"Hydrocarbon" is just a way of saying that the compound is made of hydrogen and carbon.
First, we need to understand that by seeing the ratio of moles of #H# atoms to #C# atoms in the product, we can find the ratio of #H# to #C# in the reactant.
This is due to the law of conservation of matter—hydrogen and carbon atoms can't appear or disappear out of nowhere.
(We can completely ignore the #O_2# for this question, because it's not going to help find the answer.)
Let's find the moles of #CO_2# and #H_2O# in the product.
In this equation:
#C_xH_y + O_2 -> CO_2 + H_2O#
#"47.0 mg"# of #CO_2# was produced, which equates to #"0.00107 moles"# because the molar mass of carbon dioxide is #"44.01g"#.
#"38.4 mg"# of #H_2O# was produced, which equates to #"0.00213 moles"# because the molar mass of water is #"18.02 g"#.
Now, we just need to figure out the ratio of moles of carbon to hydrogen in the products.
#"1 mole"# of #CO_2# is made of #"1 mole"# of #C# and #"2 moles"# of #O#.
Therefore, #"0.00107 moles"# of #CO_2# corresponds to #"0.00107 moles"# of #C#.
#"1 mole"# of #H_2O# is made of #"2 moles"# of #H# and #"1 mole"# of #O#.
Therefore, #"0.00213 moles"# of #H_2O# corresponds to #0.00213*2="0.00426 moles"# of #H#.
From this, we can infer that the mole ratio of #C# to #H# in our mystery hydrocarbon is:
#"0.00107 moles of carbon : 0.00426 moles of hydrogen"#
Then, in order to get a whole number ratio, divide the bigger number by the smaller number:
#0.00426/0.00107=3.98#, but that could be rounded up to #4#.
The ratio is now #"1 mole of carbon : 4 moles of hydrogen"#
#∴# the empirical formula is #CH_4#, also known as methane!
