When a mixture of a metal carbonate, MCO3, and its oxide, MO, is heated it releases carbon dioxide g — Stoichiometry Chemistry Question
Problem Context
When a mixture of a metal carbonate, MCO3, and its oxide, MO, is heated it releases carbon dioxide gas and is converted completely to the metallic oxide, MO.
If a 0.6500 g sample of MCO3 and MO forms 0.1575 L of carbon dioxide gas at 25.0 °C and a barometric pressure of 700.0 mm Hg, determine the number of moles of CO2 formed.
Model Answer
n = PV/RT = (700.0 mm Hg/760 mm Hg/atm) (0.1575 L) / [(0.0821 L atm mol–1 K–1)(298 K)] = 0.00593 mol CO2
Alternate approach 0.6500 g – 0.3891 g = 0.2609 g CO2
0.2609 g CO2 / 44.0 g/mole = 0.00593 mole CO2
When the 0.3891 g of MO resulting from the process in 1.a. is titrated with 0.500 M HCl, 38.60 mL are required. Determine the number of moles of MO in 0.3891 g.
Model Answer
(0.500 mol/L) (0.03860 L) = 0.01930 mol HCl
(0.01930 mol HCl) (1 mol MO / 2 mol HCl) = 0.00965 mol MO
Determine the atomic mass of the metal M and give its symbol.
Model Answer
(0.3891 g MO) / 0.00965 mol MO = 40.32 g/mol MO
40.32 g/mol MO – 16.0 g/mol O = 24.32 g/mol M M = Mg2+
Determine the mole percentages of MCO3 and MO in the original sample.
Model Answer
Parts 1a-1c give three constraints on the number of moles of MgO and MgCO3 present. From 1a, the total mass of the sample was 0.6500 g, so (40.32 g/mol)×(n[MgO]) + (84.32 g/mol)×(n[MgCO3]) = 0.6500 g.
From 1b, the total number of moles of magnesium = 0.00965 mol = n(MgO) + n(MgCO3). From 1c, 0.00593 mol CO2 implies n(MgCO3) = 0.005928 mol. Thus n(MgO) = 0.00965 mol – 0.005928 mol = 0.00372 mol. Since one has three linear equations in two unknowns, one may use any two of the constraints to solve:
n(MgO) = 0.00372 mol, n(MgCO3) = 0.00593 mol
Mol % MgO = 0.00372 mol MgO / 0.00965 moles total = 0.385 × 100% = 38.5 mol% MgO
Mol % MgCO3 = 0.00593 mol MgCO3 / 0.00965 moles total = 0.615 × 100% = 61.5 mol% MgCO3
Sketch or describe the normal vibrations for the CO2 molecule.
Identify the vibrations that are infrared active and outline your reasoning.
Model Answer
The asymmetric stretch and bend are IR active because of change in dipole moment; the symmetric stretch is IR-inactive because it does not produce a change in dipole moment.