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Calculating volume occupied by 1 equivalent of gas

  1. Apr 8, 2015 #1
    Its given in my textbook that equivalence of gas = (volume of gas) / (Volume of 1 eq. gas at NTP)
    How do i calculate volume occupied by 1 equivalent of gas?
    Example: What would be the volume occupied by 1 eq. of CO2 (Its given 11.2l in my book)
    Is there a formula for this?
     
  2. jcsd
  3. Apr 8, 2015 #2

    epenguin

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    The volume occupied by 1 MOLE of gas at NTP is usually given as 22.4 L.
    Obvious relation to the 11.2 L you mention.
    Best you have a shot at telling us why that relation.
     
  4. Apr 8, 2015 #3
    let me start with a simpler molecule O2
    1 mole of oxygen combines with 2 moles of hydrogen. So 1 mole of hydrogen combines with 1/2 moles of oxygen.
    So 1/2 moles of oxygen becomes 1 equivalence of oxygen. (From definition of equivalence)
    1/2 mole occupies 11.2 liters.
    Hence volume of 1 eq. oxygen is 11.2
    But carbon dioxide does not combine with hydrogen(nor oxygen). Then how can I find its eq. volume?
     
  5. Apr 8, 2015 #4

    epenguin

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    It combines with other things doesn't it? It forms carbonates.
     
  6. Apr 9, 2015 #5
    but how do i use that to find its equivalent volume?
     
  7. Apr 9, 2015 #6

    epenguin

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    I'm sorry but on second thoughts I don't know. One reason is I don't need to and no one needs to. At school we were fed a lot with "equivalent weights" made an insistent drill just because many students had great difficulty grasping it. But I think this is no longer taught, is regarded as an outdated unnecessary pedantry; it is just much easier to think always about moles. And "equivalent volumes" even we never had.

    On a second look, on the face of it 1 mole of CO2 forms 1 mole of H2CO3 which in theory could react with 1 mole of bivalent metal, say Ca, or two of monovalent, say Na, to form 1 mole of H2. Which would occupy 22.4 L at NTP. So I'd call that 1 "equivalent volume".

    But equivalents are defined in terms of reactions so maybe, check your book, they are talking about reactions involving just bicarbonates? E.g theoretically e.g. 2 moles of NaHCO3 reacting with metal gives 1 mole H2, I.e. 1 mole gives ½ mole H2 = 11.2 L.

    So I guess equivalents depend on what reaction you're talking about, moles are just moles and a more straightforward concept.
     
  8. Apr 9, 2015 #7
    I have to calculate the percentage of NaHCO3 in a mixture of NaCl, Na2CO3 and NaHCO3 given that the mixture produced 56ml of carbon dioxide on heating and 1.6g of the mixture required 25ml of 1N HCl solution for nuetralisation.
    I can easily find it using the relation eq. of NaHCO3=eq. of CO2
    And eq. of CO2=56/11200
    So weight of sodium bicarbonate is 56/11200 * 84 = 0.42g
    Only the 11.2 part confuses me.

    Also, its not really that easy to use the concept of moles because you have to balance the reaction to obtain a relation like "n moles of A combines with m moles of B..."
    Balancing is not necessary if you use equivalents. You always have to follow a simple equation "milli.Eq of A=milli.Eq of B". This is what my teacher told me.
    Secondly, concept of equivalence forms the basis of volumetric calculations and titrations (like iodometry). So it has to be taught.
     
  9. Apr 9, 2015 #8

    Borek

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    You got it reversed. If you have balanced the reaction, equivalents follow the stoichiometry. And to get equivalents you need to balance the reaction first. Solving using moles and stoichiometry is the most universal approach.

    Nope. Again, it is balanced reaction that forms the basis. Equivalents sometimes make the calculations easier, but sometimes they are the best tool to confuse students. Your problem with solving the question is a perfect example of such a case - I bet just following the stoichiometry you would solve the problem long ago.
     
