Chemistry - Molar Mass

Question

Analysis of a volatile liquid shows it contatins 62.04% carbon. 10.41% hydrogen and some oxygen by mass. At 150.0 degrees Celsius and 760.0 mm Hg, 500.0 mL of the vapor has a mass of 1.673 g. What is the molar mass of the compound and its true formula?

Answer 1

Hello,

To begin, it is important to understand key terms. A volatile liquid is a liquid that will readily turn into a gas. So we can assume that the molecule we are looking for will become a gas. This will be important to us a little later. Let's try to find the empirical formula first, the lowest whole ratio of the molecule.

First we need to find the percent of oxygen because the others were already given to us! Since we are dealing with percents, we know it needs to be subtracted from 100%.

Thus we get:

100% - 62.04% - 10.41% = 27.55%

Next, we need to turn our percentages into moles but we have percentages. What we can actually do is make the assumption that we have 100g of the substance in questions and just turn the % into g.

Carbon: 62.04g ; Hydrogen: 10.41g ; Oxygen: 27.55 g

Now let's break it down into moles!

C: (62.04) | (1 mol) = 5.165 mol of Carbon

| (12.011g)

H: (10.41) | (1 mol) = 10.327 mol of Hydrogen

| (1.008g)

O: (27.55) | (1 mol) = 1.722 mol of Oxygen

| (16.0g)

Next, divide each by the lowest # of mols, which in this case is oxygen.

C: (5.165/1.722) = 2.999 ~ 3 moles of carbon

H: (10.327/1.722) = 5.997 ~ 6 moles of hydrogen

O: (1.722/1.722) = 1 --> 1 mole of oxygen

We use this to make our formula. We know that we can only have these 3 atoms in our formula so it should be something like this C_xH_yO_z with the variables just telling us that we know some number needs to go there. We found those numbers and can just plug them in from what we found.

C₃H₆O --> We left the 1 out of the oxygen because when we see no number we assume 1.

Part one complete! Now let's find out the molar mass! We know that molar mass is in g/mol but we do not have mols. When looking at the information given and knowing we are working with gasses (because of the volatile liquid) we can plug into PV = nRT. (P = pressure in atm, V = liters, n = moles, R = ideal gas constant 0.0821 (L*atm)/(mol*K), and T = temperature in Kelvin)

We need to convert mmHg to atm. Luck for us the given number is the conversion factor itself! (760 mmHg)*(1 atm/760 mmHg) = 1 atm.

Next let's convert mL to L. 500ml/1000 = 0.5 L and Celsius to Kelvin: 150C + 273 = 423K.

Now we plug in everything!

(1 atm)(0.5 L) = n(0.0821 L*atm/mol*K)(423K)

n = 0.01439 moles

We know molar mass is in grams per mole so we simply divide the given grams by our moles we found.

1.673g/0.0144 mols = 116.2 g/mol

Almost done! We know the molar mass of our substance but it is not the same molar mass as our empirical formula!

C₃H₆O = 58 g/mol so we just need to find the ratio of our empirical formula molar mass and the molar mass we found with the ideal gas law.

116.2/58 = ~2 so we just need to multiply all of the subscripts in our formula by 2 to get our answer.

C₆H₁₂O₂ Which if we add up it gives us approximately the g/mol we found earlier.

Let me know if you have questions!

Answer 2

Assume 100 g unknown:

62.04 g C x 1 mol/12 g = 5.17 moles C

10.41 g H x 1 mol/1 g = 10.41 moles H

100 g - 62.04 g - 10.41 g = 27.55 g O x 1 mol/16 g = 1.72 moles O

Divide all by 1.72 to obtain:

3.0 moles C

6.0 moles H

1.0 moles O

Empirical formula = C₃H₆O (molar mass = 58 g/mole)

From PV = nRT and solving for n (moles):

n = PV/RT = (1 atm)0.5 L)/(0.0821)(423) = 0.014397 moles

1.673 g/0.014397 moles = 116.2 g/mole = molar mass

116.2/58 = 2.0 empirical formula units in 1 molecular formula

Molecular formula = C₆H₁₂O₂