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Answer:
177.277amu
Explanation:
the total occuring isotopes for Hafnium is =6.
First isotope had an atomic weight of 173.940amu
Second isotope =175.941amu
Third isotope =176.943amu
Fourth isotope=177.944amu
Fifth isotope. =178.946amu
sixth isotope .179.947amu
<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>
Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu
Average atomic weight= 1063.661amu /6 = 177.2768333amu
= 177.277amu to 3 decimal places.
Answer:
Explanation:
To determine the molecular formula of the compound, the empirical formula must be determined first. To determine the empirical formula, the percentage of each constituent is divided by its molar mass. This is shown below
Carbon = 60/12 = 5
Oxygen = 32/16 = 2
Hydrogen = 8/1 = 8
The next step is to divide each ratio by the smallest value. The smallest value is 2. It becomes
Carbon = 5/2 = 2.5
It is approximated to 3
Oxygen = 2/2 = 1
Hydrogen = 8/2 = 4
Therefore, the empirical formula is
C3H4O
From the given relative molecular mass of the compound, the molecular formula can be determined
