A 2.800-g mineral sample was analyzed for its calcium carbonate, CaCO3, content by titrating it with hydrochloric acid (the reaction products are calcium chloride, carbon dioxide, and water). The sample required 28.48 mL of a 0.1505 M HCl solution to react completely. Calculate the percent by mass calcium carbonate in the mineral sample.
CaCO3 + 2HCl ==> H2O + CO2 + CaCl2
How many moles HCl were used? M x L = ??
How many moles CaCO3 is equivalent. Use the equation to convert moles HCl to moles CaCO3.
How many grams CaCO3 is that many moles? g = moles CaCO3 x molar mass CaCO3
Now what is the percent CaCO3? That will be (g CaCO3/mass sample)*100 = ??%
To calculate the percent by mass of calcium carbonate in the mineral sample, we need to determine the amount of calcium carbonate and the total mass of the sample.
First, we can find the moles of hydrochloric acid (HCl) used in the titration:
moles of HCl = volume of HCl solution (in L) × molarity of HCl
Given that the volume of HCl solution used is 28.48 mL and the molarity of the HCl solution is 0.1505 M:
volume of HCl solution (in L) = 28.48 mL ÷ 1000 mL/L = 0.02848 L
moles of HCl = 0.02848 L × 0.1505 M = 0.004289 mol
Next, we need to determine the moles of calcium carbonate (CaCO3) reacted with the HCl. The balanced chemical equation for the reaction tells us that 1 mole of CaCO3 reacts with 2 moles of HCl. Therefore, the moles of CaCO3 can be calculated as half the moles of HCl used:
moles of CaCO3 = 0.004289 mol ÷ 2 = 0.002145 mol
Now we can calculate the molar mass of CaCO3:
molar mass of CaCO3 = (atomic mass of Ca) + (3 × atomic mass of C) + (3 × atomic mass of O)
Looking up the atomic masses of the elements, we have:
molar mass of CaCO3 = 40.08 g/mol + (3 × 12.01 g/mol) + (3 × 16.00 g/mol) = 100.09 g/mol
Finally, we can calculate the mass percent of calcium carbonate in the mineral sample:
percent mass CaCO3 = (moles of CaCO3 × molar mass of CaCO3) / mass of the sample × 100%
Given that the mass of the sample is 2.800 g:
percent mass CaCO3 = (0.002145 mol × 100.09 g/mol) / 2.800 g × 100% ≈ 0.767%
Therefore, the percent by mass of calcium carbonate in the mineral sample is approximately 0.767%.