Emperical Formula Help!?
A mass spectrometer analysis finds that a molecule has a composition of 48% Cd, 20.8% C, 2.62% H, 27.8% 0. Determine the emperical Formula.
Work would be greatly appreciated. I think its Cd C4 H6 O4... Those would be below the letters but I dont know how to do that.
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Let us assume 100 g of the compound is present. This means:
48g Cd, 20.8g C, 2.62g H, 27.8g O
Let us determine moles present:
Cd: 48 g / 112.4 g/mol = 0.427 mol
C: 20.8 g / 12.011 g/mol = 1.732 mol
H: 2.82 g / 1.008 g/mol = 2.7976 mol
O: 27.8 g / 16.00 g/mol = 1.7375 mol
Divide through by lowest value:
Cd: 0.427 mol / 0.427 mol = 1
C: 1.732 mol / 0.427 mol = 4.06
H: 2.7976 mol / 0.427 mol = 6.55
O: 1.7375 mol / 0.427 mol = 4.07
That 6.55 isn't wonderful, however your answer is the best one. Cadmium is divalent, so we can see the empirical formula as Cd(C2H3O2)2
C2H3O2^- is the acetate ion, so we have cadmium acetate as the compound.
Good job on the solution. Onward!
Update: That 6.55 comes from me making a copying error. It's 2.62, not 2.82 for the H. 2.62 yields 6.08, not 6.55.
Oops!