  10. Apr 9, 2015 #9
    I'm not sure. Its given in my textbook that for the reaction aA+bB→cC+dD
    eq. of A = eq. of B = eq. of C = eq. of D
    I don't see the stochiometric coefficients in the relation.

    Reference : Modern Approach to Chemical Calculations by R.C.Mukherjee (MSc,PhD)


    Note: Last time i specified the reference in this thread https://www.physicsforums.com/threads/concept-of-reduced-mass.805644/#post-5057261 @Orodruin abandoned it. I even sent him a message and i got no reply.
     
  11. Apr 9, 2015 #10

    Borek

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    Because they are already incorporated into the equivalents. The question remains, how do you know what an equivalent is?

    Equivalents were modern somewhere in the mid-20th century if not earlier. As of today they were abandoned even by IUPAC (which is rather conservative and doesn't adapt quickly to changes).
     
  12. Apr 9, 2015 #11
    I have not used balanced equations for solving volumetric calculation questions using equivalence. If you want, i will show how i solve such problems. I know what equivalence is. It depends on the situation. Like for sulphuric acid, n factor is 2. So its eq. wt. becomes Molecular wt./2
     
  13. Apr 9, 2015 #12
    one more thing. IUPAC has abandoned it. But in engineering entrance exams in India, in the chemistry section, you can still find questions based on equivalence directly or indirectly.
     
  14. Apr 9, 2015 #13

    Borek

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    So, where did you got equivalence information from?
     
  15. Apr 9, 2015 #14
    You don't have to write the balanced equations. In the following example, i have not even thought about it.
    Example question:
    7.35g of a dibasic acid was dissolved in water and diluted to 250ml and 25ml of this solution was neutralized by 15ml of 1N NaOH solution. Calculate the molecular wt. of the acid.
    solution:
    equivalent of acid = ##\frac{7.35}{E}##
    So 250ml contains ##\frac{7350}{E}## milli.eq. of acid
    Hence 25ml contains ##\frac{735}{E}## milli.eq. of acid
    Now milli.eq. of acid = milli.eq. of base
    So ##\frac{735}{E}=15##
    So E=49.
    So molecular weight = eq.wt × basicity = 49×2 = 98.
     
  16. Apr 9, 2015 #15

    Borek

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    So basically what you are saying is: I can solve any problem with equivalents, but I don't know how to find what the equivalent is.

    If you don't see a problem here, and if you don't see contradiction, I can't help you further. You were told how to find the answer. Stubbornly refusing to accept stoichiometry as a universal approach you are just wasting our time.

    EOT
     
  17. Apr 9, 2015 #16
    I have many entrance exams coming and right now i have to learn techniques and i have lots of time to understand the concepts.
    My parents complain that i waste time by using PF. But i wanted to prove them wrong. If you want, help me. Else don't.
    By the way, you are not wasting time. You are building a scientist or an engineer out of a student like me. So its not a waste.
     
  18. Apr 9, 2015 #17

    epenguin

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    So reduce both experiments to the same number of g of mixture. Say it contains x moles of NaHCO3 and y moles of Na2CO3.

    Then the moles of CO2 which you calculate from the volume are x + y.

    The moles of HCl used to titrate are x + 2y.

    Simultaneous equations don't get much simpler. Just with moles. It looks like I answered your question previously but that answer wasn't needed to solve the problem. Likewise easily solve that of #14 or iodometric titrations.

    Except for the question you seem to know how to translate moles into equivalents or vice versa if required by exam systems putting you through these hoops. I agree with
     
    Last edited: Apr 10, 2015
  19. Apr 9, 2015 #18
    One question.
    How do you use mole concept in complexometric titrations like EDTA titration for estimation of hardness of water? We are not taught complexometric reactions and we were made to use equivalence.
     
  20. Apr 10, 2015 #19

    Borek

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    Chelating agents (EDTA being one of them) always react with cations in a stoichiometric ratio 1:1, something like

    Ca2+ + EDTA4- ↔ CaEDTA2-
     
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