Question

PART IV. MISCIBILITY OF TWO LIQUIDS

1. For the first two liquids in Table IV-B (the solvents in the miscibility tests), use the molecular structures provided in the Introduction to identify each liquid as polar or nonpolar. Put an X in the appropriate column.

2. For the remaining liquids in Table IV-B, determine if they are primarily polar, primarily nonpolar, or both polar and polar based on your observations in Table III-A and based on the polarity of the solvent. Remember that “like dissolves like,” therefore:

• If the liquid is miscible only in the polar solvent, it is primarily polar.

• If the liquid is miscible only in the nonpolar solvent, it is primarily nonpolar.

• If the liquid is miscible in both solvents, it is both polar and nonpolar.

3. Filling out Table IV-C is a two-step process:

a. Based on the molecular structures in the Introduction, list ALL interparticle forces present for each liquid. Remember that since hydrogen bonds are dipole-dipole forces, do not list both.

b. Based on your answers in Table IV-B, indicate the MOST IMPORTANT interparticle force(s) for each liquid. For example, if it is primarily nonpolar, London dispersion forces are most important.
STRUCTURAL FORMULAS FOR MOLECULAR COMPOUNDS water H-O-H hexane CH3 -CH2-CH2-CH2-CH2-CH; 0 O acetone 3-pentanone CH3 -C-CH CH3-CH-C-CH2-CH ethylene glycol CHCH2 | 0-H O-H methanol CH3 -O-H 2-pentanol CH3-CH-CH2-CH2-CH; O-H 1 -CH glycerin CH, -CH- 11 O-H O-H 1 O-H CH-CH2-CH2-CH2-CH2-CH2-CH2-CH; 1-octanol O-H CH2-0-C-CH2-CH2-CH-CH2-CH2-CH-CH2-CH=CH-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CHz ÇH.-o-é-CH-C: HocH-CH-C olive oil* CH-0-C-CH-CH2-CH -CH -CH -CH -CH-CH=CH-CH -CH -CH-CH-CH-CH -CH-CH, 0 CHM-0-C-CH2-CH2-CH-CH2-CH2-CH2-CH-CH=CH-CH2-CH,-CH2-CH-CH-CH2-CH2-CH, paraffin wax** CH3-CH2-CH-CH -CH-CH2-CH-CH2-CH-CH-CH2-CH2-CH-CH2-CH-CH2-CH-CH2-CH-CH; * Olive oil is a mixture of many compounds, but the primary component is a triglyceride containing oleic acid, which is shown here. ** Paraffin wax is a mixture of many compounds, similar to the one shown here, with between 20 and 40 carbon atoms in a chain.

Table IV-B. Classification of Liquids

Liquid	Polar?	Nonpolar?
water
hexane
methanol
1-octanol
acetone
3-pentanone
2-pentanol
ethylene glycol

Table IV-C. Interparticle Forces for Liquids

Liquid	LDF Present?	DDF Present?	HB Present?	Most Important Force(s)
water
hexane
methanol
1-octanol
acetone
3-pentanone
2-pentanol
ethylene glycol

Key:           LDF = London dispersion force

                  DDF = dipole-dipole force

                  HB = hydrogen bonding

Additional Notes: • The structures in this table are “condensed” structural formulas, which means the hydrogen atoms attached to each carbon are grouped with the carbon and nonbonding electrons are not shown. • For example, the full structure of ethylene glycol is shown to the right. • In addition, correct bond angles are not shown

Answer 1

# Table IV-B

Water $\rightarrow$ Polar

Hexane $\rightarrow$ Non Polar

Methanol $\rightarrow$ Polar

1- Octanol $\rightarrow$ Non Polar

Acetone $\rightarrow$ Polar

3- Pentanone $\rightarrow$ Non Polar

2-Pentanol $\rightarrow$ Polar

Ethylene glycol $\rightarrow$ Non Polar

# Table IV-C

Water $\rightarrow$ HB

Hexane $\rightarrow$ LDF

Methanol $\rightarrow$ HB

1- Octanol $\rightarrow$ LDF

Acetone $\rightarrow$ HB and DDF (DDF is most important force)

3- Pentanone $\rightarrow$ LDF

2-Pentanol $\rightarrow$ DDF and HB (HB is most important force)

Ethylene glycol $\rightarrow$ HB