I was wondering about drawing amines. So could an amine have an ethyl group attached to it, but then have a methyl attched to the ethyl? Like this:
........CH3
H3C-CH-NH-CH3
(sorry about the dots, it kept formatting it wrong!)
If that's possible, how would that amine be named?
Also, could anyone explain how to do this problem?
A compound has the empirical formula CHCl. A .256 L flask, at 373 K and 750. torr, contains .800 g of the gaseous compound. Determine the molar mass and the molecular formula of the compound.
Thanks!
........CH3
H3C-CH-NH-CH3
(sorry about the dots, it kept formatting it wrong!)
If that's possible, how would that amine be named?
Also, could anyone explain how to do this problem?
A compound has the empirical formula CHCl. A .256 L flask, at 373 K and 750. torr, contains .800 g of the gaseous compound. Determine the molar mass and the molecular formula of the compound.
Thanks!
-
What you've drawn is methylisopropylamine, a secondary amine.
In essence, use the gas law to determine the number of moles present, then determine the molar mass from that:
n = PV/RT = (750 torr) (0.256 L) / (62.36367 L Torr /K mol) (373 K) = 0.0082539 moles
(0.800 g) / (0.0082539 moles) = 96.92 g/mol
The molar mass of CHCl is 48.4719 g/mol.
(96.92 g/mol) / (48.4719 g/mol) = 2.00
So the empirical formula gets multiplied by 2 to arrive at the molecular formula:
C2H2Cl2
In essence, use the gas law to determine the number of moles present, then determine the molar mass from that:
n = PV/RT = (750 torr) (0.256 L) / (62.36367 L Torr /K mol) (373 K) = 0.0082539 moles
(0.800 g) / (0.0082539 moles) = 96.92 g/mol
The molar mass of CHCl is 48.4719 g/mol.
(96.92 g/mol) / (48.4719 g/mol) = 2.00
So the empirical formula gets multiplied by 2 to arrive at the molecular formula:
C2H2Cl2