MISE - Physical Basis of Chemistry



NAMES:

Cory Redding

Jeff Bowes

MISE - Physical Basis of Chemistry

Third Experiment

Molecular Weight of a Volatile Liquid

Lab Report – Submit electronically (digital drop box) by Sunday, November 6, 6:01PM

Note: When submitting to digital Drop Box label you files with your name first and

then a brief description of what the document is.

Purpose: The Purpose of this experiment is to determine the molecular weight of a

compound that is a volatile liquid. By vaporizing a measured amount of the

sample, and determining the pressure, volume, and temperature of the sample,

the approximate molecular weight may be determined by presuming

ideal gas behavior.

Background:

• # 2 - Pressure, Gas Laws, Ideal Gas Law, Partial Pressures - Some Examples

• Gases - Experiments and Relationships - PowerPoint Summary

(PowerPoint summary of our discussion of gases.)

These handouts are on Bb.

Reference: For some procedural aspects of this lab:

Catalyst - The Prentice-Hall Custom Laboratory Program for Chem 53 at the

University of Pennsylvania Department of Chemistry, Pearson Custom

Publishing, 2005, pages 57-65.

Introduction:

In Experiment # 2, you had the opportunity to determine the empirical formula of

a binary compound. In order to determine the molecular formula of a compound

- given the empirical formula - you need the molecular weight of the compound

(in addition to the empirical formula weight).

In this experiment, you will utilize one of many experimental procedures to determine the

molecular weight of a compound. This method, first developed by Dumas, exploits the

ideal gas law in order to determine the molecular weight of a compound that can be

vaporized (without significant decomposition). Specifically - via the ideal gas law - you

will determine the molecular weight of a substance by determining the pressure exerted,

volume occupied, temperature, and mass of a sample of its vapor. The example below

will hopefully enable you to understand how this is accomplished.

Presuming that the vaporized volatile liquid follows the ideal gas law (PV = nRT),

we may determine the moles of the contained vapor via:

n = moles of contained vapor = , i.e., its measured pressure (P), volume (V),

and temperature (T). Of course, the gas constant (R) is tabulated and equals

0.0821 L• atm•mol-1•K-1. Then, the molecular weight of the gas (MW in g/mol)

can be determined from the measured mass of the contained vapor (g in grams)

divided by the calculated number of moles.

MW = = = .

Before you carry out the determination of the molecular weight (MW) of the volatile liquid,

a worked example may help. Please refer to the Appendix at the end of this experiment.

Data/CALCULATIONS

Note: Data and Calculations may be completed on an Excel spreadsheet.

Unknown Letter: __C___________

Molecular Weight of a Volatile Liquid

(If you performed more than one trial, adjust your data table accordingly.)

Weight of flask + foil cap + rubber band 90.8276 g

Temperature of water bath when (excess) vapor ceased to escape 89.9 ºC

Weight of flask + foil cap + rubber band + condensed vapor 91.2679 g

Weight of condensed vapor .4403 g

Determination of full volume of flask:

Weight of Erlenmeyer flask filled with water 241.63 g

Weight of clean, dry, and empty Erlenmeyer flask 90.79 g

Weight of water that completely filled Erlenmeyer flask 150.84 g

Temperature of water in the filled Erlenmeyer flask 24.2 ºC

Calculated volume of Erlenmeyer flask ___.1508_____________ L

Barometric pressure 751.5 mm Hg (xxx.x mm Hg)

Before you carry out the determination of the molecular weight (MW) of the volatile liquid,

a worked example may help. Please refer to the Appendix at the end of this experiment.

Table 1: (Revised**) Density of pure liquid water at various temperatures:

|Temperature (ºC) |density (d) in g/mL |Temperature (ºC) |density (d) in g/mL |

|1ºC |0.999890 |15ºC |0.999099 |

|2ºC |0.999940 |16ºC |0.998943 |

|3ºC |0.999960 |17ºC |0.998774 |

|4ºC |0.999970 |18ºC |0.998595 |

|5ºC |0.999970 |19ºC |0.998405 |

|6ºC |0.999950 |20ºC |0.998203 |

|7ºC |0.999910 |21ºC |0.997992 |

|8ºC |0.999860 |22ºC |0.997770 |

|9ºC |0.999790 |23ºC |0.997538 |

|10ºC |0.999720 |24ºC |0.997296 |

|11ºC |0.999620 |25ºC |0.997044 |

|12ºC |0.999520 |26ºC |0.996783 |

|13ºC |0.999400 |27ºC |0.996512 |

|14ºC |0.999270 |28ºC |0.996232 |

** Density data from 1ºC to 14ºC (at 0.1013 Mbar (1 atm)) source:

NIST - National Institute of Standards & Technology website:

Determination of the volume occupied by the vapor in the Erlenmeyer flask:

(Using the mass of water that completely filled the Erlenmeyer flask and the above

table of water densities as needed – determine the volume (in L) occupied by the vapor

in the Erlenmeyer flask. Also, explain your reasoning. Then, place your result

in the data table at the appropriate spot. Show work below.

