A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O

Question

Explanation / Answer

First, determine the number of moles and the molecular mass of the gas:

n = PV/RT = [(0.987 atm)(1.00 L)] / [(0.08206 atmL/moleK)(393 K)] = 0.0306 moles

MM = (2.30 g)/(0.0306 moles) = 75.2 g/mole

Second, determine the number of moles of C, H, and O in 100 g of the molecule:

C = (64.9 g C) x [(1 mole C)/(12.01 g C)] = 5.40 moles C

H = (13.5 g H) x [(1 mole H)/(1.01 g H)] = 13.4 moles H

O = (21.6 g O) x [(1 mole O)/(16.0 g O)] = 1.35 moles O

Now find the empirical formula of the gas:

C = (5.40 moles)/(1.35 moles) = 4

H = (13.4 moles)/(1.35 moles) = 10

O = (1.35 moles)/(1.35 moles) = 1

empirical formula = C4H10O

empirical formula mass = 74.14 g/mole

The empirical formula mass matches well with the molecular molar mass, therefore the gas is C4H10O.

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