An organic compound containing carbon, hydrogen and oxygen only was found to contain 64.9% carbon and 13.5% hydrogen. If the molecular mass of the compound is 74g/mol.

Question

I) calculate the percentage of oxygen in the compound
Ii) determine the empirical formula and molecular formula
III) write the structural formula of the three possible isomers of the compound
iv) if the compound in (Ii) above gives a fragrant smell on heating with ethanoic acid in the presence of concentrated H2SO4, name the functional group present in the compound

Answer 1

I) To find the percentage of oxygen in the compound, we can start by subtracting the percentages of carbon and hydrogen from 100%:

100% - 64.9% (carbon) - 13.5% (hydrogen) = 21.6% (oxygen)

Therefore, the compound contains 21.6% oxygen.

II) To determine the empirical formula, we need to find the ratio of the different elements present in the compound.

First, we assume that we have a 100g sample of the compound. From the given percentages, we can determine the mass of each element:

64.9g carbon and 13.5g hydrogen (since C = 64.9% of 100g and H = 13.5% of 100g)

Next, we need to convert the mass of each element to moles:

1 mole of carbon has a molar mass of 12g/mol, so 64.9g of carbon is equal to 5.41 moles of carbon
1 mole of hydrogen has a molar mass of 1g/mol, so 13.5g of hydrogen is equal to 13.5 moles of hydrogen

Now, we need to find the ratio of the moles of each element. Dividing the number of moles by the smallest number of moles (5.41), we get:

Carbon: 5.41/5.41 ≈ 1
Hydrogen: 13.5/5.41 ≈ 2.5

Multiplying these ratios by 2 to get whole numbers, we get:

Carbon: 2
Hydrogen: 5

Therefore, the empirical formula of the compound is CH2O.

To determine the molecular formula, we need to know the molecular mass of the compound (given as 74g/mol). The empirical formula mass of CH2O is:

(1 * 12) + (2 * 1) + (1 * 16) = 12 + 2 + 16 = 30g/mol

Dividing the molecular mass by the empirical formula mass, we get:

74g/mol / 30g/mol ≈ 2.47

This means that the molecular formula is approximately 2.47 times the empirical formula. So, the molecular formula of the compound is C2.47H4.94O.

III) To determine the possible isomers of the compound, we need to consider the different ways the atoms can be arranged in the molecular formula: C2.47H4.94O.

The possible isomers are:

1. Dimethyl ether: CH3-O-CH3
2. Formaldehyde: O=CH2
3. Methyl alcohol (methanol): CH3-OH

IV) If the compound gives off a fragrant smell when heated with ethanoic acid in the presence of concentrated H2SO4, it suggests the presence of an alcohol functional group (OH).

Therefore, the functional group present in the compound is the alcohol functional group (OH).