Question

A 1.268 g sample of a metal carbonate (MCO3) was treated with 100.00mL of 0.1083 M sulfuric acid (H2SO4), yielding CO2 gas and an aqueous solution of the metal sulfate (MSO4). The solution was boiled to remove all the dissolved CO2 and was then titrated with 0.1241 M NaOH. A 71.02 mL volume of NaOH was required to neutralized the excess H2SO4.
What is the identity of the metal M?

Answer 1

The answer is M = Ba (Barium)

MCO3 + H2SO4 => MSO4 + CO2 + H2O

2 NaOH + H2SO4 => Na2SO4 + 2 H2O

Moles of NaOH = volume x concentration

= 71.02/1000 x 0.1241 = 0.0088136 mol

Moles of excess H2SO4 = 1/2 x moles of NaOH

= 1/2 x 0.0088136 = 0.0044068 mol

Initial moles of H2SO4 = volume x concentration

= 100/1000 x 0.1083 = 0.01083 mol

Moles of reacted H2SO4 = initial moles of H2SO4 - excess moles of H2SO4

= 0.01083 - 0.0044068 = 0.0064232 mol

Moles of MCO3 = moles of reacted H2SO4 = 0.0064232 mol

Molar mass of MCO3 = mass/moles

= 1.268/0.0064232 = 197.4 g/mol

Molar mass of M = moles mass of MCO3 - molar mass of CO3

= 197.4 - 60.0 = 137.4 g/mol

From the periodic table, the element with molar mass of 137.4 g/mol is Ba (Barium)

Thus M = Ba (Barium)