### Determine molecular formula using the Ideal Gas LawTen Examples

Example #1: An unknown compound was determined to have the following percentage composition:

C: 55.8%, O: 37.2%, H: 7.0%.

A 4.50 g sample is taken and found to occupy 1.17 L at STP. Determine the empirical formula.

Solution:

1) Let's work with a 100 g sample, that way we can easily convert the percentages to grams:

55.8 g / 12.011 g/mol = 4.64574 mol C

37.2 g / 16.00 g/mol = 2.325 mol O

7 g / 1.008 g/mol = 6.9444 mol H

2) The mole ratio of C : H : O is:

4.64574 : 6.9444 : 2.325

3) Divide through by 2.325:

1.998 : 2.987 : 1

Within experimental error, this is a 2:3:1 lowest whole-number ratio

4) The empirical formula is:

C2H3O

5) We find moles of the compound that are present in the sample with PV = nRT:

(1.00 atm) (1.17 L) = (n) (0.08206 L atm mol¯11) (273 K)

n = 0.05223 mol

6) Calculate the molar mass:

4.50 g / 0.05223 mol = 86.2 g/mol

7) The empirical formula weight is:

2(12) + 3 + 16 = 43 g/mol

8) Determine scaling factor needed to determine molecular weight:

86 / 43 = 2

9) We need to double our empirical formula to get the actual molecular formula:

C4H6O2

Example #2: A gas containing carbon and hydrogen analyzed as 82.6% carbon and 17.4% hydrogen. A 15.2 g sample of a gas occupied a volume of 975 mL with a pressure of 9.5 atm at 161 °C. Determine the molecular formula of the compound.

Solution:

1) Assume 100 grams of the compound is present. Then, convert masses to moles:

C ---> 82.6 g / 12.011 g/mol = 6.877 mol
H ---> 17.4 g / 1.008 g/mol = 17.262 mol

2) Divide through by smallest:

C ---> 6.877 mol / 6.877 mol = 1
H ---> 17.262 mol / 6.877 mol = 2.5

3) The empirical formula is C2H5 and its 'empirical formula weight' is 29.0615 grams per mole of C2H5.

4) Use PV = nRT to determine the molecular weight:

(9.5 atm) (0.975 L) = (n) (0.08206 L atm mol¯11) (434 K)

n = 0.26008 mol

15.2 g / 0.26008 mol = 58.4 g/mol

5) Determine the molecular formula:

58.4 / 29.06 = 2

The molecular formula is C4H10

Example #3: A sample of gas contains 0.305 g carbon, 0.407 g oxygen, and 1.805 g chlorine. A different sample of the same gas weighing 1.72 g occupies 2000. milliliters and exerts a pressure of 992 torr at 1560 °C. What is the molecular formula of the gas?

Solution:

1) Empirical formula:

carbon ---> 0.305 g / 12.011 g/mol = 0.02539 mol
oxygen ---> 0.407 g / 16.00 g/mol = 0.0254375 mol
chlorine ---> 1.806 g / 35.45 g/mol = 0.05094 mol

Divide by smallest value to get a 1:1:2 ratio. Empirical formula is COCl2

2) Molecular weight:

PV = nRT

(992 torr/760 torr/atm) (2.00 L) = (n) (0.08206 L atm mol¯11) (1833 K)

n = 0.017355 mol

1.72 g / 0.017355 mol = 99.1 g/mol

3) Molecular formula:

From the empirical formula of COCl2, we determine the "empirical formula weight" to be 98.9 g/mol

99.1 / 98.9 = 1

The molecular formula is COCl2, same as the empirical formula. Its name, by the way, is phosgene.

Example #4: A gaseous compound containing xenon and fluorine, XeFx (where x = some number), was placed in a sealed flask at room temp. and the pressure was measured at 24 mm Hg. H2 gas was added until the pressure reached 100 mm Hg, with temperature constant.

An electrical discharge was passed through the gas, converting the compound to xenon and hydrofluoric acid, HF. The HF was removed using solid KOH and the flask cooled to room temperature. The pressure of the flask, now containing only xenon and unreacted H2, was 52 mm Hg. What is the empirical formula of XeFx?

Solution #1:

1) Write a chemical equation for what happens:

XeFx + x2H2 ---> Xe + xHF

2) Determine pressure of Xe:

From the equation above, the XeFx : Xe molar ratio is 1:1

24 torr of XeFx produces 24 torr of Xe

3) Determine pressure of unreacted hydrogen:

52 − 24 = 28 torr

4) Determine pressure of reacted hydrogen:

76 − 28 = 48 torr

(the 76 comes from 100 − 24)

5) Determine formula for XeFx:

Xe to H2 ratio ---> 24 to 48

the hydrogen above is in molecular form, we need it in atomic form:

Xe to H ratio ---> 24 to 96

96 / 24 = 4

XeF4

Solution #2:

Pressure is proportional to number of moles.

