How Chemists Identify an Unknown Compound

A bottle turns up in a storeroom with its label washed off. A new product comes out of a reaction flask. A white powder is seized at a border crossing. In each case the question is the same: what is this? Chemists answer it with a logical sequence of tests, each one narrowing the possibilities, until only one structure fits all the evidence.

This guide walks through that sequence the way a working analyst would approach it, then applies it to a real-style example from start to finish.

Step 0: Safety first

Treat every unknown as hazardous until you know otherwise. Wear eye protection and gloves, work in a fume cupboard, and never smell or taste a sample. Use the smallest amount that will do the job: modern instruments need milligrams, not grams. See lab safety rules and risk assessment in chemistry.

Step 1: Observe and record

Simple observations cost nothing and often cut the list of candidates dramatically.

• State and appearance: solid, liquid or gas? Crystalline or amorphous? Colour?
• Colour clues: most organic compounds and most salts of s-block metals are colourless or white. A coloured compound often contains a transition metal ion (blue for many copper(II) salts, green for nickel(II), pink for cobalt(II) in water) or an extended conjugated organic system. See transition metals.
• Solubility: does it dissolve in water? In dilute acid? In an organic solvent such as hexane? Ionic solids tend to dissolve in water; nonpolar organic compounds prefer nonpolar solvents. See polar vs nonpolar molecules.
• Behaviour on heating: a small sample heated gently may melt, decompose, give off water (a hydrate), or burn with a sooty flame (typical of aromatic compounds).

Step 2: Is it pure?

Identifying a mixture as if it were one substance leads to nonsense. Before going further, check purity.

• Melting point: a pure crystalline solid melts sharply over 1–2 °C. Impurities lower the melting point and broaden the range.
• Boiling point: a pure liquid boils at a steady temperature.
• Chromatography: a pure substance gives a single spot in thin-layer chromatography in several different solvents, or a single peak in gas chromatography or HPLC.

If the sample is a mixture, separate it first (recrystallisation, distillation or column chromatography) and identify each component.

Step 3: Inorganic or organic?

This decides which tests come next.

Clue	Suggests inorganic salt	Suggests organic compound
Melting point	usually high (hundreds of °C)	usually below 300 °C
Solubility	often dissolves in water	often dissolves in organic solvents
Heating in air	may decompose but doesn’t char	chars or burns, often with smoke
Flame test	may give a characteristic colour	usually no metal colour

For an ionic compound, the job is to identify the cation and the anion with qualitative tests. That route is covered in identifying ions: a qualitative analysis flowchart. The rest of this guide follows the organic route, which relies more on instruments.

Step 4: Which elements are present, and in what ratio?

Combustion analysis burns a weighed sample in excess oxygen. Carbon ends up as CO₂ and hydrogen as H₂O, which are trapped and weighed. Nitrogen, sulfur and halogens are measured by related methods, and oxygen is usually found by difference.

The mass percentages give the empirical formula, the simplest whole-number ratio of atoms. See percent composition and empirical vs molecular formula.

Step 5: What is the molar mass?

The molecular ion peak (M⁺) in a mass spectrum gives the relative molecular mass. Divide it by the empirical formula mass to get the multiplier, and hence the molecular formula.

The mass spectrum holds other clues too:

• An M+2 peak about one-third the height of M⁺ signals one chlorine atom (³⁵Cl : ³⁷Cl is roughly 3 : 1).
• An M+2 peak about equal to M⁺ signals one bromine atom (⁷⁹Br : ⁸¹Br is roughly 1 : 1).
• Fragment peaks show pieces that break off: a peak at 15 below M⁺ suggests loss of CH₃; a peak at m/z 43 is common for CH₃CO⁺ or C₃H₇⁺.

Step 6: How many rings and double bonds?

From the molecular formula, calculate the degree of unsaturation (also called double bond equivalents):

DBE = (2C + 2 + N − H − X) ÷ 2

where C, N, H and X are the numbers of carbon, nitrogen, hydrogen and halogen atoms (oxygen and sulfur are ignored).

Each ring or double bond counts as 1; a triple bond counts as 2. A DBE of 4 or more with at least six carbons strongly suggests a benzene ring (one ring plus three C=C).

Step 7: Which functional groups?

