The essence of Problem 33 on combustion product calculations
In Problem 33 of the second part of the USE, it is required to mathematically determine the molecular formula of an unknown organic substance based on data regarding its combustion in oxygen. The condition usually provides the volume of released CO₂ and the mass of H₂O, and sometimes additionally the mass of the starting substance.
The nature of the combusted compound is determined by the composition of the combustion products:
- only CO₂ and H₂O — a hydrocarbon or an oxygen-containing compound;
- additionally N₂ is released — a nitrogen-containing compound (amine, amino acid);
- releasing HCl, HBr, HI — a halogen derivative (halogens do not yield simple substances during combustion, only hydrogen halides);
- formation of sodium or potassium carbonate — the starting substance was a salt of a carboxylic acid.
Solution algorithm
- Find the amounts of combustion products: n(CO₂), n(H₂O), and if necessary n(N₂), n(HHal).
- Proceed to the amounts of element atoms in the starting substance.
- Perform a check for the presence of oxygen using the mass of the starting substance.
- Obtain the empirical formula.
- Use information from the condition (properties, reactions) to derive the exact structural formula.
- Write the required reaction equation — this yields the third point.
Transition from products to the number of atoms
The "one box — candy inside" principle:
- n(C) = n(CO₂) — one carbon atom per CO₂ molecule;
- n(H) = 2·n(H₂O) — two hydrogen atoms per water molecule, the amount is doubled;
- n(H) += n(HHal) — when hydrogen halide is released, each mole of it adds one mole of hydrogen and one mole of halogen;
- n(N) = 2·n(N₂) — nitrogen is released as a diatomic molecule, so the number of atoms is twice as large.
The main "traps" are doubling for water and for molecular nitrogen.
Check for the presence of oxygen
Hydrocarbons and oxygen-containing compounds burn via outwardly identical schemes: yielding CO₂ and H₂O. Oxygen atoms in the products can originate both from the substance itself and from atmospheric oxygen, making it impossible to directly calculate n(O) in the initial molecule.
The mass of the starting substance is used for verification:
- the mass of carbon is calculated: m(C) = n(C) · 12;
- the mass of hydrogen is calculated: m(H) = n(H) · 1;
- if present — the mass of nitrogen, halogen;
- the remainder m(substance) − Σm(elements) is attributed to oxygen.
If the remainder is zero, it is a hydrocarbon. If the remainder is positive (e.g., 1.6 g), then n(O) = m(O)/16, and oxygen is included in the formula. This check is performed always.
Deriving the empirical formula and determining the class
The ratio of atom amounts yields the empirical formula. For example, n(C):n(H) = 1:2 corresponds to the general formula CₙH₂ₙ — these are alkenes or cycloalkanes. The ratio CₙH₂ₙ₊₂ corresponds to alkanes, and CₙH₂ₙ₋₂ to alkynes or dienes. The class of the compound is the main intermediate result of the calculation.
The specific isomer is determined using additional information from the condition: characteristic reactions, the number of halogenation products, the type of oxidation, and molecular symmetry.
Examples of using condition clues
Example 1. Hydrocarbon C₄H₈ has a cyclic structure; upon reaction with HBr, two bromo derivatives are formed. The formula CₙH₂ₙ corresponds to alkenes or cycloalkanes; the indication of a cycle points to cycloalkanes. Cyclobutane would yield one product, so methylcyclopropane fits: HBr opens the strained three-membered ring, hydrogen goes to the more hydrogenated carbon according to Markovnikov's rule, and bromine goes to the less hydrogenated one, yielding two isomeric bromobutanes.
Example 2. Alkane C₅H₁₂ upon chlorination yields only one monochloro derivative. This is possible only with complete molecular symmetry — all hydrogen atoms are equivalent. This condition is met by 2,2-dimethylpropane (neopentane): radial symmetry makes all 12 hydrogen atoms identical, so substitution at any of them leads to a single product.
Example 3. Hydrocarbon C₅H₁₀ upon oxidation with acidified KMnO₄ yields a carboxylic acid and a ketone. Cycloalkanes C₅–C₆ are resistant to oxidation, meaning it is an alkene. The formation of a ketone requires a tertiary carbon atom at the double bond, and the formation of a carboxylic acid requires at least two carbon atoms on one side of the double bond. The conditions are met by 2-methylbut-2-ene: upon oxidation, it yields acetic acid and acetone. Hydrobromination according to Markovnikov's rule gives 2-bromo-2-methylbutane.
Key points
- Problem 33 is solved by combining calculations based on combustion products with the analysis of clues from the condition.
- The number of carbon atoms equals n(CO₂), hydrogen equals 2·n(H₂O), and nitrogen equals 2·n(N₂).
- Halogens are released during combustion solely in the form of hydrogen halides HHal.
- Checking for oxygen via the mass of the starting substance is mandatory; otherwise, it is impossible to distinguish a hydrocarbon from an oxygen-containing compound.
- The empirical formula determines the compound class according to the general formula of the homologous series.
- A single monochlorination product of an alkane indicates complete symmetry of the carbon skeleton.
- Oxidation of an alkene to an acid and a ketone indicates a tertiary carbon atom at the double bond.
- The third point for the problem is awarded for a correctly written reaction equation involving the derived substance.