The stimulant amphetamine contains only carbon, nitrogen, and hydrogen. Combustion analysis of a 42.92 mg sample of amphetamine gives 37.187 mg water and 125.75 mg of carbon dioxide. If the molar mass of amphetamine is less than 160 g/mole, what is its molecular formula?
My friend needs help on this AP Chem packet question.
I only went up to Honours, so I'm not sure about this.
Please help,
X
My friend needs help on this AP Chem packet question.
I only went up to Honours, so I'm not sure about this.
Please help,
X
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mass of H = (2H/H2O) x 37.187g = 0.1111 x 37.187 = 4.132g. %H = (4.132/42.92) x 100 = 9.63%
mass of C = (C/CO2) x 125.75 = 0.2727 x 125.75 = 34.295g %C = (34.295/42.92) x 100 = 79.90%
Hence % N = 100 - (9.63 + 79.90) = 10.47%
H= 9.63/1 = 9.63, C = 79.9/12 = 6.66 and N = 10.47/14 = 0.75
Mole proportion: H = 9.63/0.75 = 13 , C = 6.66/0.75 = 9 and N = 0.75/0.75 = 1
Empirical formular = C9H13N
To determine the molecular formula, (C9H13N)x = 160 i,e 135x = 160 hence x = 1.2 = 1
So the molecular formula is C9H13N
mass of C = (C/CO2) x 125.75 = 0.2727 x 125.75 = 34.295g %C = (34.295/42.92) x 100 = 79.90%
Hence % N = 100 - (9.63 + 79.90) = 10.47%
H= 9.63/1 = 9.63, C = 79.9/12 = 6.66 and N = 10.47/14 = 0.75
Mole proportion: H = 9.63/0.75 = 13 , C = 6.66/0.75 = 9 and N = 0.75/0.75 = 1
Empirical formular = C9H13N
To determine the molecular formula, (C9H13N)x = 160 i,e 135x = 160 hence x = 1.2 = 1
So the molecular formula is C9H13N