Question

An ammonium alkanoic on analysis give the following result percent of carbon = 60.5 and Hydrogen =6.65 if 0.309 of the salt products 0.0313g of nitrogen determine the empirical formula of the salt

Answer 1

Assuming percentag es into grammes;

$C=60.5g$

$H=6.65g$

$N=32.85g$

Converting to moles;

$60.5gC\times$ $1molC\over 12.0g/mol C$ $=5.042moles C$

$6.65gH\times$ $1mol\over 1g/molH$ $=6.65molH$

$32.85gN\times$ $1molN\over 14g/mol$ $=2.35mol N$

Dividing by the smallest number of results;

$5.04\over 2.35$ $=2$ ,$6.65\over 2.35$ $=2.85=3$ ,$2.35\over2.35$ $=1$

From the formula; the compound is;

$C_2H_3N_1$

Molar mass of the compound;

$41\times 0.0313\over 0.309$ $=4.15=4$

$(C_2H_3N_1)_4=C_8H_12N_4$