Respuesta :
Answer: The empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]
Explanation:
Let the mass of the compound be 100 g
Given values:
% of C = 59.0%
% of H = 7.15%
% of O = 26.20%
% of N = 7.65%
Mass of C = 59.0 g
Mass of H = 7.15 g
Mass of O = 26.20 g
Mass of N = 7.65 g
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
To formulate the empirical formula, we need to follow some steps:
- Step 1: Converting the given masses into moles.
Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Molar mass of N = 14 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of C}=\frac{59.0g}{12g/mol}=4.917 mol[/tex]
[tex]\text{Moles of H}=\frac{7.15g}{1g/mol}=7.15 mol[/tex]
[tex]\text{Moles of O}=\frac{26.20g}{16g/mol}=1.6375 mol[/tex]
[tex]\text{Moles of N}=\frac{7.65g}{14g/mol}=0.546 mol[/tex]
- Step 2: Calculating the mole ratio of the given elements.
Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.546 moles
[tex]\text{Mole fraction of C}=\frac{4.917}{0.546 }=9[/tex]
[tex]\text{Mole fraction of H}=\frac{7.15}{0.546 }=13[/tex]
[tex]\text{Mole fraction of O}=\frac{1.6375}{0.546 }=2.99\approx 3[/tex]
[tex]\text{Mole fraction of N}=\frac{0.546}{0.546 }=1[/tex]
- Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O : N = 9 : 13 : 3 : 1
The empirical formula of the compound becomes [tex]C_9H_{13}O_3N_1=C_9H_{13}O_3N[/tex]
To calculate the molecular formula, the number of atoms of the empirical formula is multiplied by a factor known as valency that is represented by the symbol, 'n'.
[tex]n =\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex] .....(2)
We are given:
Mass of molecular formula = 183 g/mol
Mass of empirical formula = 183 g/mol
Putting values in equation 2, we get:
[tex]n=\frac{183g/mol}{183g/mol}=1[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{1\times 9}H_{1\times 13}O_{1\times 3}N_{1\times 1}=C_9H_{13}O_3N[/tex]
Hence, the empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]
