Question

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

Short answer

The possible structures of the ketones that react with hydroxylamine to form an oxime with \(13.9 \%\) nitrogen are determined by iteratively substituting the alkyl groups (R1 and R2) in the ketone's empirical formula. By calculating the molar mass of the oxime (\(100.872 g/mol\)) and determining the leftover mass for R1R2C fragment (\(69.854 g/mol\)), possible structures can be found as long as the total molar mass does not exceed \(69.854 g/mol\). The possible ketones will have varying R1 and R2 groups (either identical or different alkyl groups) satisfying the given conditions.

Step-by-step solution

Derive the empirical formula of the oxime

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 \(R1R2C=NOH\).

Determine molar mass of the oxime

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 \(7.2 \times 14.01 = 100.872g/mol\).

Determine the empirical formula of the ketone

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.

Determine the possible structures of the ketone

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.

Key concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms of each element present in a compound. To find it, we often start by determining the percent composition of each element. Here, we're given that the oxime contains 13.9% nitrogen. To simplify our calculations, we assume we have a 100g sample of the oxime. This means 13.9g of the sample would be nitrogen.

Next, we convert the mass of nitrogen to moles by using nitrogen's molar mass, which is 14.01g/mol. By dividing 13.9g by 14.01g/mol, we find there are approximately 0.992 moles of nitrogen. Given that the stoichiometry of the reaction results in one mole of nitrogen per mole of oxime, the empirical formula of the oxime can be represented as \( R_1R_2C=NOH \). This formula indicates the presence of nitrogen, along with unspecified alkyl groups \( R_1 \) and \( R_2 \), forming the ketone part.

Molar Mass Calculation

Molar mass is a crucial concept in chemistry that tells us the mass of one mole of a substance. For complex compounds such as oximes, calculating the molar mass can be intricate. Here, the molar mass helps identify how the molecular weight divides among the component atoms.

Given that \( 1 \) mole of nitrogen (13.9g) is present in 100g of oxime, the calculation becomes straightforward. The total molar mass of the oxime is calculated by multiplying 7.2 by nitrogen's molar mass (14.01g/mol), resulting in approximately 100.872g/mol.

This mass includes nitrogen, its bonded hydrogen, and the carbon-oxygen structure remaining from the ketone. Calculating molar mass is crucial as it gives direction on what mass allocation can be expected across different groups within the molecular structure.

Alkyl Groups

Alkyl groups are chains or branches of carbon atoms that attach to molecules impacting their physical and chemical properties. In ketones, alkyl groups fill the places left by R1 and R2. These groups can significantly affect both a molecule’s function and structure.

In the case of ketone oximes, identifying potential alkyl groups involves using the calculated leftover molar mass after accounting for nitrogen, hydrogen, and oxygen in the structure. Here, R1 and R2 together need to account for a molar mass of 69.854g/mol. The goal is to identify different combinations of alkyl groups within this mass constraint.

By substituting various alkyl groups, starting from smallest like methyl (15g/mol), and moving to larger groups like ethyl, propyl, analysis can be done to determine feasible ketone structures. It's an essential step for understanding how different ketones can form oximes containing a specified nitrogen percentage.

Stoichiometry

Stoichiometry involves calculating the reactants and products in chemical reactions. It uses the principle of the conservation of mass; in reactions, matter is neither created nor destroyed.

In the context of forming oximes from ketones, stoichiometry helps in identifying how reactants combine and the proportions in which reactants convert to products. It also gives insight into how much product is expected when a given amount of reactant is used.

For example, knowing 1 mole of nitrogen is present per mole of oxime uses stoichiometry to determine that 100g of oxime aligns with the determined empirical formula, \( R_1R_2C=NOH \). This stoichiometric thinking layers the calculations for empirical formula and molar mass and guides understanding how different alkyl components integrate into the ketone's formation.