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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????
    Swear I already answered this question on your other post 🤔
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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????
    For each compound you have calculate the mr and divide the weight you have of each compound by it's mr and then divide all the compounds by the smallest number.For example the mr of CO2 is 12+16+16=44 So you do 2.64 divided by 44 and then h2o has an mr of 18 so you do 18/1.08 and then do the same thing for sulphur dioxide then divide all the numbers by the smallest number you get.I think this is how you do it,but this is just a guess.
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    (Original post by Anonymous1502)
    For each compound you have calculate the mr and divide the weight you have of each compound by it's mr and then divide all the compounds by the smallest number.For example the mr of CO2 is 12+16+16=44 So you do 2.64 divided by 44 and then h2o has an mr of 18 so you do 18/1.08 and then do the same thing for sulphur dioxide then divide all the numbers by the smallest number you get.I think this is how you do it,but this is just a guess.
    But how do you work out L??????
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    (Original post by saarahtalha)
    But how do you work out L??????
    After you do the calculation the ratio of products produced is 2:2:1 (co2,h20,so2) But Im not sure what the name of compound L would be.
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    You can work out the amount of carbon, hydrogen etc in each of the products by doing Mr of element / total Mr of compound multiplied by mass of product - so for Carbon, 12/44 x 1.92 = 0.72g carbon. This has all come from L. Do this for the other compounds and then calculate empirical formula in the normal way.
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    (Original post by PuffyPenguin)
    You can work out the ratio of carbon to hydrogen to sulfur in L from the other ratio I think
    How?
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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????


    You can work out the amount of carbon, hydrogen etc in each of the products by doing Mr of element / total Mr of compound multiplied by mass of product - so for Carbon, 12/44 x 1.92 = 0.72g carbon. This has all come from L. Do this for the other compounds and then calculate empirical formula in the normal way. I got C2H4S
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    (Original post by PuffyPenguin)
    You can work out the amount of carbon, hydrogen etc in each of the products by doing Mr of element / total Mr of compound multiplied by mass of product - so for Carbon, 12/44 x 1.92 = 0.72g carbon. This has all come from L. Do this for the other compounds and then calculate empirical formula in the normal way. I got C2H4S
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