Combined techniques

Matt’s exam tip

Focus on your working when analysing data for structure questions. Show clearly what each piece of data tells you before drawing the final structure. In a six-mark question, the structure itself may only be worth one mark – you can still earn five marks just by interpreting the data well.

Technique Summary Table

Elemental analysis

Gives empirical formula (which can be used along with M_r to determine molecular formula).

Example Find the empirical formula

For the compound with a composition by mass of C 52.2%, H 13.0% and O 34.8%.

Ex

Worked Example

Determine the empirical formula step by step:

1. Convert % to masses (assume 100 g): C = 52.2 g, H = 13.0 g, O = 34.8 g
2. Convert to moles: C = 52.2 ÷ 12.0 ≈ 4.35, H = 13.0 ÷ 1.0 ≈ 13.0, O = 34.8 ÷ 16.0 ≈ 2.18
3. Divide by smallest value (2.18): C = 2, H = 6, O = 1
4. Empirical formula = C₂H₆O

Mass spectrometry

Determines molecular mass and possible fragments.

Example Distinguishing isomers of C₄H₁₀

Hydrocarbons A and B both have molecular formula C₄H₁₀ (same molecular ion peak at 58), however they have different fragment patterns in their spectra, showing different structures.

Fragment peaks at 15 and 43 show a CH₃ fragment and C₃H₇ fragment. However, no fragment at 29 means no C₂H₅ group. This means the likely possible structure is CH₃CH(CH₃)CH₃.

The extra peak at m/z 29 for Hydrocarbon B means it has a C₂H₅ group in its structure (as well as a CH₃ and C₃H₇ group). This would indicate CH₃CH₂CH₂CH₃ as its structure.

IR spectroscopy

Shows presence of functional groups.

Key absorbances you should instantly be able to recognise!

• C=O: ~1700 cm⁻¹
• O–H (broad): ~3200–3600 cm⁻¹

Example Ethanoic acid IR spectrum

Ethanoic acid (CH₃COOH) has two absorbances in its IR spectra that help identify it. One for the O–H bond (2500 to 3000) and one for the C=O bond (1680 to 1750).

NMR spectroscopy

Provides detailed information about C or H environments in a molecule.

Example Ethanol ¹H NMR

Ethanol (CH₃CH₂OH) in ¹H NMR:

• 3 peaks in the H NMR shows 3 unique H environments.
• CH₃ group shows a triplet (next to CH₂).
• CH₂ group shows a quartet (next to CH₃).
• OH appears as a singlet (no splitting).

Summary

• Elemental analysis determines empirical formula.
• Mass spectrometry gives molecular mass and fragments for structural clues.
• IR identifies key functional groups from absorption frequencies.
• NMR provides detailed hydrogen or carbon environments for structure determination.
• Combining all methods allows accurate deduction of molecular structure.