1mL=1g (of water) therefore 150.8mL= 150.8 g

D=m/v

.997296 g/mL=150.8g/V

V=150.8g/.997296g/mL

V=151.21mL

V=.1512L

Volume of vapor .1512 L

Determination of molecular weight of the volatile liquid:

(Assuming ideal gas behavior, use the pressure of the vapor (P), its volume (V), and

its temperature (T) to determine the moles of vapor in the Erlenmeyer flask.

Then, using the mass of vapor which occupied the Erlenmeyer flask, determine the

molecular weight (MW in g/mole) of the volatile liquid. If you performed more than

one trial, show all work for each trial and then compute the average molecular weight.) Show work below.

mw= gRT/PV

Convert:

(751.5mmHg)/(760mmHg)= .9888atm

(89.9oC) + 273=362.9K

MW=(.4403g)(.0821 L*atm*mol-1*K-1)(362.9K)/ (.9888atm)(.1512 L)

mW=87.744

(Average) Molecular Weight 87.744 g/mole

Determination of Empirical and Molecular Formula of volatile liquid:

The table below lists the pure volatile liquids that were distributed in lab. The first column

gives the letter designation of the compound and the remaining columns give the

appropriate elemental mass percents. Using the data for your particular volatile

liquid, please determine the empirical formula and the molecular formula for your

particular sample. Show all work.

|Sample Letter |mass % carbon (C) |mass % hydrogen (H) |mass % oxygen (O) |

|“A” (cyclohexane) |85.60 % |14.40 % |None |

|“B” (ethyl acetate) |54.52 % |9.17 % |36.31 % |

|“C” (2-propanol) |59.94 % |13.44 % |26.62 % |

Sample c

To determine the empirical formula:

C= (59.94g)/(12g/mol)= 4.995 mol

H= (13.44g) / (1 g/mol) = 13.44mol

O= (26.62g) / (16 g/mol)= 1.66mol

Find the ratio:

4.995mol C/1.66mol O= 3.00

13.44 mol H / 1.66 mol O= 8.10

1.66 mol O/ 1.66mol O= 1

C3H8O

TO DETERMINE THE MOLECULAR FORMULA:

MW=

C=3*12= 36

H=1* 8= 8

O= 1*16= 16

Total Molecular Weight: 60

Average molecular weight / empirical weight= molecular formula

87.744 g/mol / 60 g/mol = 1.46

We rounded down to 1 to determine that the molecular formula is the same as the empirical formula in this case… C3H8O.

Had we rounded up to 2, the molecular formula would have been C6H16O2, which when simplified is still the empirical formula of C3H8O.

Empirical Formula C3H8O Molecular Formula C3H8O

Conclusion Questions

1. Why is the barometric (i.e., atmospheric) pressure considered to be the pressure

of the vapor, i.e., how does the experimental procedure ensure this? Explain carefully.

The experimental procedure ensured that the barometric pressure was the same as the pressure of the vapor because we poked a hole in the foil to allow the slow, constant escape of the excess vapor. By allowing the vapor to slowly escape, we avoided air from the room from rushing into the flask. The temperature was also kept constant so the system could equilibrate. Had the temperature been higher, the liquid would have heated too quickly and air from the room could have rushed in.

2. Why isn't it necessary to weigh the amount of liquid initially put into the flask?

Explain.

It is not necessary to weigh the amount of liquid initially put into the flask because a large amount of the liquid is excess. We are concerned only with the vapor that fills the flask after the excess liquid has evaporated and exited the flask. At the end of the experiment, we allowed the remaining vapor to cool and condense into a liquid. It was necessary to weigh the cooled vapor (liquid) to determine the weight of the vapor that filled the flask.

Extra –

It is sometimes stated that the above method (Dumas method) relies on the

presumption that the investigated gas follows the ideal gas law (PV = nRT).

