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 = 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 : 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,
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,
Molecular mass = n × Empirical mass
334 = n × 334
⇒ n = 1
The molecular formula = [tex]C_{21}H_{22}N_2O_{2}[/tex]
