Question

An organic compound is burnt in an O2 flux. The released carbon dioxide is collected in a solution of barium hydroxide. Calculate the percentage of carbon in this compound if 0.1698 g produced 0.5019 g of barium carbonate.

Answer 1

Solution:

Lets show the reaction

CO2(g) + Ba(OH)2(aq) --- > BaCO3(s)+ H2 (g)

Mole ratio between CO2 : BaCO3 is 1 : 1

Calculation of moles of CO2

n CO2 = n BaCO3 x 1 mol CO2 / 1 mol BaCO3

= ( 0.5019 g BaCO3 / 197.34 g per mol BaCO3 ) x 1 mol CO2 / 1 mol BaCO3

= 0.002543 mol BaCO3 x 1 mol CO2 / 1 mol BaCO3

= 0.002543 mol CO2

Mol ratio CO2 to C is 1 ; 1

So moles of C = 0.002543 mol C

Mass of C = 0.002543 mol C x molar mass of C

= 0.002543 mol C x 12 g /mol

= 0.03052 g C

Mass percent of C in given organic compound

= 0.03052 g C / 0.1698 g x 100

= 18.0 %

So the mass of percent of C in organic compound = 18.0 %