Identifying an Unknown Liquid with Lewis Structures and IMFs
An advanced chemistry class is identifying an unknown colorless liquid. The teacher provides the following data: the liquid is miscible with water, has a sharp odor, boils at about 118 degrees Celsius, has a molar mass near 60 grams per mole, and its 0.10 M aqueous solution has a pH of 2.9 and conducts electricity weakly. Students propose three candidates: ethanol (CH3CH2OH), acetic acid (CH3COOH), and formaldehyde (HCHO). One student argues that the unknown must be ionic because its solution conducts electricity. Another argues that acetic acid cannot be the answer because it contains a double bond and 'double bonds make molecules nonpolar.' The teacher wants students to use Lewis structures, intermolecular forces, and acid behavior to resolve the disagreement rather than simply looking up the answer.
1. Evaluate the three candidates against the data and identify the best match. Use Lewis structures, polarity, hydrogen bonding, and acid dissociation in your reasoning.
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Acetic acid is the best match. Ethanol has a molar mass of about 46 g/mol, is neutral in water, and boils near 78 degrees Celsius, so it fails the molar mass, boiling point, and pH data. Formaldehyde has a molar mass of about 30 g/mol, is a gas at room temperature, and has no acidic O-H hydrogen, so it fails the boiling point and acidity data.
Acetic acid has a molar mass of about 60 g/mol. Its Lewis structure contains a polar C=O bond and a polar O-H bond, giving strong hydrogen bonding and a boiling point near 118 degrees Celsius. The molecule is polar overall, which explains miscibility with water. The 0.10 M solution has pH 2.9 because acetic acid is a weak acid: it partially ionizes, producing a small concentration of hydronium ions. A rough check with Ka about 1.8 x 10^-5 gives hydronium concentration near 1.3 x 10^-3 M and pH near 2.9, consistent with the data. The weak conductivity also fits partial ionization rather than complete dissociation. The teacher should require students to connect each structural feature to a measured property, not just match a name.
2. Critique the student's claim that conductivity proves the unknown is ionic. What does weak conductivity indicate for acetic acid?
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Conductivity shows that mobile charged particles are present, but it does not prove that the solute is an ionic compound. An ionic solid such as NaCl dissociates essentially completely in water, producing a high concentration of ions and strong conductivity at the same molarity. Acetic acid is a molecular compound, yet it ionizes partially:
CH3COOH + H2O <-> CH3COO- + H3O+
Only a small fraction of molecules form ions, so the solution contains few charge carriers and conducts weakly. The observation is therefore consistent with a weak electrolyte, not proof of an ionic lattice. The student should compare conductivity at equal concentration with a known strong electrolyte and a known nonelectrolyte, and should note that conductivity depends on ion concentration, ion charge, and mobility. The pH data independently support partial acid ionization. The teacher can use this to reinforce that a single property rarely identifies a substance; multiple lines of evidence are stronger.
3. The student says the C=O double bond makes the molecule nonpolar. Use the Lewis structure and molecular geometry to correct this claim.
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The claim confuses the presence of a double bond with the absence of a dipole. A C=O bond is polar because oxygen is more electronegative than carbon, so the bond has a dipole pointing toward oxygen. Molecular polarity depends on the vector sum of all bond dipoles and lone-pair contributions, not on whether a bond is single or double.
In acetic acid, the carbonyl carbon is trigonal planar with bond angles near 120 degrees. The C=O dipole, the C-O single-bond dipole, and the O-H dipole do not cancel. The methyl group is also not symmetric enough to cancel the polar carbonyl end. The result is a net molecular dipole, which is why acetic acid is polar and mixes with water. The student should draw the Lewis structure, mark electronegativity differences, and then consider geometry. A symmetrical molecule such as CO2 has polar C=O bonds but no net dipole because the bond dipoles cancel; acetic acid is not symmetrical, so the same reasoning does not apply.