1) If the volume of a flask is 165 mL and the mass of the vapor of the flask is

Question

1) If the volume of a flask is 165 mL and the mass of the vapor of the flask is 0.8569 grams, what is the density of the gas in this flask?

At 19 oC, what is the molecular weight of this gas? (See section 5.4 in Chang)

2) The percent composition of a volatile compound is 52.13% C, 13.15% H, 34.72% O. This material was determined to have 0.3167 grams in 170 mL volume. The temperature was recorded at 27 oC and the pressure was 29.53 inches Hg.

What is the molecular formula for this compound (remember that percent composition only gives empirical formulas!) ?

Explanation / Answer

1 Density = mass in grams over volume in liters

Density = .8569 grams over .165 Liters = 5.19 grams per liter

You don't give the pressure in the flask, so I will assume one atmosphere pressure

Molecular wt. gas = grams gas times R times T over P times V

R= gas constant .082 T = Kelvin temp l9 C + 273 = 292 degrees K

V = .165 Liters

Molecular wt gas = .8569 grams times .082 times 292 degrees K over 1.0 atm times .165 Liters

Molec wt gas = 124.3

Prep for number 2. One atm pressure = 29.92 inches of Hg column on a mercury barometer.

29.53 inches mercury in problem over 29.92 inches Hg = .987 atm pressure.

2 Molecular wt gas = .3167 grams times .082 times 300 degrees K over .,987 atm times .170 Liters volume

Molecular wt. gas = 46.43

Next assume you have l00 grams of this compound to work with, therefore you will have

52.13 grams Carbon, l3.15 grams H and 34.72 grams Oxygen

Then divide each of these element masses by their respective mass per mole to convert them into moles.

52.13 grams C over l2 grams/mole = 4.344 moles C

l3.15 grams H over l.008 grams/mole = 13.05 moles H

34.72 grams O over l6 grams/mole = 2.17 moles O

Next try to convert these mole values into whole number ratios by dividing all three values by the smallest value

4.34 moles C over 2.17 = 2 moles C

13.05 moles H over 2.17 = 6.01 moles H

2.17 moles O over 2.17 = one mole O so the empirical (lowest ratio between atoms is C2H6O

and the total mass of all the atoms in this empirical or simplest formula = 46

We calculated the actual molecular wt. of this substance above to be 46, so the empirical or simplest formula must also be the actual molecular formula.

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