Experimental reasoning and analytical data
Analytical Chemistry and Laboratory Techniques
This independent foundation chapter develops identification tests, instrumental analysis, separation methods, accurate titration technique and the treatment of measurement results - the practical skills that decide marks in the written and assessed practical components.
Qualitative tests
- Flame tests: Li crimson, Na yellow, K lilac, Ca brick-red, Cu blue-green. Clean the nichrome wire between samples.
- Adding NaOH then excess NH3: Cu2+ gives a blue precipitate soluble in excess NH3; Fe2+ green; Fe3+ brown; Al3+ and Pb2+ white, soluble in excess NaOH.
- Halides: acidify with dilute HNO3, then add AgNO3. AgCl is white and soluble in dilute NH3, AgBr cream and sparingly soluble, AgI yellow and insoluble.
- Carbonates fizz with acid and turn limewater milky; sulfates give a white precipitate with BaCl2 after acidification, which excludes carbonate and sulfite interference.
Instrumental methods
- Infrared: very broad carboxylic-acid O–H often around 2500–3300 cm−1; alcohol O–H commonly occupies a different higher region, roughly 3200–3600 cm−1 depending on hydrogen bonding; C=O about 1680–1750 cm−1, C–O about 1000–1300 cm−1.
- Mass spectrometry: the molecular ion peak gives Mr; fragmentation patterns support structure.
- NMR: number of environments, integration ratios and splitting pattern.
- Chromatography: Rf = distance moved by spot ÷ distance moved by solvent front; retention-time comparison with suitable standards under controlled conditions can support identification in GC/HPLC; different compounds can coelute.
Quantitative technique
- Read the appropriate meniscus using the actual instrument graduations and specified uncertainty; 0.05 cm3 is a common school-burette convention, not a universal resolution; use the average of concordant titres and discard the rough one.
Worked example
25.0 cm3 of hydrochloric acid requires 23.40 cm3 of 0.1000 M NaOH. n(NaOH) = 0.1000 × 0.02340 = 2.340 × 10−3 mol. The equation is 1:1, so n(HCl) = 2.340 × 10−3 mol and c = 2.340 × 10−3/0.0250 = 0.0936 mol dm−3.
Traps that cost marks
- Failing to acidify before the silver nitrate test, so carbonate gives a false precipitate.
- Quoting flame colours without the correct metal, or contaminating the wire between tests.
- Including the rough titre in the average.
- Quoting Rf greater than 1 or measuring the solvent front incorrectly.
- Reporting more significant figures than the burette reading supports.