Find all possible structures of the ketone that reacts with hydroxylamine to form an oxime which contains \(13.9 \%\) nitrogen.

From the definition of the percentage composition in chemistry, it's known that the percent composition of an element in a compound is the mass percent of each element. Given the percentage composition of nitrogen, we can derive the empirical formula by assuming we have 100g of the compound. This means, nitrogen would weigh 13.9g. We can then determine the moles of nitrogen as \( moles = \frac{mass}{molar mass} \). The molar mass of Nitrogen (N) is \(14.01g/mol\). The mole of Nitrogen in the compound is therefore: \[ \frac{13.9g}{14.01g/mol} = 0.992mol \] Assuming it forms one mole of oxime, the stoichiometry of the reaction requires one mole of N in the oxime, thus the empirical formula of the oxime is \(R1R2C=NOH\).

Given that Nitrogen forms 1 mole and 13.9g in \(100g\) of the oxime, it's possible to calculate the molar mass of the whole oxime. Since in \(1\) mole (or \(100g\)) of the oxime there is 13.9g (or \(1\) mole) of Nitrogen, it can be stated that: \[ 1 \, mole \, of \, Nitrogen = \frac{100g \, of \, oxime}{13.9g \, of \, Nitrogen} \] That simplifies to \(1 \, mole \, of \, oxime = 7.2 \, mole \, of \, Nitrogen\). According to the above relation, the molar mass of oxime is \(7.2\) times the molar mass of Nitrogen, which means \(7.2 \times 14.01 = 100.872g/mol\).

Subtract the molar mass of Nitrogen and Hydrogen from the oxime's molar mass. The leftover mass belongs to the fragment R1R2C (=O) from the ketone. The molar weight of \(NH\) in the empirical formula is \(14.01g/mol + 1.008 g/mol = 15.018 g/mol\). This value should be subtracted from the total molar mass of the oxime: \[ 100.872 g/mol - 15.018 g/mol = 85.854 g/mol \] The molecular weight of oxygen is \(16 g/mol\) so the mass left for R1R2C is \(85.854 g/mol - 16 g/mol = 69.854 g/mol\). This value corresponds to the molar mass of the carbon atoms and the attached groups in the ketone structure.

The groups attached can be alkyl groups (R1 and R2) which either can be identical or different. The goal is to split the leftover molar mass of \(69.854 g/mol\) to possible structures (R1 and R2 groups). The smallest alkyl group (methyl) has a molar mass of \(15 g/mol\). This means, for the smallest possible ketone, \(R1\) and \(R2\) are both methyl groups. From here, the alkyl groups for R1 and R2 can be iteratively substituted with ethyl, propyl and so on, as long as the total molar mass does not exceed \(69.854 g/mol\). By continuing this substitution, the list of possible ketones that react with hydroxylamine to form an oxime which contains 13.9% nitrogen can be compiled.