If 0.158 g of a white, unknown solid carbonate of a group 2A metal (M) is heated and the resulting CO2 is transferred to a 285 ml sealed flask and allowed to cool to 25 degrees Celsius, the pressure in the flask is 69.8 mmHg. What is the identity of the carbonate?
using the ideal gas law equation we can find the number of moles of CO₂ formed PV = nRT where P - pressure - 69.8 mmHg x 133 Pa/mmHg = 9 283 Pa V - volume - 285 x 10⁻⁶ m³ n - number of moles R - universal gas constant - 8.314 Jmol⁻¹K⁻¹ T - temperature in Kelvin - 25 °C + 273 = 298 K substituting these values in the equation 9283 Pa x 285 x 10⁻⁶ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 298 K n = 1.067 x 10⁻³ mol
decomposition of metal carbonate is as follows MCO₃ ---> MO + CO₂ stoichiometry of MCO₃ to CO₂ is 1:1 therefore number of moles of MCO₃ heated = number of CO₂ moles formed number of MCO₃ moles = 1.067 x 10⁻³ mol molar mass = mass / number of moles molar mass = 0.158 g / 1.067 x 10⁻³ mol = 148 g/mol since carbonate molar mass is known - (molar mass of C x 1 C atom) + (molar mass of O x 3 O atoms) 12 + 16x 3 = 12 + 48 = 60 then mass of metal M - 148 - 60 = 88 group II metal with molar mass of 88 is Ra - Radium
