Question

4)

LUI Uue Udte LOJ LULU 4) Analysis of a sample of nitroglycerine yielded 0.7061 g of nitrogen and 2.4210 g of ox Solution: nitroglycerine yielded 0.6053 g of carbon, 0.0841 g of hydrogen, 1. The empirical formula of nitroglycerine is ...

Answer 1

Mass of carbon yielded from the sample w_C = 0.6053 g

Atomic weight of carbon, M_C = 12.0 g/mol

So, number of moles of carbon atoms in 0.6053 g carbon is, n_C = w_C / M_C = 0.6053 g / 12.0 g/mol = 0.0504 mol

Mass of hydrogen yielded from the sample w_H = 0.0841 g

Atomic weight of hydrogen, M_H = 1.0 g/mol

So, number of moles of hydrogen atoms in 0.0841 g hydrogen is, n_H = w_H / M_H = 0.0841 g / 1 g/mol = 0.0841 mol

Mass of nitrogen yielded from the sample w_N = 0.7061 g

Atomic weight of nitrogen, M_N = 14.0 g/mol

So, number of moles of nitrogen atoms in 0.7061 g nitrogen is, n_N = w_N / M_N = 0.7061 g / 14.0 g/mol = 0.0504 mol.

Mass of oxygen yielded from the sample w_O = 2.4210 g

Atomic weight of oxygen, M_O = 16.0 g/mol

So, number of moles of oxygen atoms in 2.4210 g oxygen is, n_O = w_O / M_O = 2.4210 g / 16.0 g/mol = 0.1513 mol

Ratio of number of moles of carbon, hydrogen, nitrogen and oxygen present in the sample is

n_c : n_H : n_N : n_O = 0.0504 : 0.0841 : 0.0504 : 0.1513 = 1 : 1.7 : 1 : 3 = 1 : 2 : 1 : 3

So, the empirical formula of nitroglycerine is CH₂NO_{3 .}