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?

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  1. 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

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  2. How do you get a ratio?

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  3. 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.

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