Question

A student collects 350 mL of a vapor at a temperature of 67°C. The atmospheric pressure at the time of collection is 0.900 atm. The student determines that the mass of the vapor collected is 0.79 g. Which of the following is most likely to be the chemical analyzed?

acetone
ethanol
ethyl acetate
pentane.

Answer 1

The answer to your question is: Pentane (0.011 moles)

Step-by-step explanation:

Data

V = 350 ml = 0.35 l

T = 67°C = 340 °K

P = 0.9 atm

mass = 0.79 g

R = 0.082 atm L/mol°K

Formula

PV = nRT

n = PV / RT

n = (0.9)(0.35) / (0.082)(340)

n = 0.315 / 27.88

n = 0.0112

Now

MW acetone = 58g

MW ethanol = 46g

MW ethyl acetate = 88 g

MW pentane = 72 g

For acetone

58 g ------------ 1 mole

0.79 g -------- x

x = 0.014 moles

For ethanol

46g --------------- 1 mole

0.79g ------------ x

x = 0.17 moles

For ethyl acetate

88 g ------------- 1 mole

0.79 g ----------- x

x = 0.0089 moles

For pentane

72 g -------------- 1 mole

0.79 g ------------ x

x = 0.011 moles

The substance is pentane