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Doubts regarding basic chemistry concepts

  1. Jun 17, 2017 #1
    Doubts regarding the law of reciprocal proportions: What i have understood by reading the definition is that the ratio of the masses in which two elements A and B combine with a fixed mass of a third element C, will be a whole number multiple of the ratio of the masses in which A and B combine with each other.
    My question is that is order important here? Let A and C combine in the ratio a:c1 and let B and C combine in the ratio b:c2=((b*c1)/c2):c1.(by mass). Then according to the law of reciprocal proportions, a:((b*c1)/c2)=k (a:b) where k ia whole number. If we reverse the order in which we consider the mass of the elements in the ratio, will we get the same result then also? (Is ((b*c1)/c2):a=j (b:a) where j is a whole number in that case also).

    Doubts regarding equivalent weight:I do understand the definition of equivalent weight but cannot make sense of it and understand the significance of the definition. What is the meaning of "parts by mass"
    in it's definition and what is the use of equivalent weight?
     
  2. jcsd
  3. Jun 17, 2017 #2

    symbolipoint

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    Your concept is good but your numerical description using symbols is confusing. Maybe try again.

    Combining ratios of elements really does make use of ratio IN MOLES. We use then a ratio of MASS to MOLES to make a conversion into PARTS BY MASS, if we want it; but finding the result as simple ratio between whole numbers takes more work/effort.
     
  4. Jun 18, 2017 #3
    Here is the numerical part with proper editing.
    Let A and C combine in the ratio [itex]\frac{a}{c1}[/itex] and let B and C combine in the ratio [itex]\frac{b}{c2}=\frac{b*c1}{c1*c2}[/itex].(by mass). Then according to the law of reciprocal proportions, [itex]\frac{a*c2}{b*c1}[/itex]=k [itex]\frac{a}{b}[/itex] where k is whole number. If we reverse the order in which we consider the mass of the elements in the ratio, will we get the same result then also? (Is [itex]\frac{b*c1}{a*c2}[/itex]=j [itex]\frac{b}{a}[/itex] where j is a whole number in that case also). I had done some calculations and i found out that this is not true. So is this violating the law of reciprocal proportions?
     
  5. Jun 18, 2017 #4

    symbolipoint

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    Many people learned compounds being made according to mole ratios, and reactions occurring along mole ratios of the reactants. Then people were taught about "molecular" weights and using them to convert moles to masses, and then people might have also been taught about gravimetric factors. Remember the word, "stoichiometry"?

    If given reactants or products in grams or kilograms, they can be converted to moles, using their formula weights or molecular weights. If given reactants or products in moles, they can be converted to their masses using their formula weights or molecular weights.
     
  6. Jun 19, 2017 #5
    If a*c2/b*c1 = k*a/b, it follows that the reciprocal b*c1/a*c2 = (1/k)*b/a. If k is a whole number, 1/k is not, unless k =1. So yes, order matters. A better statement of the principle would be that the ratio will be a whole number or reciprocal whole number (or perhaps, even more generally, the ratio of two whole numbers) multiple of the ratio in which A and B combine.
     
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