Calculate the atomic mass(average) of chlorine using the followi
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    Some Basic Concepts of Chemistry
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    Calculate the atomic mass(average) of chlorine using the following data:

       %Natural Abundance   Molar mass 
    35Cl   75·77    34·9689
    37Cl   24·23  36·9659

              
                                                                
                                                                  

    Fractional abundance of 35Cl = 0·7577:
    Molar mass = 34.9689
    Fractional abundance of 37Cl = 0.2423
                        Molar mass = 36.9659
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    Calculate the amount of carbon dioxide that could be produced when
    (i) 1 mole of carbon is burnt in air.
    (ii) 1 mole of carbon is burnt in 16g of dioxygen.
    (iii) 2 moles of carbon are burnt in 16g of dioxygen.

    The balanced equation for the combustion of carbon in dioxygen/air is as represented below:
    C(s)     +   O2 (g)        CO2(g)
    1 mol         1 mol            1 mol
    (12g)            (32 g)         (44 g)

    (i) Since in air, the combustion is complete, therefore when 1 mole of carbon is burnt, the amount of CO2 formed is 44g.

    (ii) Since 16g of dioxygen is available, therefore it can combine only with 0.5 mole of carbon. Thus, dioxygen is the limiting reactant and O2 produced is 22g.

    (iii) Here again, dioxygen is the limiting reactant. 16g of O2 can combine with 0.5 mole of carbon. Thus CO2 produced is 22g.
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    Answer

    Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0·375 molar aqueous solution. Molar mass of sodium acetate is 82·0245g mol–1.

     

    1000 mL of the solution contain sodium acetate = 0·375 mole,
         
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    Answer

    Calculate the mass percentage of different elements present in sodium sulphate (Na2SO4).

    Molecular mass of sodium =23
    Molecular mass of sulphur =32
    Molecular mass of oxygen =16
    Thus,

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    Answer
    Determine the empirical formula of an oxide of iron which has 69·9% iron and 30.1% dioxygen by mass.

    Element

    Percentage

    Atomic Mass

    Atomic ration

    Simplest ratio

    Simplest whole number ratio

    Iron (Fe)

    69.9

    55.85

    69.9/55.85=1.25

    1.25/1.25 =1

    2

    Oxygen (O)

    30.1

    16.00

    30.1/16.00=1.88

    1.88/1.25=1.5

    3

    Empirical formula Fe2O3
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    Answer
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