Page 11 - Focus SPM KSSM F4 2020 - Chemistry
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Chemistry Form 4  Chapter 3 The Mole Concept, Chemical Formula and Equation
          SPM    Highlights                                              SPM Tips

                                                        For molecular elements such as oxygen, make sure
         Which substance has a different mass with 1 mole of   you read the question carefully to understand what
         glucose, C H O ?                               is required in the question. Use RAM to calculate
                     6
                   12
                 6
         [Relative atomic mass: H,1; C, 12; O, 16]
         A  3 moles of propanol, C H OH                 the number of atoms of oxygen, O. Use RMM to
                                                        calculate the number of molecules of oxygen,
                              7
                             3
         B  4 moles of propane, C H 6                   O .
                            3
         C  6 moles of ethane, C H 6                      2
                           2
         D  10 moles of water, H O
                                                         5.  We  can  compare  the  number  of  particles  in
                           2
         Examiner’s tip                                    substances by just comparing the number of
         Students need to know the molar mass and the mass   moles of the substances. See Example 3.10.
         of each substance, including glucose.
            Substance    Mass(g) = number of moles ×
                                molar mass              EXAMPLE 3.10
          1 mole C H O  6(12) + 12(1) + 6(16) = 180      How many times the number of atoms in 7 g of
                 6  12  6
          3 moles       3 × [3(12) + 7(1) + 16 + 1] = 3 ×   nitrogen gas is larger than that in 7 g of iron?  Chapter
          C H OH        60 = 180
            3  7                                         [Relative atomic mass: N, 14; Fe, 56]
          4 moles C H   4 × [3(12) + 6(1)] = 4 × 42 = 168
                  3  6                                   Solution                                     3
          6 moles C H   6 × [2(12) + 6(1)] = 6 × 30 = 180
                  2  6                                   We only need to compare the number of moles
          10 moles H O  10 × [2(1) + 16] = 10 × 18 = 180  of both samples.
                   2
         Answer: B                                       The number of moles of nitrogen atoms
                                                         = mass ÷ molar mass
          4.  Substances  with  the  same  number  of  moles   = 7 ÷ 14
            have  equal  number  of  particles  even  though   = 0.5 mol
            the particles are different in size and mass.   The number of moles of iron atoms
                                                         = mass ÷ molar mass
                                                         = 7 ÷ 56
                                  28 g                   = 0.125 mol
                     12 g
                0.5 mole of magnesium  0.5 mole of iron  Number of moles of nitrogen atoms  =   0.5    = 4
                                                           Number of moles of iron atoms  0.125
         Figure 3.5  Both blocks of magnesium and iron have 0.5 x
                 6.022 × 10  atoms                       Therefore, 7 g of nitrogen gas has 4 times more
                         23
                 [Relative atomic mass: Mg, 24; Fe, 56]  of atoms than that in 7 g of iron.
        EXAMPLE  3.9
                                                       Number of moles and volume of gases
         What is the mass of gold that has the same      1.  The  number  of  moles  of  a  gas  can  also  be
         number of atoms as 4 g of oxygen?                 determined through its volume. This is not true
         [Relative atomic mass: O, 16; Au, 197]            of a solid or liquid substance.
         Solution                                        2.   All  gases  with  the  same  volume  under  the
         The number of oxygen atoms, O                     same pressure and temperature, have the same
         = mass ÷ molar mass                               number of particles.
         = 4 ÷ 16 = 0.25 mol                             3.   The volume of one mole of gas is known as
         0.25 mole of gold has the same number of          molar volume.
         atoms as 0.25 mole of oxygen. So, the mass of     •   Molar  volume  of  any  gas  is  22.4 dm
                                                                                                  3
         gold = number of moles × molar mass                   mol at STP or  24 dm  mol  at room
                                                                   -1
                                                                                    3
                                                                                         -1
                              = 0.25 × 197                     conditions.
                              = 49.25 g                    •   STP refers to the standard temperature of
                                                               0ºC and the pressure of 1 atm.

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 03 SPM CHEMISTRY F4.indd   39                                                               27/02/2020   11:23 AM
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