Determine the emperical formulae of a compound with the following composition by mass: 18% C, 2.5% H, 63.5% I and 16% O. If this compound has a molar mass of 400g/mol, what is the molecular formulae?
Please do show the steps and what is meant by 400g/mol.
Please do show the steps and what is meant by 400g/mol.
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Assume a 100-gram sample:
mole of C = 18.0g x (1 mol C/12.01g) = 1.50 moles
mole of H = 2.50g x (1 mol H/1.01g) = 2.48 moles
mole of I = 63.5g x (1 mol I/126.90g) = 0.500 mole
mole of O = 16.0g x (1 mol O/16.00g) = 1.00 mole
C: 1.50/.500 = 3
H: 2.48/.500 = 4.96 ≈ 5
I: .500/.500 = 1
O: 1.00/.500 = 2
Empirical Formula is : C3H5IO2
n = (400 g/mol)/199.98 g/mol = 2
Molecular Formula = C3H5IO2 x (2) = C6H10I2O4
Thank you...
mole of C = 18.0g x (1 mol C/12.01g) = 1.50 moles
mole of H = 2.50g x (1 mol H/1.01g) = 2.48 moles
mole of I = 63.5g x (1 mol I/126.90g) = 0.500 mole
mole of O = 16.0g x (1 mol O/16.00g) = 1.00 mole
C: 1.50/.500 = 3
H: 2.48/.500 = 4.96 ≈ 5
I: .500/.500 = 1
O: 1.00/.500 = 2
Empirical Formula is : C3H5IO2
n = (400 g/mol)/199.98 g/mol = 2
Molecular Formula = C3H5IO2 x (2) = C6H10I2O4
Thank you...
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Nesteve JOHN's answer is correct...