This allows one to determine the moles of the contained vapor via:

n = moles of contained vapor = , i.e., its measured pressure (P), volume (V),

and temperature (T). Of course, the gas constant (R) is tabulated and equals

0.0821 L• atm•mol-1•K-1. Then, the molecular weight of the gas (MW in g/mol)

can be determined from the measured mass of the contained vapor (g in grams) divided

by the calculated number of moles.

MW = = = . In other words, if the gas did not behave ideally, then the

calculated molecular weight would be a crude approximation at best. Please explain

why this is commonly not a big problem, i.e., why an approximate value of the

molecular weight is often sufficient when determining the actual chemical formula

of a volatile compound. What other information is commonly obtained for a compound

in the process of determining its chemical formula such that an approximate molecular

weight is often “good enough”?

The empirical formula may also be obtained for a compound in the process of determining its chemical formula by using the Law of Simplicity. Scientists often took a leap of faith by assuming that the ratio of the compound was 1:1. Once they determined one of the compounds empirical formulas by assuming they combined in a 1:1 ratio, they were able to use that information to determine the empirical formulas of other compounds formed by those elements.

Appendix - A worked Example:

A certain volatile hydrocarbon (a binary compound of carbon and hydrogen) is found to be

92.3 % carbon, by mass. In a separate experiment, utilizing the Dumas method, a

4.00 mL pure liquid sample of this hydrocarbon is vaporized in a 125 mL Erlenmeyer flask

when the barometric pressure is 768.0 torr. The empty flask - fitted with a foil cap pierced with

a pinhole - weighs 25.3478 g. After the excess gas escapes, the temperature is measured as

98.0oC. The flask and contents are subsequently cooled to 25oC and the vapor condenses to

a liquid. The weight of the flask and contents is found to be 25.6803 g. The flask is then

emptied, cleaned, and filled with water. When weighed on a triple-beam balance, the

difference in weight between the flask filled to the brim with water and the dry empty

flask – at 25ºC - is 128.12 g. Please determine the following:

• The empirical formula of this hydrocarbon.

• The volume that the vapor occupied in the Erlenmeyer flask (in L).

• The molecular formula of this hydrocarbon.

Solution to Example:

The ideal gas law in terms of mass (g = grams), molecular weight (MW),

temperature (T), volume (V), and pressure (P).

PV = nRT and n = moles = = .

n = = Solving for MW: MW = .

The first equation above may be easily used to find the molecular weight of the

hydrocarbon. Note that the ideal gas constant R = 0.0821 .

Also note that 760 torr = 1 atm. The kelvin (K) temperature is obtained by adding 273 to

the temperature in oC.

The first part of this problem, obtaining the empirical formula. A hydrocarbon contains only

carbon and hydrogen. Thus, the mass % of H = 100 % - 92.3 % = 7.70 % H. We assume

100 g of the hydrocarbon. Then, there are 92.3 g of C and 7.70 g of H present (relatively).

Determining moles of each element:

moles C = (92.3 g C)• = 7.69 moles C.

moles H = (7.70 g H)• = 7.62 moles H.

Divide to obtain the mole ratio:

= = = so, the Empirical Formula = CH.

As was presumed when we carried out this experiment, the vapor completely fills

the flask - once the excess escapes (also flushing out the air). The volume of water that

fills the flask equals the full volume of the flask and the volume occupied by the gas in the

ideal gas law (V).

V = volume of vapor = full volume of the flask = volume of water that fills it to the brim.

V = Vwater filling flask = .

Recall that density is temperature -dependent. So, we must use the density of water at

its temperature when it filled the flask. By the info in the problem, this temperature

is 25ºC. Referring to the table of densities (Table 1), the density of liquid water at 25ºC

is listed as: 0.997044 g/mL.

V = Vwater filling flask = = 128.50 mL = 0.12850 L

To determine the molecular formula, we need to find the molecular weight. The most useful

form of the ideal gas law - considering the given information - is:

MW = ; T = 98 + 273 = 371 K and V = 0.1285 L.

mass of condensed vapor = mass of gas in flask = 25.6803 g - 25.3478 g = 0.3325 g.

MW = = = 78.0 g/mole.

The EFW = Empirical Formula Weight for CH = 12.01 + 1.01 = 13.02 g/mole.

N = # of empirical formula units = = = 5.99 = 6.

Thus, in this case, since N = 6, the molecular formula is 6 x Empirical Formula.

Thus, the Molecular Formula = C6H6 . (The compound is benzene).

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