Question

strychnine a deadly poison, has a formula mass of 334 g mol- and a percent composition of 75.42% C, 6.63% H, 8.38% N and the balance oxygen. Calculate the empirical and molecular formulas of strychinie

Answer 1

Answer:

Empirical formula = Molecular formula = $C_{21}H_{22}N_2O_{2}$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 75.42

Molar mass of C = 12.0107 g/mol

% moles of C = $\frac{75.42}{12.0107}$ = 6.2794

% of H = 6.63

Molar mass of H = 1.00784 g/mol

% moles of H = $\frac{6.63}{\:1.00784}$ = 6.57842

% of N = 8.38

Molar mass of N = 14.0067 g/mol

% moles of N = $\frac{8.38}{14.0067}$ = 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 = $\frac{9.57}{15.999}$ = 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 = $C_{21}H_{22}N_2O_{2}$

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 = $C_{21}H_{22}N_2O_{2}$