A hydrocarbon is burnt completely to give 3.447g of #sf(CO_2)# and 1.647g of #sf(H_2O)#. What is the empirical formula ?

Your compound's empirical formula is #C_3H_7#.

I'll show you an alternative approach to getting the number of moles of carbon and of hydrogen the hydrocarbon sample contained.

This method uses the percent composition of carbon and of hydrogen in carbon dioxide and water, respectively.

So, you know that the combustion reaction produces 3.447 g of carbon dioxide. If you use the molar mass of carbon and that of carbon dioxide, you can determine the percent composition of carbon in #CO_2#.

#(12.0cancel("g/mol"))/(44.0cancel("g/mol")) * 00 = "27.27% C"#

This means that, for every 100 g of #CO_2#, you get 27.27 g of carbon. This means that the mass of carbon in the original hydrocarbon was

#3.447cancel("g"CO_2) * "27.27 g C"/(100cancel("g"CO_2)) = "0.9400 g C"#

Do the same for water and hydrogen, but keep in mind that you get 2 hydrogen atoms for every molecule of water.

#(2 * 1.01cancel("g/mol"))/(18.02cancel("g/mol")) * 100 = "11.21% H"#

Since every 100 g of water will contain 11.21 g of hydrogen, your hydrocarbon contained

#1.647cancel("g"H_2O) * "11.21 g H"/(100cancel("g"H_2O)) = "0.1846 g H"#

Now use the molar masses of carbon and hydrogen to determine how many moles of each you have

#"For C": (0.9400cancel("g"))/(12.0cancel("g")/"mol") = "0.07833 moles C"#

#"For H": (0.1846cancel("g"))/(1.01cancel("g")/"mol") = "0.1828 moles H"#

Once again, divide both numbers by the smallest one to get the mole ratio carbon and hydrogen have in the hydrocarbon

#"For C": (0.07833cancel("moles"))/(0.07833cancel("moles")) = 1#

#"For H": (0.1828cancel("moles"))/(0.07833cancel("moles")) = 2.33#

You sample contained

#C_1H_2.33#

Multiply by 3 to get rid of the fractional subscript

#C_3H_7# #-># your compound's empirical formula.

The empirical formula is #C_3H_7#.

Convert moles to grams by dividing by the #M_r#:

Moles #CO_2 = 3.447/44.01=0.0783#

Moles #H_2O=1.647/18.01=0.0914#

Moles #C=0.0783#

Moles#H=0.0914xx2=0.1829#

Ratio #C:H# by moles = #0.0783:0.1829rArr#

#1:2.33rArr#

#3:7#

So the empirical formula is #C_3H_7#