Question

A hydrocarbon compound is burnt completely in air to form 17.6 g of carbon dioxide gas and 7.2 g of water. What is the molecular formula of the hydrocarbon compound? [Given that the relative atomic mass of C= 12, H= 1, O=16] ​

Answer 1

Answer: The empirical formula is
`CH_2`. I think you said molecular formula instead of empirical formula, because there is not enough information to determine the molecular formula of the hydrocarbon.

Step-by-step explanation:

All of the masses I have used are in grams.

First, calculate the molar mass of carbon dioxide and water:

Molar mass of
`CO_2 = 2*16+12=44`

Molar mass of
`H_2O=2*1+16=18`

Then, calculate the number of moles of carbon dioxide and water:

# of mol of
`CO_2=(17.6)/(44)=0.4`

# of mol of
`H_2O=(7.2)/(18)=0.4`

Third, calculate the number of moles of carbon in the carbon dioxide and the number of moles of hydrogen in the water:

# of mol of
`C=0.4*1=0.4`

# of mol of
`H=0.4*2=0.8`

Finally, divide the number of moles of carbon and the number of moles of hydrogen by the smaller number obtained in step 3 for the number of moles of each element (in this case 0.4):

For
`C=(0.4)/(0.4)=1`

For
`H=(0.8)/(0.4)=2`

Since the ratio of C (carbon) to H (hydrogen) in the hydrocarbon is 1 to 2, the empirical formula of the hydrocarbon is
`CH_2`.