Empirical Formula Lab Report
Given the unknown colourless liquid it is essential that the structure is derived. Firstly it is important to determine the percentage composition of elements to work out the empirical formula. The empirical formula was found to be C10H12O and the mass of the unknown was 148.09 m/z which when calculating the molecular weight of the empirical formula it did equal 148.09 g mol -1. This means that the empirical formula is also the molecular formula. As 12.01x10 carbons +1.008x12 hydrogens + 16= 148.09 09 g mol -1.
From this knowledge the unknown molecule must include either an alcohol, aldehyde, ketone, ether or methoxy- functional group.
Secondly it is important to calculate the degrees of unsaturation in the molecule. The degrees of unsaturation …show more content…
The most distinct peak is that of a singlet at 9.95 ppm. This has a relative intensity of one. The more deshielded a molecule the higher the chemical shift. Because the oxygen of the carbonyl bond is inductively withdrawing the electron density will reside favourably on the oxygen, which means that the electron density associated with the hydrogen is reduced which results in a high chemical shift. A chemical shift of this is only typical for alcohols or aldehydes attached directly to an aromatic ring, or nitrogen containing molecules of which is not applicable for the unknown molecule are there is no nitrogen in the molecular formula. From this peak the only possible functional group is an aldehyde as a ketone does not have a hydrogen to give a peak at this chemical shift. The relative intensity does fit with the configuration of the carbon only being bound to one hydrogen. This means that the aldehyde group must be on one end of the molecule as an aldehyde cannot be attached to more than one carbon. This peak is a singlet which indicates that the aldehyde proton is not coupled to any other proton this is because the carbon adjacent does not contain any hydrogens. This would be consistent with the aldehyde group being directly attached to the benzene ring. If the peak was between 9.4 and 10 ppm this would be consistent with the aldehyde group being bound to a linear carbon system. As this peak is between 9.7 and 10.5 ppm2 this expected for an aldehyde proton attached to an aromatic
From this knowledge the unknown molecule must include either an alcohol, aldehyde, ketone, ether or methoxy- functional group.
Secondly it is important to calculate the degrees of unsaturation in the molecule. The degrees of unsaturation …show more content…
The most distinct peak is that of a singlet at 9.95 ppm. This has a relative intensity of one. The more deshielded a molecule the higher the chemical shift. Because the oxygen of the carbonyl bond is inductively withdrawing the electron density will reside favourably on the oxygen, which means that the electron density associated with the hydrogen is reduced which results in a high chemical shift. A chemical shift of this is only typical for alcohols or aldehydes attached directly to an aromatic ring, or nitrogen containing molecules of which is not applicable for the unknown molecule are there is no nitrogen in the molecular formula. From this peak the only possible functional group is an aldehyde as a ketone does not have a hydrogen to give a peak at this chemical shift. The relative intensity does fit with the configuration of the carbon only being bound to one hydrogen. This means that the aldehyde group must be on one end of the molecule as an aldehyde cannot be attached to more than one carbon. This peak is a singlet which indicates that the aldehyde proton is not coupled to any other proton this is because the carbon adjacent does not contain any hydrogens. This would be consistent with the aldehyde group being directly attached to the benzene ring. If the peak was between 9.4 and 10 ppm this would be consistent with the aldehyde group being bound to a linear carbon system. As this peak is between 9.7 and 10.5 ppm2 this expected for an aldehyde proton attached to an aromatic