Logic of transition from mass fractions to the formula
Chemistry Unified State Exam Problem 33 in the variant with element mass fractions is solved using the same logic as the problem with combustion products: the goal is to obtain the simplest formula of the organic substance and then, using the condition data, establish the molecular formula and structure.
The key technique is to equate 100% to 100 grams of the substance. If the condition states, for example, that the substance contains 11.97% nitrogen, 51.28% carbon, 27.35% oxygen, and 9.4% hydrogen, it is sufficient to write: "Let the mass of the substance be 100 g". Then the mass fractions in percent automatically turn into element masses in grams: 11.97 g of nitrogen, 51.28 g of carbon, and so on.
Algorithm for calculating the simplest formula
- Write down: "Let the mass of the substance be 100 g".
- Convert percentages into grams for each element.
- Divide the mass of each element by its molar mass to obtain the amount of substance (in moles).
- Find the smallest of the obtained amounts and divide all values by it to get the molar ratio of the atoms.
- If necessary, multiply the ratio by an integer to obtain integer subscripts.
- Write down the simplest formula.
Example: the substance contains 72% C, 21.33% O, 6.67% H. In 100 g — 72 g of carbon (72/12 = 6 mol), 21.33 g of oxygen (21.33/16 ≈ 1.33 mol), 6.67 g of hydrogen (6.67/1 = 6.67 mol). Division by 1.33 gives the ratio C : H : O ≈ 4.5 : 5 : 1, which after doubling leads to integer subscripts.
A carboxylic acid salt may contain a metal: if the condition specifies percentages of hydrogen, carbon, oxygen, and, for example, calcium, calcium is also included in the calculation of substance amounts (m/40 for Ca).
Advantage over the combustion product problem
Calculation by mass fractions is simpler than by combustion products: fewer intermediate steps and fewer places to make a mistake. In combustion products, difficulties often arise with verifying the oxygen mass and recalculating via CO₂ and H₂O. Here, element masses are obtained directly from the condition, and the risk of an arithmetic error is minimal.
Analyzing the condition: clues about the compound class
The calculation yields only the empirical formula; the structure and class of the substance are determined by analyzing the textual part of the condition. Typical clues:
- The reaction with diamminesilver(I) hydroxide (ammoniacal silver oxide solution) is a characteristic of terminal alkynes: the triple bond is strictly located at the end of the carbon chain, not in the middle.
- Dehydration of a tertiary alcohol yields an alkene, with the carbon skeleton containing at least four carbon atoms and branching at the former tertiary atom.
- Free-radical chlorination of a liquid hydrocarbon yielding a single monochloro derivative indicates a molecule with high symmetry (e.g., neopentane); liquid state implies hydrocarbons starting from pentanes.
Clues from KMnO₄ oxidation products
Oxidation with an acidified potassium permanganate solution is a powerful source of information about structure.
- Formation of a monobasic carboxylic acid and release of CO₂ is possible for: alkenes with a terminal =CH₂ or =CHR group, terminal alkynes, and benzene homologues with a side chain (e.g., ethylbenzene is oxidized to benzoic acid with the release of CO₂; the C–C bond at the aromatic ring is preserved, and the terminal carbon atom is cleaved).
- Alkadienes typically undergo either complete oxidation to CO₂ (1,3-butadiene — oxalic acid is unstable and decomposes into CO₂), or yield a dibasic acid (isoprene gives a methyl-substituted dibasic acid). Therefore, the formation of a monobasic acid is uncharacteristic of dienes.
- Formation of a dibasic acid with carboxyl groups in adjacent positions without CO₂ release is practically an unambiguous indication of o-xylene (1,2-dimethylbenzene), which oxidizes to o-phthalic acid. Preservation of all carbon atoms without CO₂ cleavage is typical of benzene homologues in which each side chain is methyl.
- Adjacent, meta-, or para- arrangement of functional groups almost always hints at a benzene ring.
Combining calculations and condition analysis
Problem 33 only seems purely mathematical. In reality, it is a symbiosis of organic chemistry and arithmetic: the calculation gives the empirical formula, while the text of the condition narrows down the range of possible structures to a specific isomer. It is recommended to work with both parts simultaneously: preliminary conclusions from the condition (class, presence of multiple bonds, aromatic ring, metal) are confirmed by numerical results, and the calculation helps choose between several structural options.
Key points
- Equating 100% to 100 g of substance is a standard technique for transitioning from mass fractions to element masses.
- The simplest formula is obtained by dividing element masses by their molar masses and normalizing to the smallest amount.
- A carboxylic acid salt may contain a metal — its mass is also included in the calculation.
- Reaction with ammoniacal silver oxide solution indicates a terminal alkyne.
- Dehydration of a tertiary alcohol yields a branched alkene with at least 4 carbon atoms.
- Formation of a monobasic acid and CO₂ upon KMnO₄ oxidation is characteristic of alkenes, terminal alkynes, and benzene homologues.
- A dibasic acid with adjacent COOH groups without CO₂ release is a sign of o-xylene.
- Calculations and condition analysis should be used together — this increases the reliability of the answer.