Question

12.5 g of zinc carbonate sample on heating decomposes to give carbon dioxide and 6.0 g of zinc oxide. Calculate the percentage purity in the zinc carbonate sample. (Given atomic mass of Zn = 65 g)

Answer 1

Given:

Mass of impure zinc carbonate sample = 12.5 g

Mass of zinc oxide obtained = 6.0 g

Atomic mass: Zn = 65, C = 12, O = 16

Reaction:

\[\ce{ZnCO3 −> ZnO + CO2}\]

1. Molecular masses:

Molar mass of ZnCO₃ ​= 65 + 12 + 48 = 125 g/mol

Molar mass of ZnO = 65 + 16 = 81 g/mol

According to the equation:

\[\ce{125 g ZnCO3​→81 g ZnO}\]

Therefore, ZnCO₃ required to produce 6.0 g ZnO:

Mass of pure ZnCO₃​ = `125/81 xx 6` = 9.259 g

2. Percentage purity:

Percentage purity = `"Mass of pure substance"/"Mass of sample" xx 100`

= `9.259/12.5 xx 100`

= 74.07%​

The percentage purity of ZnCO₃ sample is 74.1%​.