Question

A mixture of sodium carbonate and other inert compounds was heated to a high temperature and the sodium carbonate decomposed to give sodium oxide and carbon dioxide.

Na2CO3(s) reaction arrow identifying the presence of heat Na2O(s) + CO2(g)

A 3.0805-g sample of this mixture resulted in 2.4830 g of material after being heated at a high temperature. What was the mass percent (% w/w) of sodium carbonate in the original sample?

Answer 1

Given,

Mass of the sample before heating= 3.0805 g.

Mass of the sample after heating= 2.4830 g.

Na2CO3(s) Na2O(s) + CO2(g).

Loss in mass after heating= 3.0805- 2.4830

= 0.5975 g.

This loss in mass is due to evolution of Carbon dioxide gas.

According to above equation:.

1 mole of CO2 is produced by 1 mole of Na2CO3.

0.5975/44 mole of CO2 is produced by 0.0135 mole of Na2CO3.

Now, molar mass of Na2CO3= 106 g/mol.

Mass of Na2CO3= 0.0135× 106 g

= 1.431 g.

Also, mass % of Na2CO3= mass of Na2CO3/ total mass of sample × 100

= 1.431/3.0805 × 100

= 46.45 %