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  • Saturday, 21 January 2017

    [How To]Perform a Material Balance calculation


    Dear all........!!
    Back just today with a dead basic post, which every chemical engineer is terrified about, and believe me majority of the engineers just use that word without having much clarity, i.e., Material Balance.

    I'll explain you the basic things regarding Material balance that were relevant to Pharma field.

    What is Material Balance ?

    Material Balance is nothing but just checking whether the input quantities and output quantities of a Chemical reaction is upto the calculation, simply equal or not. if not what quantity is missing and whether the chemical reaction is sustainable or not, if yes then what amount og output we can get based on the input we are taking.

    Material balance will always adhere to the basic law of conservation,

    What is Law of Conservation ?

    Law of Conservation will just state that what ever the input that we are taking may transform to another form, but finally whatever the output will be, equals to the input.

    Input Material Mass = Output Material Mass
    What is Limiting Reactant ?

    Limiting Reactant is the input raw material that which is taken in limit, and based on the quantity that we are taking as Limiting reactant, the output depends.

    What is KSM ?

    KSM  is basically abbreviated as Key Starting Material, the name itself indicates that major chemical reaction in the synthesis will start with the KSM.


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    What is Intermediate ?

    Intermediate  is the One which were produced in the N-1 stages of reaction, when reaction propagates in N stages.

    So, upto me if you were fully thorough of the definitions that i mentioned above then my remaining post will make some sense to you.

    Now i'll demonstrate a set of chemical reactions, that will be producing some product Z.

    Now, the below synthetic scheme is for a product manufacturing Z, where the input batch size is 49 Kgs of X.


    Here, i'll be not giving you any solvents, as these were just some development steps.


    KSM : X ,  Limiting Reactant, KSM : X, Final Product: Z,



    Mol Wt Weight
    A 49.5 100
    B 25 16.1
    X 70.5 60
    Y 60 60
    Z 60
    C 70
    D 18
    A* 80
    C* 45
    X* 68
    Y* 69.42


    REACTION 1





    REACTION 2





    REACTION 3





    That's it, now i'll explain you, the mechanism,

    In Reaction 1 : Two reactants A [1 mole] & B [0.7 mole]  react to form C [0.7 mole] & D [1 mole], which will be useful in the Reaction 2,

    In Reaction 2 : Here our KSM X[1 mole] enters the scene and starts reacting with of C [0.7 moles]&  A [1 mole] to form an Intermediate X* [ 1 mole], C* [ 1 mole], A* [ 0.7 moles].

    Note: You way be having a query about D, whatever that is produced will be washed away during workups.

    In Reaction 3 : Intermediate X* [1 mole] Reacts with reactant Y [ 0.85 moles] in presence of catalyst C* to produce our desired product Z [1 mole] and Z* [ 0.85 moles]  as Bi-Products.




    This is the mechanism, this will be given in the development report, now we have to start out material balance.

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    Now we need to analyse the above Reactions based on Raw Materials, i'll show you how,

    Consider Reactant A : i've taken 100 Kgs as input, and in Reaction 1, 49.5 Kgs got reacted with 17.5 Kgs of B[0.7 moles] to form C [ 45.08 Kgs] & D [ 16.56 Kgs],




    Now analysing the reaction with the input & output streams, we got to know that in Reaction 1, B is limiting reactant and A is excess reactant, the excess quantity that is left over in Reaction 1 is 54.46 Kg.

    In Reaction 2,
     KSM X enters the scene and reacts with the produced Intermediate C & Reactant A, to produce Intermediate X*[57.9 Kgs ] & C*[ 38.3 Kgs ] & A*[47.7 Kgs], D [16.6 Kgs]

    out of which Intermediate A* & D will be washed during workup's and the excess quantity of Reactant A[12.3 Kgs] & C[3.4 Kgs]

    In Reaction 3,
    Intermediate X*[57.9 Kgs] reacts with Reactant Y[60 Kgs] in presence of Intermediate C*[38.3 Kgs] to form desired product Z[51.1 Kgs],

    from the Reaction 3, the output streams that we are getting will be Intermediate C*[38.3 Kgs] & Bi-product Y*[50.2 Kgs], and the un reacted quantities will be Reactant Y[16.6 Kgs]


    So Now, Lets calculate the Input stream quantities,







    Input RM Quantity[Kg] Output Quantity[Kg]
    A 100 Z 51.1
    B 16.1 C* 38.3
    X 60 Y* 50.2
    Y 60 D 16.6
    Total 236.1 A* 47.7
    Excess A 12.3
    Excess C 3.4
    Excess Y 16.6
    Total 236.1



    That's it........!!!

    hope you all enjoyed this tutorial......!!!

    If you are having any doubts feel free to contact us............!!!!

    comments are most appreciated





    NOTE: The above mentioned equations were just own one's developed, not copied from any synthesis route, and these are the property of PHARMA ENGINEERING


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    About The Author


    Hi! I am Ajay Kumar Kalva, Currently serving as the CEO of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.
    Follow Me on Twitter AjaySpectator & Computer Innovations

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