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    Hi, im stuck on this question:

    Artists between the 13th and the 19th Centuries used a green pigment called verdigris. The artists made the pigment by hanging copper foil over boiling vinegar. A sample of verdigris has the formula [(CH3COO)2Cu]2.Cu(OH)2.xH2O. Analysis of the sample shows that it contains 16.3% water by mass. Calculate the value of x in the formula.

    I have worked out the first bit (460.5) but I dont know what to do afterwards.Apparently the ratio is 5 if that helps Any help is appreciated
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    (Original post by SilverWater)
    Hi, im stuck on this question:

    Artists between the 13th and the 19th Centuries used a green pigment called verdigris. The artists made the pigment by hanging copper foil over boiling vinegar. A sample of verdigris has the formula [(CH3COO)2Cu]2.Cu(OH)2.xH2O. Analysis of the sample shows that it contains 16.3% water by mass. Calculate the value of x in the formula.

    I have worked out the first bit (460.5) but I dont know what to do afterwards.Apparently the ratio is 5 if that helps Any help is appreciated
    The Mr of the dehydrated pigment is 460.5. And as water is 16.3%, the dehydrated pigment is 83.7% of the overall Mr. Is that helpful?
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    (Original post by Kvothe the Arcane)
    The Mr of the dehydrated pigment is 460.5. And as water is 16.3%, the dehydrated pigment is 83.7% of the overall Mr. Is that helpful?
    So if Im thinking right you would do 460.5 x 0.837 and 460.5 x 0.167 which would get 385.4385 and 75.0615 respectively, and then divide the two and you get 5.1349.... then you would round and get 5.

    is that the right process lol?
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    (Original post by SilverWater)
    So if Im thinking right you would do 460.5/ 0.837 and 460.5/0.837 which would get 385.4385 and 75.0615 respectively, and then divide the two and you get 5.1349.... then you would round and get 5.

    is that the right process lol?
    Not really. Although dividing 460.5 by 0.837 is one approach, I'm not sure what you've done after that.

    So you know 83.7% of overall is dry. 16.3% of overall is the water weight. Mr of water is 18 so 18x = water weight.
    If 83.7% of overall is 460.5 how would you work out the wet mass?
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    (Original post by Kvothe the Arcane)
    Not really. Although dividing 460.5 by 0.837 is one approach, I'm not sure what you've done after that.
    I actually meant I multiplied instead of dividing. (corrected my working out now so my approach should make abit more sense)
    (Original post by Kvothe the Arcane)
    If 83.7% of overall is 460.5 how would you work out the wet mass?
    Would you do 460.5 X 100/83.7 in order work out the additional mass from the water???? I'm not really sure tbh
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    (Original post by SilverWater)
    I actually meant I multiplied instead of dividing. (corrected my working out now so my approach should make abit more sense)


    Would you do 460.5 X 100/83.7 in order work out the additional mass from the water???? I'm not really sure tbh
    Yep 460.5 x 100/83.7 gives you total mass, including the water! which gives 550.2.

    The total mass of water is then simply 550.2-460.5 = 89.7

    89.7/18 = 5.0 so 5 water molecules per formula unit
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    (Original post by MexicanKeith)
    Yep 460.5 x 100/83.7 gives you total mass, including the water! which gives 550.2.

    The total mass of water is then simply 550.2-460.5 = 89.7

    89.7/18 = 5.0 so 5 water molecules per formula unit
    Ah thanks
 
 
 
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