To find the mass percent of dolomite [CaMg(CO3)2] in a soil sample, a geochemists titrates 12.86 g of soil with 33.56mL of 0.2516 M HCl. What is the mass percent of dolomite in the soil?PLEASE explain your answers so I can follow you! Thanks to all. All help is greatly appreciated.
There are essentially two reactions going on here Calcium carbonate with HCl and Magnesium Carbonate with HCl. Luckily both react in with the same stoichiometry So first let's see what the reactions are CaCO3 + 2HCl - CaCl2 + H2O + CO2 MgCO3 + 2HCl - MgCl2 + H2O + CO2 So for every 2 moles of HCl one mole of the carbonates is neutralised. Let's work out the number of moles of HCl use use the formula M n/V (concentration M, n number of moles, V volume in litres) 0.2516 ?/(33.56/1000) 0.2516 ?/0.03356 0.2516 * 0.03356 ? ? 0.0084 moles acid were used So looking back at the 2:1 ratio of moles you know now that this 0.0084 moles HCl neutralised 0.0042 moles of Dolomite (0.0084/2) Now work out the molecular mass of Dolomite: Ca 40.07 Mg 24.30 C 12 x 2 O 16 x 6 So molecular weight of dolomite is 184.37 g/mole You have 0.0042 moles so 184.37 * 0.0042 0.7743 g this is the amount in your sample So % wise you have 0.7743/12.86 x 100 6.02%
