Page 147 - Chemistry ICSE Class X
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Mole Concept and Stoichiometry                                                                       133
          So,      Molecular mass of K Cr O  = 78 u + 104 u + 112 u = 294 u
                                     2
                                        2
                                          7
          Thus, 294 units of K Cr O  contain 112 units of oxygen. Therefore,
                           2
                                7
                              2
                                          Mass of oxygen         112 u
          Percentage of oxygen in K Cr O  =               × 100 =      × 100 = 38.1
                                 2
                                    2
                                      7
                                         Mass of K Cr O 7        294 u
                                                  2
                                                     2
          Thus, K Cr O  contains 38.1% oxygen.
                      7
                   2
                 2
                   (c)  Al (SO )    Al S O    {  2 atoms of Al  + 3 atoms of S  + 12 atoms of O
                           4 3
                                      12
                        2
                                  2 3
                                              2 × 27 u      3 × 32 u      12 × 16 u
                                               54 u          96 u         192 u
          So,      Molecular mass of Al (SO )  = (54 + 96 + 192) u = 342 u
                                      2
                                          4 3
          Thus,    342 u of Al (SO )  contain 192 u of oxygen (O). Therefore,
                                 4 3
                             2
                                          192 u
          Percentage of oxygen in Al (SO )  =    × 100 = 56.1
                                     4 3
                                 2
                                          342 u
          Thus, Al (SO )  contains 56.1% of oxygen.
                 2
                     4 3
          EXAMPLE 5.19.  How many grams of Cr are there in 85 g of Cr S ? Atomic
                                                                       2 3
          masses are: Cr = 52 u and S = 32 u.
          Solution:   Cr S     {   2 atoms of Cr   +    3 atoms of S
                         2 3
                                      2 × 52 u            3 × 32 u
                                       104 u                96 u
                   Molecular mass of Cr S   =  (104 + 96) u = 200 u
                                      2 3
          So,        200 u of Cr S  contain  =  104 u of Cr
                               2 3
          or         200 g of Cr S  contain  =  104 g of Cr
                               2 3
          Thus,
                85 g of Cr S  would contain  =   104 g   × 85 g = 44.2 g of Cr
                        2 3
                                            200 g
          Determination of Empirical and Molecular Formula
          What is meant by empirical formula
          'ORKTKECN HQTOWNC QH C UWDUVCPEG OC[ DG FGſPGF CU HQNNQYU
              “Empirical formula (or stoichiometric formula) of a substance is the
          simplest formula which gives the lowest whole-number ratio between
          the number of atoms of different elements present in the substance.”
          What is meant by molecular formula
          /QNGEWNCT HQTOWNC QH C UWDUVCPEG OC[ DG FGſPGF CU HQNNQYU
              “A shorthand notion for the molecule of a substance in terms of
          symbols and number of atoms of each element present in it is called its
          molecular formula.”
              Molecular and empirical formulae of any compound can be determined
          from the given composition of that compound. This is illustrated below.
          How to determine the formula of an unknown compound
          The chemical formula of a compound can be determined from the
          composition of the compound. The composition of a compound is commonly
          expressed in terms of percentage of each element present in it.
              To determine the formula of an unknown compound, follow the
          following steps:
          Step 1.  Determine the percentage of each element present in the compound
                 from the mass of each element present in a certain known mass of the
                 compound.
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