a compound consists of 17.6% of hydrogen and 82.4% nitrogen. Determine the empirical formula of the compound

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 gMass of N = 82.4 gTo 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 : 1Hence, the empirical formula for the given compound is [tex]NH_3[/tex]

17,6% + 82,6% = 100%mass of N = 14gmass of H = 1g17,6% H = 17,6g H82,4% N = 82,4g N17,6g : 1g = 17,6  moles of H82,4g : 14g = 5,89 moles of NN  :  H  =  5,89  :  17,6  ||:5,89N  :  H  =  1      :   2,99≈3the empirical formula of the compound = NH₃  (ammonia)