The added H2 amount is 100 − 24 = 76 torr

Reaction produces Xe and HF (which is removed). Reaction vessel also contains unreacted H2 = 52 torr

Xe = 24 torr due to one to one ratio of XeFx reacted to Xe produced.

Unreacted H2 is 52 − 24 = 28 torr

Reacted H2 is 76 − 28 = 48 torr

The ratio is NOT 24 to 48 leading to XeF2

We must consider the hydrogen as atoms, not molecules. 48 torr of H2 equals 96 torr of H since H2 ---> 2H.

24 to 96 is the proper ratio giving XeF4 for the answer.

Example #5: A 13.4 g sample of an unknown liquid is vapourized at 85.0 °C and 100.0 kPa. The vapor has a volume of 4.32 L. The percentage composition of the liquid is found to be 52.1% carbon, 13.2% hydrogen, and 34.7% oxygen. What is the molecular formula?

Solution:

1) Determine the empirical formula of the compound:

C ---> 52.1 g / 12.011 g/mol = 4.3377 mol
H ---> 13.2 g / 1.008 g/mol = 13.095 mol
O ---> 34.7 g / 16.00 g/mol = 2.16875 mol

C ---> 4.3377 / 2.16875 = 2
H ---> 13.095 / 2.16875 = 6.038
O ---> 2.16875 / 2.16875 = 1

Empirical formula ---> C2H6O

2) Determine the moles of the compound:

PV = nRT

(100.0 kPa / 101.325 kPa/atm) (4.32 L) = (n) (0.08206 L atm / mol K) (358 K)

n = 0.14513 mol

3) Determine the molecular weight of the compound:

13.4 g / 0.14513 mol = 92.33 g/mol

4) Determine the molecular formula:

The empirical formula weight of C2H6O is 46.0684

92.33 / 46.0684 = 2

The molecular formula is C4H12O2

Example #6: A 2.135 g sample of a gaseous chlorofluorocarbon (a type of gas implicated in the depletion of stratospheric ozone) occupies a volume of 315.5 mL at 739.2 mmHg and 26.1 °C. Analysis of the compound shows it to be 14.05% C, 41.48% Cl and 44.46% F by mass. What is the molecular formula of the compound?

Solution:

1) Assume 100 g of compound is present. Convert masses to moles:

C ---> 14.05 g / 12.011 g/mol = 1.1698 mol
Cl ---> 41.48 g / 35.453 g/mol = 1.1699997 mol
F ---> 44.46 g / 18.998 g/mol = 2.340246 mol

2) Divide through by smallest:

C ---> 1.1698 mol / 1.1698 mol = 1
Cl ---> 1.1699997 mol / 1.1698 mol = 1
F ---> 2.340246 mol / 1.1698 mol = 2

3) Empirical formula is CClF2. The "empirical formula weight" is 85.46 grams per mole of CClF2.

4) Use PV = nRT to determine the molar mass (I left the units off):

(739.2 / 760) (0.3155) = (n) (0.08206) (299.1)

n = 0.0125026 mol

2.135 g / 0.0125026 mol = 170.76 g/mol

5) Determine the molecular formula:

170.76 / 85.46 = 2

The molecular formula is C2Cl2F4.

Example #7: A 4.702 g sample of a pure gas occupies 1.500 L at 150.0 kPa and 49.8 °C. The gas contains only carbon (85.63% by mass) and hydrogen. Determine the molecular formula of the gas.

Solution:

1) Use PV = nRT to determine moles of gas present:

(150.0 kPa) (1.500 L) = (n) ( ______ L kPa /mol K) (322.8 K)

Comment: the value for R varies based on the units used. Many sites across the Internet provide tables which provide the value associated with a given set of units. Here is one such table. The value we need is in the second table, seven rows down and two columns in.

