Question

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. The empirical formula of the compound was found to be C3H6O2. Show how this was calculated.

What does the empirical formula tell you about the compound?

The molar mass of the actual compound was found to be 222.27g/mol. Find the molecular formula of this compound. What does the molecular formula tell you about the compound?

Can you see what type of functional group this compound could have?

Answer 1

See explanation.

Step-by-step explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

`n_C=0.01962molCO_2*(1molC)/(1molCO_2) =0.01962molC`

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

`n_H=0.01961molH_2O*(2molH)/(1molH_2O)=0.03922molH`

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

`m_O=0.4647g-0.01962molC*(12gC)/(1molC)-0.03922molH*(1gH)/(1molH) =0.1900gO`

4. Compute the moles of oxygen by using its molar mass:

`n_O=0.1900gO*(1molO)/(16gO)=0.01188molO`

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

`C=(0.01962)/(0.01188)=1.65\\ \\H=(0.03922)/(0.01188) =3.3\\\\O=(0.01188)/(0.01188) =1`

6. Search for the closest whole number (in this case multiply by 2):

`C_3H_6O_2`

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

`M=12*3+1*6+16*2=74g/mol`

Which is about three times in the molecular formula, for that reason, the actual formula is:

`C_9H_(18)O_6`

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.