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