Question

a sample that contains only MgCO₃ and BaCO₃ weighs 0.9337 g. when it is dissolved in excess acid, 0.3485 g of carbon dioxide is liberated. what percentage of MgCO₃ did the sample contain? assume all the carbon originally present is converted to carbon dioxide.

Answer 1

The percentage of MgCO₃ in the sample can be found by using the mole concept to determine moles of CO₂ produced, and corresponding it to the total moles of carbonate. Assuming equal moles of MgCO₃ and BaCO₃, a system of equations is set up and solved to find the mass and then the percentage of MgCO₃.

Step-by-step explanation:

To find the percentage of MgCO₃ in the sample, we first use the mole concept. The molar mass of CO₂ is 44.01 g/mol, so 0.3485 g of CO₂ is equivalent to 0.3485 g / 44.01 g/mol = 0.007920 moles of CO₂. Assuming all the carbon in the sample was converted to carbon dioxide, these moles are also the total moles of carbonate ions in the original sample.

As MgCO₃ and BaCO₃ both contain one mole of CO₃²⁻ for each mole of the compound, the moles of CO₃²⁻ calculated from the CO₂ produced is equal to the sum of the moles of MgCO₃ and BaCO₃ in the sample. This gives us a total molar amount of MgCO₃ + BaCO₃ equal to 0.007920 moles.

Now let's assign x as the mass of MgCO₃ in the sample and (0.9337 - x) as the mass of BaCO₃. The molar mass of MgCO₃ is 84.31 g/mol and that of BaCO₃ is 197.34 g/mol. Using the mole concept again, we can write the equation: x / 84.31 g/mol + (0.9337 - x) / 197.34 g/mol = 0.007920 moles.

Solving for x gives us the mass of MgCO₃ in the original sample, which then allows us to calculate the requested percentage concentration by dividing the mass of MgCO₃ by the total sample mass and multiplying by 100.