Ten Examples

Determine empirical formula given percent composition | Return to Mole Table of Contents |

Determine empirical formula when given mass data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data

Determine the formula of a hydrate Hydrate lab calculations

**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 C37.2 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 : 1Within experimental error, this is a 2:3:1 lowest whole-number ratio

4) The empirical formula is:

C_{2}H_{3}O

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¯^{1}K¯^{1}) (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:

C_{4}H_{6}O_{2}

**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 C_{2}H_{5} and its 'empirical formula weight' is 29.0615 grams per mole of C_{2}H_{5}.

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

(9.5 atm) (0.975 L) = (n) (0.08206 L atm mol¯^{1}K¯^{1}) (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 = 2The molecular formula is C

_{4}H_{10}

**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 molDivide by smallest value to get a 1:1:2 ratio. Empirical formula is COCl

_{2}

2) Molecular weight:

PV = nRT(992 torr/760 torr/atm) (2.00 L) = (n) (0.08206 L atm mol¯

^{1}K¯^{1}) (1833 K)n = 0.017355 mol

1.72 g / 0.017355 mol = 99.1 g/mol

3) Molecular formula:

From the empirical formula of COCl_{2}, we determine the "empirical formula weight" to be 98.9 g/mol99.1 / 98.9 = 1

The molecular formula is COCl

_{2}, same as the empirical formula. Its name, by the way, is phosgene.

**Example #4:** A gaseous compound containing xenon and fluorine, XeF_{x} (where x = some number), was placed in a sealed flask at room temp. and the pressure was measured at 24 mm Hg. H_{2} 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 H_{2}, was 52 mm Hg. What is the empirical formula of XeF_{x}?

**Solution #1:**

1) Write a chemical equation for what happens:

XeF_{x}+^{x}⁄_{2}H_{2}---> Xe + xHF

2) Determine pressure of Xe:

From the equation above, the XeF_{x}: Xe molar ratio is 1:124 torr of XeF

_{x}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 XeF_{x}:

Xe to H_{2}ratio ---> 24 to 48the hydrogen above is in molecular form, we need it in atomic form:

Xe to H ratio ---> 24 to 96

96 / 24 = 4

XeF

_{4}

**Solution #2:**

Pressure is proportional to number of moles.The added H

_{2}amount is 100 − 24 = 76 torrReaction produces Xe and HF (which is removed). Reaction vessel also contains unreacted H

_{2}= 52 torrXe = 24 torr due to one to one ratio of XeF

_{x}reacted to Xe produced.Unreacted H

_{2}is 52 − 24 = 28 torrReacted H

_{2}is 76 − 28 = 48 torrThe ratio is NOT 24 to 48 leading to XeF

_{2}We must consider the hydrogen as atoms, not molecules. 48 torr of H

_{2}equals 96 torr of H since H_{2}---> 2H.24 to 96 is the proper ratio giving XeF

_{4}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 molC ---> 4.3377 / 2.16875 = 2

H ---> 13.095 / 2.16875 = 6.038

O ---> 2.16875 / 2.16875 = 1Empirical formula ---> C

_{2}H_{6}O

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 C_{2}H_{6}O is 46.068492.33 / 46.0684 = 2

The molecular formula is C

_{4}H_{12}O_{2}

**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 CClF_{2}. The "empirical formula weight" is 85.46 grams per mole of CClF_{2}.

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 = 2The molecular formula is C

_{2}Cl_{2}F_{4}.

**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 gcarbon ---> 85.63 g / 12.011 g/mol = 7.12930 mol

hydrogen ---> 14.37 g / 1.008 g/mol = 14.2560 molcarbon ---> 7.12930 mol / 7.12930 mol = 1

hydrogen ---> 14.2560 mol / 7.12930 mol = 2The empirical formula is CH

_{2}

4) Determine the molecular formula:

56.09 / 14.027 = 3.9987 = 414.027 is the weight of one mole of CH

_{2}CH

_{2}times 4 gives C_{4}H_{8}<--- 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 gcarbon ---> 92.3 g / 12.011 g/mol = 7.6846 mol

hydrogen ---> 7.7 g / 1.008 g/mol = 7.6389 molWithin 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 = 8The molecular formula is C

_{8}H_{8}

**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 / MMPV = (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 C

_{2}H_{2}Cl_{2}

**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 gConvert to moles:

C = 5.162

H = 10.32

O = 1.725Find lowest whole-number ratio:

C = 3

H = 6

O = 1The empirical formula is C

_{3}H_{6}OThe "empirical formula weight" (not a standard chemistry term, BTW) = 58

116.2/58 = 2

The molecular formula is C

_{6}H_{12}O_{2}

Determine empirical formula given percent composition | Return to Mole Table of Contents |

Determine empirical formula when given mass data

Determine identity of an element from a binary formula and a percent composition

Determine identity of an element from a binary formula and mass data