Infrared spectroscopy identifies bonds from their absorption frequencies:

Absorption (cm⁻¹)	Bond	Likely group
3200–3550, broad	O–H	alcohol
2500–3300, very broad	O–H	carboxylic acid
3300–3500, sharper	N–H	amine or amide
1680–1750, strong	C=O	aldehyde, ketone, acid, ester, amide
about 2200–2260	C≡N	nitrile
1000–1300	C–O	alcohol, ester, ether

Simple chemical tests can confirm the IR evidence cheaply: bromine water decolourising for C=C, 2,4-dinitrophenylhydrazine giving an orange precipitate for C=O, Tollens’ reagent giving a silver mirror for aldehydes, and sodium carbonate fizzing for carboxylic acids.

Step 8: How are the atoms connected?

NMR spectroscopy is the most powerful structure tool. ¹³C NMR shows how many different carbon environments there are. ¹H NMR shows:

• the number of hydrogen environments (number of signals)
• the ratio of hydrogens in each (integration)
• the type of environment (chemical shift)
• the number of neighbouring hydrogens (splitting, by the n + 1 rule)

At this point you can propose a structure.

Step 9: Check the proposal against everything

A proposed structure must explain every piece of data: the formula, the DBE, every IR band, every NMR signal and its splitting, and the main fragments. If one observation doesn’t fit, the structure is wrong or the sample isn’t pure. Finally, if a reference sample is available, compare melting points (a mixed melting point with no depression confirms identity) or overlay the spectra.

Worked example

An unknown colourless liquid boils at 77 °C and gives a single peak by GC. Combustion analysis gives C 54.5%, H 9.1%, and the rest oxygen.

Empirical formula. O = 100 − 54.5 − 9.1 = 36.4%.

• C: 54.5 ÷ 12.01 = 4.54
• H: 9.1 ÷ 1.008 = 9.03
• O: 36.4 ÷ 16.00 = 2.28

Dividing by 2.28 gives C 2 : H 4 : O 1, so the empirical formula is C₂H₄O (mass 44).

Molecular formula. The mass spectrum shows M⁺ at m/z 88. 88 ÷ 44 = 2, so the molecular formula is C₄H₈O₂.

DBE = (2 × 4 + 2 − 8) ÷ 2 = 1: one C=O or C=C, or one ring.

IR: strong absorption at 1740 cm⁻¹ (C=O) and at about 1240 cm⁻¹ (C–O), with no broad O–H band. That rules out a carboxylic acid and points to an ester.

Mass spectrum fragments: a strong peak at m/z 43 (CH₃CO⁺).

¹H NMR:

δ (ppm)	Integration	Splitting
4.1	2H	quartet
2.0	3H	singlet
1.3	3H	triplet

• The quartet (2H) and triplet (3H) together are the classic pattern of an ethyl group, CH₃CH₂–. The CH₂ at 4.1 ppm is shifted strongly downfield, so it’s bonded to oxygen: –O–CH₂CH₃.
• The 3H singlet at 2.0 ppm has no neighbouring hydrogens and sits next to a C=O: CH₃–C(=O)–.

Proposal: CH₃COOCH₂CH₃, ethyl ethanoate (ethyl acetate).

Check: C₄H₈O₂ ✓; one C=O ✓; ester IR bands ✓; CH₃CO⁺ fragment at 43 ✓; three NMR signals in a 2 : 3 : 3 ratio with correct splitting ✓; the boiling point of ethyl ethanoate is 77 °C ✓. Every piece of evidence fits.

Notice that the isomer methyl propanoate, CH₃CH₂COOCH₃, has the same formula and similar IR, but its NMR differs: its OCH₃ is a singlet near 3.7 ppm and its CH₂ quartet sits near 2.3 ppm, not 4.1. That’s why NMR is usually the deciding step.

Common pitfalls

• Skipping the purity check. Two overlapping sets of peaks will produce an impossible structure.
• Trusting one technique alone. IR tells you groups, not connectivity; a mass alone fits many formulas.
• Ignoring missing peaks. The absence of an O–H band was as important in the example as the presence of C=O.
• Forgetting isomers. Always ask what other structures share the formula, and which data would tell them apart.

Key takeaways

• Start with safety, observation and a purity check before any identification.
• Combustion analysis gives the empirical formula; the molecular ion in mass spectrometry gives the molecular formula.
• The degree of unsaturation counts rings and multiple bonds.
• IR identifies functional groups; NMR shows how the atoms are connected.
• A structure is only accepted when it explains every piece of evidence, including what is absent. For an overview of the instruments, see the spectroscopy overview.