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    Complete combustion of 1.80g of compound L produced 2.64g of CO2, 1.08g of water and 1.92g of sulphur dioxide
    calculate the empirical formula of L????
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    (Original post by saarahtalha)
    Complete combustion of 1.80g of compound L produced 2.64g of CO2, 1.08g of water and 1.92g of sulphur dioxide
    calculate the empirical formula of L????
    Find the moles of each product and you put them into whole number ratios, giving your MOLECULAR formula of compound L. You then simply find the common factor of the molecular formula and divide each of the molecules in the compound giving you the empirical formula.

    you may have to round off your mole figures as well
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    (Original post by saarahtalha)
    Complete combustion of 1.80g of compound L produced 2.64g of CO2, 1.08g of water and 1.92g of sulphur dioxide
    calculate the empirical formula of L????
    Empirical question.
    Divide each of the masses by the molecular mass
    2.64/44 = 0.06
    1.08/18 =0.06
    1.92/48 = 0.04
    Then just ratio it out
    3:3:2

    Count the number of total atoms which is
    3 C
    13 O
    12 H
    2 S

    So answer is C3H12O13S2
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    (Original post by AbelTesfaye_isMJ)
    Find the moles of each product and you put them into whole number ratios, giving your MOLECULAR formula of compound L. You then simply find the common factor of the molecular formula and divide each of the molecules in the compound giving you the empirical formula.

    you may have to round off your mole figures as well
    Wait I'm confused how does making the whole ratios give u the Mr of L???
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    (Original post by saarahtalha)
    Wait I'm confused how does making the whole ratios give u the Mr of L???
    In theory, yes. To which you cancel down after. But as Daniel just demonstrated, the molecular formula just happens to be the same as the empirical formula. this is because it cannot further be cancelled down since its already in its simple form
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    (Original post by AbelTesfaye_isMJ)
    In theory, yes. To which you cancel down after. But as Daniel just demonstrated, the molecular formula just happens to be the same as the empirical formula. this is because it cannot further be cancelled down since its already in its simple form
    I don't get it so whats the answer
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    The answer above is wrong. You forgot about L reacting with O2
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    This is how you do it
    Moles of CO2=2.64/44=0.06
    Moles of H2O=1.08/18=0.06
    Moles of SO2=1.92/64=0.03
    0.06:0.06:0.03
    2:2:1
    Number of C - 2
    Number of H - 4
    Number of S - 1
    Number of O - 8(not part of substance L because L reacts with O2)
    Empirical formula - C2H4S
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    (Original post by cold_eye)
    this is how you do it
    moles of co2=2.64/44=0.06
    moles of h2o=1.08/18=0.06
    moles of so2=1.92/64=0.03
    0.06:0.06:0.03
    2:2:1
    number of c - 2
    number of h - 4
    number of s - 1
    number of o - 8(not part of substance l because l reacts with o2)
    empirical formula - c2h4s
    thank you!
 
 
 
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