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    A hydrocarbon was subjected to combustion analysis. A sample of the compound gave 0.314g CO2 (carbon dioxide) and 0.129g H2O (hydrogen oxide).

    1. Calculate the masses of carbon and hydrogen in the sample.
    2. Find the number of moles of carbon and hydrogen in the sample of the compound.
    3. What is the empirical formula of the compound?
    4. If its relative molecular mass is 84, what is its molecular formula?
    5. Draw a possible skeletal formula.

    any help appreciated! rep up for grabs!
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    18
    H2O= hydrogen oxide?

    Its water, especially if it combustion of hydrocarbons

    And I'm sorry, I'd help but its chemical calculations, of which I cannot do

    I've got my book open trying to work it out...
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    Is that all the info you were given?
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    1. I've got 0.105 g of carbon and 0.086g of hydrogen
    2. 0.0825 mol of C? and 0.086 mol of H?

    Before i continue cam some check that please
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    1. What percentage of CO2 is carbon? and what percentage of H2O is hydrogen? Once you have this you can get the mass of each...
    2. Take the above answer and divide by Mr (moles = mass/Mr )
    3. What is the ratio of carbon moles to hydrogen moles?
    4. Find the empirical mass, then find the multiple you need to convert imperial into molecular
    5. Draw it :p:
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    (Original post by Loz17)
    1. I've got 0.105 g of carbon and 0.086g of hydrogen
    2. 0.0825 mol of C? and 0.086 mol of H?

    Before i continue cam some check that please
    Not quite no...

    % mass of C in CO2 = 12/(12 + 2 x 16) x 100 = 27.27%

    so mass = 0.314g/100 x 27.27 = 0.086g

    same idea for the hydrogen
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    (Original post by EierVonSatan)
    Not quite no...

    % mass of C in CO2 = 12/(12 + 2 x 16) x 100 = 27.27%

    so mass = 0.314g/100 x 27.27 = 0.086g

    same idea for the hydrogen

    Where does the (12 + 2 x 16) come from?

    Im guessing the 12 in that is the Mr?
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    (Original post by Loz17)
    Where does the (12 + 2 x 16) come from?

    Im guessing the 12 in that is the Mr?
    It's Ar/Mr (C/CO2)
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    (Original post by EierVonSatan)
    It's Ar/Mr (C/CO2)
    Oh so its 12 + 32 in effect (Ar of C and Mr of O2)

    Sorry if terminology wrong there
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    (Original post by Loz17)
    Oh so its 12 + 32 in effect (Ar of C and Mr of O2)

    Sorry if terminology wrong there
    yeah because 2 x 16 = 32 :p:
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    (Original post by EierVonSatan)
    yeah because 2 x 16 = 32 :p:
    Lol sorry i read it as 12 add 2 (14) and multiply by 16 originally
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    thanks to both of you!! heres my answers:

    1. 12/44= 27.27%. 0.314/100 *27.27 = 0.086g of carbon
    2818 = 11.1%. 0.129/100 * 11.1 = 0.0143g of hydrogen

    2. 0.086/12 = 0.0071 of carbon
    0.0143/2 = 0.00715 of hydrogen

    3. CH2

    4. C6H12

    do you think these are right?
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    (Original post by wayno9789)
    thanks to both of you!! heres my answers:

    1. 12/44= 27.27%. 0.314/100 *27.27 = 0.086g of carbon
    2818 = 11.1%. 0.129/100 * 11.1 = 0.0143g of hydrogen

    2. 0.086/12 = 0.0071 of carbon
    0.0143/2 = 0.00715 of hydrogen

    3. CH2

    4. C6H12

    do you think these are right?
    looks good :yy:
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    thanks a lot guys. much appreciated. will add rep as soon as poss
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    How did you get to answer c?

    I am in the same problem as you!
 
 
 
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