Complete combustion of 4.80 g of a hydrocarbon produced 15.6 g of CO2 and 4.80 g of H2O. What is the empirical formula for the hydrocarbon?

How do you get a ratio?

I slipped up by converting to C first. It can be done that way but it's longer. Let me start over.

Convert 15.6g CO2 to mols. mols = g/molar mass
15.6/44 = 0.354 mols CO2 = 0.354 mols C since there is 1 atom C in 1 molecule CO2.

4.80g H2O to mols.
4.80/18 = 0.2667 mols H2O and mols H will be twice that or 0.2667*2 = 0.533 mols H.

To find the ratio do what I suggested you do. Divide the smaller number by itself (that gives you a 1.00 every time), then divide the other number by the same smaller number.
We have 0.354 mols C
and 0.533 mols H.

0.354/0.354 = 1.00 C
0.533/0.354 = 1.51 H which rounds to 1.5 and that is a ratio of 2C for 3H or the empirical formula is C2H3.
However, let me point out that 4.80g C2H3 will give you 15.6 g CO2 but it will NOT give you 4.80g H2O so I think someone goofed when they constructed the problem. But the above procedure is how you work the problem.

Well, it seems we have a case of "Mystery Element and the Combustion Conundrum!"

Let's put on our detective hats and solve this case. We know that the hydrocarbon undergoes complete combustion, which means it only produces carbon dioxide (CO2) and water (H2O).

We have 15.6 g of CO2 and 4.80 g of H2O. To find the empirical formula, we need to determine the ratios of the elements involved.

First, we need to find the amount of carbon in CO2. Since the molecular mass of CO2 is 12.01 g/mol (carbon) + 2 * 16.00 g/mol (oxygen), which equals 44.01 g/mol, we divide the mass of CO2 (15.6 g) by the molar mass (44.01 g/mol). This gives us approximately 0.354 mol of carbon.

Next, let's find the amount of hydrogen in H2O. Since the molecular mass of H2O is 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen), which equals 18.02 g/mol, we divide the mass of H2O (4.80 g) by the molar mass (18.02 g/mol). This gives us approximately 0.266 mol of hydrogen.

Now, we compare the moles of carbon and hydrogen in the hydrocarbon:

Carbon: 0.354 mol
Hydrogen: 0.266 mol

To simplify the ratios, we can divide both values by the smallest number of moles (in this case, 0.266 mol).

After dividing, we get approximately:
Carbon: 1.33 mol
Hydrogen: 1 mol

Now that we have the ratios, let's convert them to whole numbers by multiplying each by a factor of 3:

Carbon: 4 mol
Hydrogen: 3 mol

Therefore, the empirical formula of the hydrocarbon is C4H3.

Case closed, my dear Watson!

To determine the empirical formula for the hydrocarbon, we need to find the ratio of carbon and hydrogen atoms in the compound.

First, let's calculate the number of moles of carbon dioxide (CO2) and water (H2O) produced.

Moles of CO2 = mass of CO2 / molar mass of CO2
Molar mass of CO2 = 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol
Moles of CO2 = 15.6 g / 44.01 g/mol = 0.354 mol

Moles of H2O = mass of H2O / molar mass of H2O
Molar mass of H2O = 2 * 1.01 g/mol + 16.00 g/mol = 18.02 g/mol
Moles of H2O = 4.80 g / 18.02 g/mol = 0.266 mol

Since the hydrocarbon contains only carbon and hydrogen, and the combustion reaction produced equal moles of CO2 and H2O, we can assume that the number of carbon atoms in the hydrocarbon is equal to the number of moles of CO2, and the number of hydrogen atoms is equal to the number of moles of H2O.

Therefore, the empirical formula for the hydrocarbon is CH2.

Note: The empirical formula gives the simplest, most reduced ratio of atoms present in a compound. If you want to find the molecular formula, which gives the actual number of atoms in the compound, you would need additional information, such as the molecular weight or the density of the hydrocarbon.

Convert to mols C and mols H.

15.6 x (12/44) = ?
4.80 x (2*1/18) = ?
Find the ratio of the two. The easy way to do that is to divide the smaller number by itself, then divide the other number by the same small number. I get 1:1.5 which in small whole numbers is 2:3