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To determine the molecular formula of the compound with a molar mass of [tex]\( 96.69 \, \text{g/mol} \)[/tex], we need to compare this molar mass with the calculated molar masses of the possible molecular formulas provided.
First, let's identify the molar masses of the elements involved:
- The atomic mass of Nickel (Ni) is [tex]\( 58.69 \, \text{g/mol} \)[/tex].
- The atomic mass of Fluorine (F) is [tex]\( 18.998 \, \text{g/mol} \)[/tex].
Now, let's calculate the molar masses of the possible molecular formulas provided:
1. For [tex]\( \text{NiF}_2 \)[/tex]:
[tex]\[ \text{Molar mass of} \ \text{NiF}_2 = \text{{molar mass of Ni}} + 2 \times \text{{molar mass of F}} = 58.69 \, \text{g/mol} + 2 \times 18.998 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \ \text{NiF}_2 = 58.69 \, \text{g/mol} + 37.996 \, \text{g/mol} = 96.686 \, \text{g/mol} \][/tex]
2. For [tex]\( \text{NiF} \)[/tex]:
[tex]\[ \text{Molar mass of} \ \text{NiF} = \text{{molar mass of Ni}} + \text{{molar mass of F}} = 58.69 \, \text{g/mol} + 18.998 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \ \text{NiF} = 77.688 \, \text{g/mol} \][/tex]
3. For [tex]\( \text{Ni}_2\text{F}_4 \)[/tex]:
[tex]\[ \text{Molar mass of} \ \text{Ni}_2\text{F}_4 = 2 \times \text{{molar mass of Ni}} + 4 \times \text{{molar mass of F}} = 2 \times 58.69 \, \text{g/mol} + 4 \times 18.998 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \ \text{Ni}_2\text{F}_4 = 117.38 \, \text{g/mol} + 75.992 \, \text{g/mol} = 193.372 \, \text{g/mol} \][/tex]
4. For [tex]\( \text{Ni}_3\text{F}_6 \)[/tex]:
[tex]\[ \text{Molar mass of} \ \text{Ni}_3\text{F}_6 = 3 \times \text{{molar mass of Ni}} + 6 \times \text{{molar mass of F}} = 3 \times 58.69 \, \text{g/mol} + 6 \times 18.998 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \ \text{Ni}_3\text{F}_6 = 176.07 \, \text{g/mol} + 113.988 \, \text{g/mol} = 290.058 \, \text{g/mol} \][/tex]
Comparing the calculated molar masses to the given molar mass ([tex]\( 96.69 \, \text{g/mol} \)[/tex]):
- The molar mass of [tex]\( \text{NiF}_2 \)[/tex] is [tex]\( 96.686 \, \text{g/mol} \)[/tex], which is very close to [tex]\( 96.69 \, \text{g/mol} \)[/tex].
- The molar mass of [tex]\( \text{NiF} \)[/tex] is [tex]\( 77.688 \, \text{g/mol} \)[/tex], which is significantly different.
- The molar mass of [tex]\( \text{Ni}_2\text{F}_4 \)[/tex] is [tex]\( 193.372 \, \text{g/mol} \)[/tex], which is also significantly different.
- The molar mass of [tex]\( \text{Ni}_3\text{F}_6 \)[/tex] is [tex]\( 290.058 \, \text{g/mol} \)[/tex], which is very different as well.
Therefore, the molecular formula of the compound with a molar mass of [tex]\( 96.69 \, \text{g/mol} \)[/tex] is [tex]\( \text{NiF}_2 \)[/tex].
First, let's identify the molar masses of the elements involved:
- The atomic mass of Nickel (Ni) is [tex]\( 58.69 \, \text{g/mol} \)[/tex].
- The atomic mass of Fluorine (F) is [tex]\( 18.998 \, \text{g/mol} \)[/tex].
Now, let's calculate the molar masses of the possible molecular formulas provided:
1. For [tex]\( \text{NiF}_2 \)[/tex]:
[tex]\[ \text{Molar mass of} \ \text{NiF}_2 = \text{{molar mass of Ni}} + 2 \times \text{{molar mass of F}} = 58.69 \, \text{g/mol} + 2 \times 18.998 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \ \text{NiF}_2 = 58.69 \, \text{g/mol} + 37.996 \, \text{g/mol} = 96.686 \, \text{g/mol} \][/tex]
2. For [tex]\( \text{NiF} \)[/tex]:
[tex]\[ \text{Molar mass of} \ \text{NiF} = \text{{molar mass of Ni}} + \text{{molar mass of F}} = 58.69 \, \text{g/mol} + 18.998 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \ \text{NiF} = 77.688 \, \text{g/mol} \][/tex]
3. For [tex]\( \text{Ni}_2\text{F}_4 \)[/tex]:
[tex]\[ \text{Molar mass of} \ \text{Ni}_2\text{F}_4 = 2 \times \text{{molar mass of Ni}} + 4 \times \text{{molar mass of F}} = 2 \times 58.69 \, \text{g/mol} + 4 \times 18.998 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \ \text{Ni}_2\text{F}_4 = 117.38 \, \text{g/mol} + 75.992 \, \text{g/mol} = 193.372 \, \text{g/mol} \][/tex]
4. For [tex]\( \text{Ni}_3\text{F}_6 \)[/tex]:
[tex]\[ \text{Molar mass of} \ \text{Ni}_3\text{F}_6 = 3 \times \text{{molar mass of Ni}} + 6 \times \text{{molar mass of F}} = 3 \times 58.69 \, \text{g/mol} + 6 \times 18.998 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of} \ \text{Ni}_3\text{F}_6 = 176.07 \, \text{g/mol} + 113.988 \, \text{g/mol} = 290.058 \, \text{g/mol} \][/tex]
Comparing the calculated molar masses to the given molar mass ([tex]\( 96.69 \, \text{g/mol} \)[/tex]):
- The molar mass of [tex]\( \text{NiF}_2 \)[/tex] is [tex]\( 96.686 \, \text{g/mol} \)[/tex], which is very close to [tex]\( 96.69 \, \text{g/mol} \)[/tex].
- The molar mass of [tex]\( \text{NiF} \)[/tex] is [tex]\( 77.688 \, \text{g/mol} \)[/tex], which is significantly different.
- The molar mass of [tex]\( \text{Ni}_2\text{F}_4 \)[/tex] is [tex]\( 193.372 \, \text{g/mol} \)[/tex], which is also significantly different.
- The molar mass of [tex]\( \text{Ni}_3\text{F}_6 \)[/tex] is [tex]\( 290.058 \, \text{g/mol} \)[/tex], which is very different as well.
Therefore, the molecular formula of the compound with a molar mass of [tex]\( 96.69 \, \text{g/mol} \)[/tex] is [tex]\( \text{NiF}_2 \)[/tex].
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