Respuesta :
Answer : The empirical of the compound is, [tex]C_{11}H_{12}N_2[/tex]
Explanation :
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 76.71 g
Mass of H = 7.02 g
Mass of N = 16.27 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of N = 14 g/mole
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{76.71g}{12g/mole}=6.39moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.02g}{1g/mole}=7.02moles[/tex]
Moles of N = [tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.27g}{14g/mole}=1.16moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{6.39}{1.16}=5.5[/tex]
For H = [tex]\frac{7.02}{1.16}=6.0\approx 6[/tex]
For N = [tex]\frac{1.16}{1.16}=1[/tex]
The ratio of C : H : N = 5.5 : 6 : 1
To make in a whole number we are multiplying the ratio by 2, we get:
The ratio of C : H : N = 11 : 12 : 2
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_{11}H_{12}N_2[/tex]
Therefore, the empirical of the compound is, [tex]C_{11}H_{12}N_2[/tex]
