Answer :
17,6% + 82,6% = 100%
mass of N = 14g
mass of H = 1g
17,6% H = 17,6g H
82,4% N = 82,4g N
17,6g : 1g = 17,6 moles of H
82,4g : 14g = 5,89 moles of N
N : H = 5,89 : 17,6 ||:5,89
N : H = 1 : 2,99≈3
the empirical formula of the compound = NH₃ (ammonia)
mass of N = 14g
mass of H = 1g
17,6% H = 17,6g H
82,4% N = 82,4g N
17,6g : 1g = 17,6 moles of H
82,4g : 14g = 5,89 moles of N
N : H = 5,89 : 17,6 ||:5,89
N : H = 1 : 2,99≈3
the empirical formula of the compound = NH₃ (ammonia)
Answer: The empirical formula for the given compound is [tex]NH_3[/tex]
Explanation:
We are given:
Percentage of H = 17.6 %
Percentage of N = 82.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of H = 17.6 g
Mass of N = 82.4 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{17.6g}{1g/mole}=17.6moles[/tex]
Moles of Nitrogen = [tex]\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{82.4g}{14g/mole}=5.88moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 5.88 moles.
For Hydrogen = [tex]\frac{17.6}{5.88}=2.99\approx 3[/tex]
For Nitrogen = [tex]\frac{5.88}{5.88}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of H : N = 3 : 1
Hence, the empirical formula for the given compound is [tex]NH_3[/tex]