Respuesta :
Answer:
Empirical formula = Molecular formula = [tex]C_{21}H_{22}N_2O_{2}[/tex]
Explanation:
[tex]Moles =\frac {Given\ mass}{Molar\ mass}[/tex]
% of C = 75.42
Molar mass of C = 12.0107 g/mol
% moles of C = [tex]\frac{75.42}{12.0107}[/tex] = 6.2794
% of H = 6.63
Molar mass of H = 1.00784 g/mol
% moles of H = [tex]\frac{6.63}{\:1.00784}[/tex] = 6.57842
% of N = 8.38
Molar mass of N = 14.0067 g/mol
% moles of N = [tex]\frac{8.38}{14.0067}[/tex] = 0.59828
Given that the strychnine contains C, H, N and O. So,
% of O = 100% - % of C - % of H - % of N = 100 - 75.42 - 6.63 - 8.38 Â = 9.57 %
Molar mass of O = 15.999 g/mol
% moles of O = [tex]\frac{9.57}{15.999}[/tex] = 0.59816
Taking the simplest ratio for C, H, N and O as:
6.2794 : 6.57842 : 0.59828 Â : 0.59816
= 21 : 22 : 2 : 2
The empirical formula is = [tex]C_{21}H_{22}N_2O_{2}[/tex]
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus, Â
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 12*21 + 1*22 + 14*2 + 16*2 = 334 g/mol
Molar mass = 334 g/mol
So, Â
Molecular mass = n × Empirical mass
334 = n × 334
⇒ n = 1
The molecular formula = [tex]C_{21}H_{22}N_2O_{2}[/tex]
