Question

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?

Answer 1

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