Experiment No. 5: Determination of Partial Molar VolumesJanrick Nichol O. See, Camille V. TrinidadBiochem 156.1 (TEJ): Sir Jayson Vedad

I. AbstractIn this experiment, the partial molar volumes of two substances, namely, ethanol and water, were

determined at 25 ºC by varying the amount of substances added to another substance which volume was kept constant. The number of moles of the added substance was plotted against the total volume of the solution and the slope of the graph was obtained through the use of linear regression. The slope of the graph obtained, numerically amounts to the partial molar volume so it was found that the partial molar volume of ethanol is 55.637 mL/mol while the partial molar volume of water is 15.743 mL/mol.

II. Keywordspartial molar volume, ethanol, water, binary solution, thermodynamic properties

III. IntroductionThe properties of solutions vary from the

pure forms of their components in different degrees because of multiple factors. The quantities measured for solutions with variations from their pure forms are called partial molar quantities. Partial molar quantities are defined as the partial derivatives of the thermodynamic properties of a solution with respect to the amount of components. An example of a partial molar quantity is the partial molar volume. The partial molar volume of a component in a solution is the effective volume contribution of that component in the solution (Barrow, 1981).

Some solutions may have properties that are similar to the pure form of each component, meaning to say that each molecule in the solution has approximately the same environmental characteristics as that of when each is present in the environment of a pure liquid. Such is the case for ideal solutions. An example of an ideal solution is the binary solution of toluene and benzene. Because toluene differs from benzene by only a methyl substituent, the volume of their solution is approximately the sum of the volumes of the individual components. That is, the partial molar volume has a negligible deviation from the actual volume of the component. This will be different for two components with different sizes of molecules, or for solvent-solute interactions wherein the solute is strongly-ionizable.

Figure 1: Two figures showing the effect of mixing a component with small molecules and another component with large

molecules. Mixing causes the small molecules to fit between the spaces of larger ones.

Phenomena observed because of partial molar volume include the difference in the volume

change of a solution to the amount of component placed in the solution. Partial molar volumes of saltwater are calculated in the study of oceanography and aquatic environmental science. The concept of partial molar volume is also applied in the field of Biochemistry for the careful calculation of the molecular weight of proteins and nucleic acids using ultracentrifugation.

IV. ExperimentalA. Determination of the Partial Molar Volume of Ethanol

Four 50 mL volumetric flasks were cleaned and prepared. 30.0 mL of distilled water from a buret was added to each flask. Then, ethanol of varying volumes (18.0, 20.0, 22.0, 24.0 mL) was added to each of the volumetric flasks. The volume of the resulting solution was determined by marking the upper meniscus of the solution and refilling the volumetric flask with distilled water of noted volumes until it reached the mark. The moles of ethanol added in each trial were computed and plotted against the volume of the solution to get the partial molar volume of ethanol in water.

Table 1. The volumes of water and ethanol placed in each flask in determining the partial molar volume of ethanol

B. Determination of the Partial Molar Volume of Water

The same procedure was employed in the determination of the partial molar volume of ethanol water. Instead of filling each flask with 30.0 mL of water, 30.0 mL of ethanol was used and the volume of added water was varied. The volume of the resulting solution was determined by marking the upper meniscus of the solution and refilling the volumetric flask with distilled water of noted volumes until it reached the mark. The partial molar volume of water in ethanol was computed by graphing the moles of water added against the total volume of solution in each trial.

Chem 156.1 (Determination of Partial Molar Volumes) Page 1

Vol. of Water 30 mL 30 mL 30 mL 30 mLVol. of

Ethanol 18 mL 20 mL 22 mL 24 mL

Table 2. The volumes of water and ethanol placed in each flask in determining the partial molar volume of water

V. ResultsA. Determination of Partial Molar Volume of Ethanol in Water

Table 3. The data collected from varying the volume of ethanol in 30.0 mL of water

Flask No. 1 2 3 4

VolH20 (mL) 30.0 30.0 30.0 30.0

VolC2H5OH

(mL)18.0 20.0 22.0 24.0

MassC2H5OH

(g)14.38 16.00 17.60 19.20

MolesC2H5OH 0.313 0.347 0.382 0.417

Volsolution

(mL)47.4 49.0 51.2 53.1

Computation:ρC2H5OH, 25@ C = 0.79991 g/mLMWC2H5OH = (12.0107 X 2) + (1.0079 X 5) + (15.9994) + 1.0079MWC2H5OH = 46.0682 g/mol

