Question

When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 86.18 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

- Empirical:

`C_3H_7`

- Molecular:

`C_6H_(14)`

Step-by-step explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

`n_C=21.71gCO_2*(1molCO_2)/(44gCO_2)*(1molC)/(1molCO_2)=0.493molC`

Moreover, the moles of hydrogen:

`n_H=10.37gH_2O*(1molH_2O)/(18gH_2O)*(2molH)/(1molH_2O)=1.152molH`

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

`C=(0.4934)/(0.4934)=1\\H=(1.15222)/(0.4934)=2.335\\CH_(2.335)`

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

`C_3H_7`

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

`C_6H_(14)`

Which is hexane.

Best regards.