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    10cm3 of a hydrocarbon of formula CxHy was burnt in excess oxygen. At the end of the reaction o.o45g of liquid water had been formed and the total volume of gas left was 80cm3, of which 30cm3 was unburnt oxygen. Deduce the values of x and y in the formula for the hydrocarbon

    I have tried this question so many times but I just can't figure out how to get the values of x and y - I know how to use the moles=mass/Mr equation as well as the n=volume/24.... Any help would be much appreciated!
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    I tried a bit...but couldn't get the result though:/
    I calculated the vol of water formed by mdv formula. Taking the density of H20 as 1g /cm3 at room temp. So it's .045/1= .045 cm3

    Total vol of reactants = total vol of products

    So I calculated the total vol of products:

    Out of 80cm3 gas, 50cm3 is CO2

    So total vol of products= 50+0.045 cm3= 50.045cm3
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    Hey there! (: Today, in my class, we learnt it! The above given info couldn't lead me to any answer...so just ignore it pls...so sorry for posting it..


    We need to find the Emperical formula of hydrocarbon.
    So...we need the mass of carbon and hydrogen...as usual.

    We know that
    -hydrogen is present in water
    - carbon is present in CO2
    - mass of H2O given is 0.045g
    -vol of CO2 is 50cm3
    - 1 mole of any gas at room temp occupies 24 dm3
    - we assume that by the time products are formed the temp. is 25degrees Celsius.
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    1) To find the mass of H from the mass of H2O

    Find rfm of hydrogen in H2O and rfm of H2O
    (Rfm - relative formula mass)

    We can conclude:
    There's 2g of hydrogen in 18g of water (rfm values)
    So in 0.045 g of water what's the amount of hydrogen?
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    2) Find the mass of C from mass of CO2


    
 a) but we have volume of CO2, 50cm3 i.e. 0.05 dm3, so, find the moles of CO2 


    If 1 mole of CO2 has 24 dm3
    then x moles of CO2 has 0.05 dm3
    x= 2.08 * 10^-3 



    b) Use "m.n.Rfm" ratio to find mass of CO2 from the moles of CO2
    (Note m.n.Mr ratio is similar to above mentioned ratio) 


    m= n*rfm = [2.08 * 10^-3] * 44
    = 0.09152g of CO2 in reaction.


    
c) Finally find the mass of carbon from mass of CO2 in reaction (very similar to step 1 where the mass of hydrogen was found from water)

    12g of carbon in 44g CO2 (rfm of C, rfm of CO2 )
    x g of carbon in 0.09152 g CO2
    = 0.025g of carbon in CO2
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    3) The last step is the usual way of calculating empirical formula

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    I wonder if the answers and method could be confirmed by someone...
 
 
 
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