Answer
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Hint: We need to know that the molecular formula specifies the numbers of atoms present in the molecule of a chemical substance. And the molecular formula is the same as that of the empirical formula. The molecular formula mainly contains the chemical symbols of the corresponding element and it has numeric subscripts to express the number of atoms present in a molecule. Therefore, the molecular formula will show the actual number of atoms present in a molecule.
Complete answer:
The molecular formula of the given compound is not equal to \[N{H_4}\]. Hence, option (A) is incorrect.
\[{N_2}{H_3}\]is not the empirical formula of the given compound. Hence, option (B) is incorrect.
The molecular formula of the given compound is not equal to \[{N_2}{H_6}\]. Hence, option (A) is incorrect.
According to the question, gaseous compounds of nitrogen and hydrogen contain \[12.5\% \] hydrogen by mass.
Hence, \[100g\] of compounds consist \[12.5g\] H and \[100 - 12.5 = 87.5g\] of nitrogen atoms.
The atomic mass of hydrogen and nitrogen is equal to \[1g/mol and 14g/mol\] respectively.
So, the number of moles of \[H = \dfrac{{Given Weight}}{{mol.wt}}\]
Substitute the values in above equation,
The number of moles of \[H = \dfrac{{12.5g}}{{1g/mol}} = 12.5mol\]
The number of moles of \[N = \dfrac{{87.5g}}{{14g/mol}} = 6.25mol\]
Hence, the mole ratio of nitrogen and hydrogen is, \[N:H = 6.25:12.5. = 1:2\]
Therefore, the empirical formula of a given compound is equal to \[N{H_2}\].
The molecular formula of a compound can be found by dividing molecular weight with empirical formula weight. Thus,
\[Molecular\ Formula = \left( {N{H_2}} \right) \times \dfrac{{Molecular\ Weight}}{{empirical\ formula\ weight}}\]
Substitute the values in above equation will get,
\[Molecular\ Formula = \left( {N{H_2}} \right) \times \dfrac{{32g/mol}}{{16g/mol}}\]
Therefore, the molecular formula of the given compound is \[{N_2}{H_4}\].
Hence, option (D) is correct.
Note:
We have to know that the molecular formula represents the number of each type of atom present in the molecule. But in the case of an empirical formula, it represents the simplest total number ratio of atoms present in the compound. The molecular formula can be found by using empirical formula, by dividing the molecular mass of the compound with its empirical formula mass.
Complete answer:
The molecular formula of the given compound is not equal to \[N{H_4}\]. Hence, option (A) is incorrect.
\[{N_2}{H_3}\]is not the empirical formula of the given compound. Hence, option (B) is incorrect.
The molecular formula of the given compound is not equal to \[{N_2}{H_6}\]. Hence, option (A) is incorrect.
According to the question, gaseous compounds of nitrogen and hydrogen contain \[12.5\% \] hydrogen by mass.
Hence, \[100g\] of compounds consist \[12.5g\] H and \[100 - 12.5 = 87.5g\] of nitrogen atoms.
The atomic mass of hydrogen and nitrogen is equal to \[1g/mol and 14g/mol\] respectively.
So, the number of moles of \[H = \dfrac{{Given Weight}}{{mol.wt}}\]
Substitute the values in above equation,
The number of moles of \[H = \dfrac{{12.5g}}{{1g/mol}} = 12.5mol\]
The number of moles of \[N = \dfrac{{87.5g}}{{14g/mol}} = 6.25mol\]
Hence, the mole ratio of nitrogen and hydrogen is, \[N:H = 6.25:12.5. = 1:2\]
Therefore, the empirical formula of a given compound is equal to \[N{H_2}\].
The molecular formula of a compound can be found by dividing molecular weight with empirical formula weight. Thus,
\[Molecular\ Formula = \left( {N{H_2}} \right) \times \dfrac{{Molecular\ Weight}}{{empirical\ formula\ weight}}\]
Substitute the values in above equation will get,
\[Molecular\ Formula = \left( {N{H_2}} \right) \times \dfrac{{32g/mol}}{{16g/mol}}\]
Therefore, the molecular formula of the given compound is \[{N_2}{H_4}\].
Hence, option (D) is correct.
Note:
We have to know that the molecular formula represents the number of each type of atom present in the molecule. But in the case of an empirical formula, it represents the simplest total number ratio of atoms present in the compound. The molecular formula can be found by using empirical formula, by dividing the molecular mass of the compound with its empirical formula mass.
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