MassC2H5OH = VolC2H5OH x ρC2H5OH

Mass in Trial1 = 18 mL x ρC2H5OH

= 14.38 gMass in Trial2 = 20 mL x ρC2H5OH

= 16.00 gMass in Trial3 = 22 mL x ρC2H5OH

= 17.60 gMass in Trial4 = 24 mL x ρC2H5OH

= 19.20 g

MolC2H5OH = VolC2H5OH x ρC2H5OH / MWC2H5OH

Mol in Trial1 = 18 mL x ρC2H5OH / MWC2H5OH

= 0.313 molesMol in Trial2 = 20 mL x ρC2H5OH / MWC2H5OH

= 0.347 molesMol in Trial3 = 22 mL x ρC2H5OH / MWC2H5OH

= 0.382 molesMol in Trial4 = 24 mL x ρC2H5OH / MWC2H5OH

= 0.417 moles

0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.4444

46

48

50

52

54

f(x) = 55.6367562180795 x + 29.8814931694555R² = 0.997293507111764

Partial Molar Volume of Ethanol

Moles of Ethanol

Vol

ume

of S

oluti

on (m

L)

Figure 2. Graph of moles of ethanol of Flask 1 to 4 against the total volume of the solution.

B. Determination of Partial Molar Volume of Water in Ethanol

Table 4. The data collected from varying the volume of water in 30.0mL of ethanol

Flask No. 1 2 3 4VolH20

(mL)18.0 20.0 22.0 24.0

VolC2H5OH

(mL)30.0 30.0 30.0 30.0

Mass H20

(g)17.95 19.94 21.93 23.93

Moles H20 0.996 1.11 1.22 1.33

Volsolution

(mL)47.5 49.4 51.3 52.7

Computations:ρH2O,25@ C (in g/cc) = 997.045 kg/m3

( 1000g1kg )( 1m100cm )3

ρH2O = 0.997045 g/mLMWwater = (1.0079 x 2) + 15.9994

= 18.0152 g/mol

MassH2O= VolH2O x ρH2O

Mass in Flask 1 = 18 mL x ρH2O

= 17.95 gMass in Flask 2 = 20 mL x ρH2O

= 19.94 gMass in Flask 3 = 22 mL x ρH2O

= 21.93 gMass in Flask 4 = 24 mL x ρH2O

= 23.93 g

MolH2O = VolH2O x ρH2O / MWH2O

Mol i Flask 1 = 18 mL x ρH2O / MWH2O

= 0.996 molesMol in Flask 2 = 20 mL x ρH2O / MWH2O

= 1.11 molesMol in Flask 3 = 22 mL x ρH2O / MWH2O

= 1.22 molesMol in Flask 4 = 24 mL x ρH2O / MWH2O

= 1.33 moles

Chem 156.1 (Determination of Partial Molar Volumes) Page 2

Vol. of Water 18mL 20mL 22mL 24mLVol. of

Ethanol 30mL 30mL 30mL 30mL

0.9 1 1.1 1.2 1.3 1.444

46

48

50

52

54

f(x) = 15.7426575236124 x + 31.9005466425152R² = 0.995866959121643

Partial Molar Volume of Water

Moles of Water

Vol

ume

of S

oluti

on (m

L)

Figure 3. Graph of moles of water of Flask 1 to 4 against the total volume of the solution.

VI. DiscussionPartial molar quantities describe intensive

quantities that reveal how the properties of a solution change with concentration. A good example of a partial molar quantity is the partial molar Gibbs free energy. The partial molar Gibbs free energy is a quantity that results from dividing the change in the free energy of a system to the number of moles of a component of the system. This quantity simply points to the incremental change in Gibbs free energy of a system after the addition of a mole of the component. It also provides a measure of the “driving force” of chemical processes. Another name for the partial molar Gibbs free energy is the chemical potential (µ). The mathematical formula for the chemical potential µ is given by:

µ1 = (∂G∂n1)T, P, n’

Where:µ1 = chemical potential of component one of solution.G = Gibbs free energy of the systemn1 = number of moles of component oneT = TemperatureP = Pressuren’ = amount of all other substances present

Partial molar volume is another partial molar quantity. It describes the change in the total volume of the solution per change in the number of moles of component in a solution. Components that make up a solution are not always additive in terms of the volume that they occupy in the solution. If the solution is binary (has two components) and both the components are very similar in their chemical structure, e.g. a toluene-benzene solution, then the volume occupied by the two components at a constant temperature and pressure is approximately additive. The molecules of toluene simply resemble that of benzene and hence, the environment of a molecule of toluene in the solution is the same as when it is present in an

environment of its pure form. Note that for binary solutions, the sum of the mole fractions of component A and component B of the solution is equal to unity.

