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29 June, 18:19

Complete combustion of 7.80 g of a hydrocarbon produced 25.1 g of CO2 and 8.55 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary

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  1. 29 June, 21:50
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    The empirical formula is C3H5

    Explanation:

    Step 1: Data given

    Mass of the compound = 7.80 grams

    Mass of CO2 = 25.1 grams

    Molar mass of CO2 = 44.01 g/mol

    Mass of H2O = 8.55 grams

    Molar mass of H2O = 18.02 g/mol

    Molar mass C = 12.01 g/mol

    Molar mass H = 1.01 g/mol

    Molar mass O = 16.0 g/mol

    Step 2: Calculate moles CO2

    Moles CO2 = mass CO2 / molar mass CO2

    Moles CO2 = 25.1 grams / 44.01 g/mol

    Moles CO2 = 0.570 moles

    Step 3: Calculate moles C

    For 1 mol CO2 we have 1 mol C

    For 0.570 moles CO2 we have 0.570 moles C

    Step 4: Calculate mass C

    Mass C = 0.570 moles * 12.01 g/mol

    Mass C = 6.846 grams

    Step 5: Calculate moles H2O

    Moles H2O = 8.55 grams / 18.02 g/mol

    Moles H2O = 0.474 moles

    Step 6: Calculate moles H

    For 1 mol H2O we have 2 moles H

    For 0.474 moles H2O we have 2*0.474 = 0.948 moles H

    Step 7: Calculate mass H

    Mass H = 0.948 moles * 1.01 g/mol

    Mass H = 0.957 grams

    Step 8: Calculate mol ratio

    We divide by the smallest amount of moles

    C: 0.570 moles / 0.570 = 1

    H: 0.948 moles / 0.570 = 1.66

    This means for 1 mol C we have 1.66 moles H OR for 3 moles C we have 5 moles H

    The empirical formula is C3H5
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