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    Ive had trouble with a question, i know the answer ( AL2 O3) but i dont know how to get there. Here is it, can someone explain how to get to there answer.

    Determine the empirical formula of the compound formed when 5.40g of aluminium reacts with 4.80g of oxygen
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    (Original post by Yasiin2000)
    Ive had trouble with a question, i know the answer ( AL2 O3) but i dont know how to get there. Here is it, can someone explain how to get to there answer.

    Determine the empirical formula of the compound formed when 5.40g of aluminium reacts with 4.80g of oxygen

    Do you know the steps to getting an empirical formula?

    Hint you'll need the relative atomic mass of aluminium and oxygen to do this.

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    first u divide the masses of aluminium and oxygen (5.40 and 4.80) by their relative atomic mass ( 27 and 16) then u divide the two numbers u get by the smallest of the 2 and thats it
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    (Original post by Yasiin2000)
    first u divide the masses of aluminium and oxygen (5.40 and 4.80) by their relative atomic mass ( 27 and 16) then u divide the two numbers u get by the smallest of the 2 and thats it
    Yeah so if you've done that you'll get for the ratio
    1:1.5 right ? However because you have that decimal and its 1.5 you'll have to multiply by 2 which will give the bove answer you got

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    (Original post by GabbytheGreek_48)
    Yeah so if you've done that you'll get for the ratio
    1:1.5 right ? However because you have that decimal and its 1.5 you'll have to multiply by 2 which will give the bove answer you got

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    wait i dont understand why you multiplied by 2, and why u chose 2
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    (Original post by Yasiin2000)
    wait i dont understand why you multiplied by 2, and why u chose 2
    to get rid of the decimal? when we are talking about an empirical formula, the numbers in the compound are always whole you wouldn't say AlO3/2 you would say Al2O3. Simplest whole number ratio, so if you don't multiply by 2 you are left without a whole number ratio and if you multiply by anything greater than 2 it is no longer the simplest ratio, hence no longer the empirical formula
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    (Original post by k.russell)
    to get rid of the decimal? when we are talking about an empirical formula, the numbers in the compound are always whole you wouldn't say AlO3/2 you would say Al2O3. Simplest whole number ratio, so if you don't multiply by 2 you are left without a whole number ratio and if you multiply by anything greater than 2 it is no longer the simplest ratio, hence no longer the empirical formula
    and as per usual, tsr saves me. thanks guys
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    (Original post by Yasiin2000)
    and as per usual, tsr saves me. thanks guys
    No problem yeah the multiplying stumped me too but when it's a decimal like that (like it's not too close to the next number up or down like .9 or.1) you usually multiply

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