XA + XB = 1.0

When a binary solution is composed of two structurally-different molecules, e.g. water and ethanol, the approximation that the components are additive becomes invalid. The explanation for this lies in the fact that the volume occupied by water molecules depend on the identity of the molecules that are surrounding them. Ethanol-water mixtures contain molecules that have different sizes, and they interact by hydrogen bonds. When a large amount of absolute ethanol is mixed with a mole of water, a large amount of ethanol will also surround each water molecule. The same happens if one mole of ethanol is added to a large amount of water. The attraction of ethanol and water molecules causes molecular packing, resulting to a lower volume of solution than what can be deduced additively. Hence, the partial molar volume is not necessarily equal to one mole of the volume of the pure component. The mathematical formula for partial molar volume is given by:

V1 = (∂V∂n1)P, T, n’

Where:V1 = partial molar volume of component oneV = total volume of solution n1 = number of moles of component oneT = Temperature P = Pressuren’ = amount of all other substances present

Remembering that the change in a given variable a divided by the change in another given variable b is equal to the slope of the plot of both variables, we can say that the partial molar volume of component one is equal to the slope of the plot of the total volume of the solution as the number of moles of component one is being changed (pressure, temperature and amount of all other substances present made constant).

According to Atkins (2006), a significant use of partial molar volume is in computing for the total volume of a solution. Given a solution composed by two components A and B, when the composition of a solution is changed by the addition of dnA of A and dnB of B, the change in the total volume of the solution is:

dV =( ∂V∂nA )P,T,nB ∙ (dnA) + (

∂V∂nB )P,T,nA ∙ (dnB)

= VA dnA + VB dnB

Chem 156.1 (Determination of Partial Molar Volumes) Page 3

If we integrate this equation, we can get the total volume of the solution.

V=∫0

nA

EQ¿(V ,¯ )(A ) d n(A )+∫0

nB

EQ ¿(V ,¯ )(B ) d n(B )

= VA∫0

nA

d n(A ) + VB∫0

nB

dn(B )

V = VA nA + VB nB

The values of partial molar volume quantities can be negative. This is because partial molar volume is not extensive. A negative partial molar volume in a solution corresponds to a decrease in the total volume of the solution. This happens to conditions such as when MgSO4 is mixed with water. When 1 mol of MgSO4 is added to a large volume of water, the solution contracts as a result. This is because the salt breaks the open structure of water as its ions become hydrated.

Figure 4: A graph showing the change in the total volume of a solution corresponding to the change in the amount of

component nA present in solution. Tangent line I shows a positive slope (positive partial molar volume), and tangent line II

shows a negative slope (negative partial molar volume).4

There are some factors that affect partial molar volume. A change in temperature and pressure results in a change in the density of liquids, which ultimately affects the total volume of the solution. Partial molar volume changes as a consequence of altering these state functions. The components of the solution also affect partial molar volume. Molecular and ionic interactions are dependent on the amount and nature of components present in the environment. Packing, repulsion, hydrogen bonding, ionization, and other forms of molecular interaction can determine how much deviation the partial molar volume will have as compared to its actual molar volume.

Figure 5: The partial molar volumes of water and ethanol that corresponds to different mole fractions of ethanol at 250C. This graph shows how partial molar volumes vary with a change in

the amount of components making up the solution.3

Analysis of Experimental ResultsPartial Molar Volume of 95% Ethanol at 25 0 C

Utilizing the Linear Regression function, the calculated slope was determined to be equal to 55.63 mL. The y-intercept of the graph is equal to 29.88 mL. The slope derived from Linear Regression is equal to the calculated value of the partial molar volume of 95% ethanol at 250C. For analysis, the experimenters have determined the linear equation for this condition. The linear equation for the experimental total volume of the solution is:

Vtotal = (∂Vtotal∂n(ethanol )

)P, T, n sub(water) ∙ nethanol + (Vwater)

y = m x + b

From this equation, we can say that if ethanol were not added to the solution, the total volume would just be equal to the volume of water, or the y-intercept. The actual volume of water is 30.0 mL. However, the calculated y-intercept is 29.88 mL. We can calculate the percent error of the experiment using their difference.