(150.0 kPa) (1.500 L) = (n) (8.31447 L kPa /mol K) (322.8 K)

n = 0.0838329 mol

2) Use the provided mass of the sample to determine the molecular weight:

4.702 g / 0.0838329 mol = 56.09 g/mol

3) Determine the empirical formula of the substance using the percent composition data. Start by assuming 100 g of the substance is present. Therefore:

carbon ---> 85.63 g
hydrogen ---> 14.37 g

carbon ---> 85.63 g / 12.011 g/mol = 7.12930 mol
hydrogen ---> 14.37 g / 1.008 g/mol = 14.2560 mol

carbon ---> 7.12930 mol / 7.12930 mol = 1
hydrogen ---> 14.2560 mol / 7.12930 mol = 2

The empirical formula is CH2

4) Determine the molecular formula:

56.09 / 14.027 = 3.9987 = 4

14.027 is the weight of one mole of CH2

CH2 times 4 gives C4H8 <--- the molecular formula

Example #8: Determine the molecular formula of a substance that contains 92.3%C and 7.7%H by mass. This same substance when in the gas phase has a density of 0.00465 g/mL at STP.

Solution:

1) Use percent composition data to determine the emirical formula. Star by assuming 100 g of the substance is present. Therefore:

carbon ---> 92.3 g
hydrogen ---> 7.7 g

carbon ---> 92.3 g / 12.011 g/mol = 7.6846 mol
hydrogen ---> 7.7 g / 1.008 g/mol = 7.6389 mol

Within reasonable error, this is a 1:1 molar ratio. The empirical formula is CH.

2) Use the Ideal Gas Law to get moles of the gas present:

PV = nRT

(1.00 atm) (0.0010 L) = (n) (0.08206 L atm / mol K) (273 K)

n = 0.000044638 mol

Comment: I changed 1.0 mL to 0.0010 L

3) Use the mass part of the density to get the molar mass:

0.00465 g / 0.000044638 mol = 101.7 g/mol

4) Determine the molecular formula:

104.2 / 13.019 = 8

The molecular formula is C8H8

Example #9: A compound has the empirical formula CHCl. A 256-mL flask, at 100 °C and 0.967 atm, contains 0.800 g of the gaseous compound. Calculate the molar mass of this compound, and determine its molecular formula.

Solution:

1) I will use the Ideal Gas Law, but I'm going to modify it first. P, V, n, R, and T have their usual meanings while m stands for mass in grams and MM stands for the molar mass (which is what I will solve for). We start with these equations:

PV = nRT and n = m / MM

PV = (m / MM) RT

(MM) PV = mRT

MM = mRT / PV

MM = [(0.800 g) (0.08206 L atm / mol K) (373 K)] / [(0.967 atm) (0.256 L)]

MM = 98.92 g/mol

2) Determine the molecular formula:

The weight of CHCl is 48.472.

98.92/ 48.472 = 2

The molecular formula is C2H2Cl2

Example #10: Compound Z is an organic compound which contains only carbon, hydrogen, and oxygen. Elemental analysis shows that the compound is 62.0% carbon and 10.4% hydrogen. An experiment was conducted to determine the molecular weight. A sample of compound Z was added to a premassed flask. The flask with the liquid was immersed in a beaker of boiling water. The flask was allowed to remain in the boiling water until all the liquid had vaporized and the flask was filled with vapor at room pressure. The flask was then removed from the boiling water, allowed to cool to room temperature, and the mass of the flask filled with vapor was determined. The flask was then cleaned and filled with water and massed again. Determine the empirical formula and the molecular formula from the data given.

Mass of the flask = 158.762 g
Temperature of the boiling water = 99.2 °C
Room pressure = 75.47 cm Hg
Mass of flask with vapor = 159.882 g
Mass of flask filled with water = 455.031 g
Water temperature = 23.0 °C

Solution:

Note: this technique for determining the molecular weight is called the Method of Dumas.

1) The mass of the vapor filling the flask:

159.882 g − 158.762 g = 1.120 g

2) The volume of the flask:

455.031 g − 158.762 g = 296.269 g

(296.269 g) (1.00 g/mL) = 296.269 mL

You could refine this by getting the density of water at 23 °C and using it to convert from 296.269 g to the proper volume.

3) The molecular weight of compound Z

PV = nRT

(754.7 mmHg / 760 mmHg/atm) (0.296269 L) = (n) (0.08206 L atm / mol K) (372.2 K)

n = 0.0096325 mol

1.120 g / 0.0096325 mol = 116.3 g/mol

We take the temperature of the vapor to be the temperature of the boiling water the flask is sitting in. That's the source of the 372.2 K.

4) Calculate empirical formula, then molecular formula:

Assume 100 g of Z is present. Therefore:

C = 62.0 g
H = 10.4 g
O = 27.6 g

Convert to moles:

C = 5.162
H = 10.32
O = 1.725

Find lowest whole-number ratio:

C = 3
H = 6
O = 1

The empirical formula is C3H6O

The "empirical formula weight" (not a standard chemistry term, BTW) = 58

116.2/58 = 2

The molecular formula is C6H12O2