72 H12=12

72/84 x 100% = 85.7%
12/84 x 100% = 14.3%

3. C6=72 H12=12
72 + 12 = 84
1 mole C6H12 = 84g