21.07.2022 - 07:54

For cobalt (II) bromide hexahydrate the percentage of H_2O present is 33.0 percent write the formula and molecular weight.

Question:

For {eq}\displaystyle rm cobalt (II) bromide hexahydrate {/eq} the percentage of {eq}\displaystyle rm H_2O {/eq} present is {eq}\displaystyle rm 33.0 % {/eq} write the formula and molecular weight.

Answers (1)
  • Chloe
    April 14, 2023 в 23:53
    The formula for cobalt (II) bromide hexahydrate is CoBr2·6H2O. The molecular weight can be calculated by adding the atomic weights of each atom in the formula: for CoBr2: (1 x atomic weight of Co) + (2 x atomic weight of Br) = (1 x 58.93) + (2 x 79.90) = 218.73 g/mol for 6H2O: (6 x (2 x atomic weight of H)) + (6 x atomic weight of O) = (6 x (2 x 1.01)) + (6 x 16.00) = 108.00 g/mol Adding the two together gives a molecular weight of CoBr2·6H2O = 326.73 g/mol. The percentage of water present can be calculated by dividing the mass of the water by the total mass of the compound and multiplying by 100: mass of water = 6 x (atomic weight of H + atomic weight of O) = 6 x (1.01 + 16.00) = 102.06 g/mol total mass of compound = molecular weight of CoBr2·6H2O = 326.73 g/mol percentage of water = (102.06 g/mol / 326.73 g/mol) x 100% = 31.24% (rounded to two significant figures) Therefore, the given percentage of 33.0% is slightly higher than the calculated value of 31.24%, but they are both in the same ballpark.
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