Question

Gravimetric Analysis of Mn in Ore.

A manganese-containing ore is digested and chemically treated to isolate manganese as MnO₂(s). Heating the MnO₂ at 1000 °C in air converts it entirely to Mn₃O₄(s).
Suppose a 1.48-g ore sample yields 0.135 g of Mn₃O₄(s) by this process.

Calculate the percent Mn in the ore sample.

Calculate the percent MnO₂ in the ore sample.

Answer 1

we have reaction:

3MnO₂ + O₂    -------> Mn₃O₄

0.135g

molar mass of MnO₂ =   86.936g/mol it has 54.94g Mn in it

1g MnO2 has 54.94g/86.936 = 0.632g Mn

molar mass of Mn₃O₄ = 228.812 g/mol

from the above reaction

3 mol MnO₂ gives 1 mol Mn₃O₄

moles of Mn₃O₄ = 0.135g/228.812 g/mol = 0.00059 mol

moles of MnO₂ = 3 x moles of Mn₃O₄ = 3 x 0.00059mol = 0.00177 mol

mass of MnO₂ = moles x molar mass = 0.00177 mol x 86.936g/mol = 0.1539 g

0.1539g sample will have = 0.632g Mn x 0.1539g = 0.097258g

% Mn = 0.097258g/1.48g x 100 = 6.57%

total mass of sample = 1.48g % MnO2 = mass MnO₂/Mass of ore x 100 = 10.4%

Ans = 10.4% MnO₂