Percent Error=( theoretical volume−calculated volume

theoretical volume ) x

100%

= (30.0−29.8830.0

) x 100%

=0.4% error

Chem 156.1 (Determination of Partial Molar Volumes) Page 4

Partial Molar Volume of Water at 25 0 C Utilizing the Linear Regression function,

the calculated slope was determined to be equal to 15.74 mL. The y-intercept of the graph is equal to 31.90 mL. The slope derived from Linear Regression is equal to the calculated value of the partial molar volume of water at 250C. For analysis, the experimenters have again determined the linear equation for this condition. The linear equation for the experimental total volume of the solution is:

Vtotal = (∂Vtotal∂n(water )

)P, T, n sub(ethanol) ∙ nwater + (Vethanol)

y = m x + b

From this equation, we can say that if water were not added to the solution, the total volume would just be equal to the volume of ethanol, or the y-intercept. The actual volume of ethanol is 30.0 mL. However, the calculated y-intercept is 31.90 mL. We can calculate the percent error of the experiment using their difference.

Percent Error=( theoretical volume−calculated volume

theoretical volume ) x

100%

= (30.0– 31.9030.0

) x 100%

= -6.33% error

The percent error computed for the determination of the partial molar volume of both water and 95% ethanol at 250C are due to several factors. Ethanol is a volatile organic liquid, with a boiling point of about 780C. The vaporization of ethanol in solution may have decreased the number of moles of ethanol present in solution, resulting to the observed difference. But still, the error is very low.

VII. Conclusion and RecommendationsThe total volume of a solution cannot

always be approximated as the addition of the individual volumes of its components. Other factors must be considered, such as temperature and molecular interactions.

Determination of the partial molar volumes of the components in a binary solution is very feasible and has a low percent error upon correct experimentation and proper laboratory techniques.

In order to obtain better results in determining the partial molar volume of ethanol in water and vice versa, it is recommended that the experiment should be conducted in relatively lower

temperatures to decrease the vaporization if ethanol. Temperature must also be held constant. Variation of temperature may lead to erratic results.

Creating and reading marks on the volumetric flask is subject to the thickness of the marker and to human judgment. Experimenters suggest using markers that create finer marks on the glassware.

The experimenters recommend conducting the experiment using other types of binary solutions like aromatic hydrocarbon components (example: benzene and toluene) and water-ionic compound tandems (example: sodium chloride and water) so as to evaluate the variation of partial molar volumes for different systems. Also, the flammability of ethanol is a threat to the safety of the experiment; hence, the need to be careful must be emphasized to future experimenters.

VIII. References

[1] Barrow, G.M. (1981). Physical Chemistry for the Life Sciences (2nd Ed.). New York, NY: McGraw-Hill, Inc.

[2] Colby College. Partial Molal Volume. Retrieved August 15, 2011, from http://www.colby.edu/chemistry/PChem/lab/PartMolalV.pdf

[3] de Paula, J. & Atkins, P. (2006). Physical Chemistry (8th Ed.). USA: W.H. Freeman and Company.

[4] EveryScience.com. Partial Molar Volume. Retrieved: August 15, 2011, from http://www.everyscience.com/Chemistry/Physical/Mixtures/a.1265.php.

[5] Lesk, A. M. (1982). Introduction to Physical Chemistry. Englewood Cliffs, NJ: Prentice-Hall, Inc.

[6] Salzman, W.R. (n.d.). Mixtures; Partial Molar Quantities; Ideal Solutions. Retrieved August 15, 2011, from http://www.chem.arizona.edu/~salzmanr/480a/480ants/mixpmqis/mixpmqis.html.

[7] Woodbury, G. (1997). Physical Chemistry. Pacific Grove, CA: Brooks/Cole Publishing Company.

I hereby certify that I have given substantial contribution to this report.

Chem 156.1 (Determination of Partial Molar Volumes) Page 5

Janrick Nichol O. See

Camille V. Trinidad

Chem 156.1 (Determination of Partial Molar Volumes) Page 6