Question

Dolomite, [CaMg(CO₃)₂] is found in a soil sample. A geochemist titrates 24.65 g of soil with 57.85 mL of 0.3315M HCl. What is the mass % of dolomite in the soil? ( Write answer to two decimal places, ex. 15.77)

Following is the acid/base reaction which occurs during the titration: CO₃^-2_(aq) + HCl_(aq) --> HCO₃^-_(aq) + Cl^-_(aq)

Answer 1

Solution :-

Balanced reaction equation is as follows

CO₃^-2_(aq) + HCl_(aq) --> HCO₃^-_(aq) + Cl^-_(aq)

now lets calculate the moles of the HCl reacted

moles of HCl = molarity * volume in liter

                        = 0.3315 mol per L * 0.05785 L

                       = 0.019177 mol HCl

now using the mole ratio lets calculate the moles of the CO3^2-

mole ratio is 1 : 1

therefore moles of the CO3^2- are same as moles of HCl

so moles of CO3^2- = 0.019177 mol

now lets calculate the moles of the dolomite using the mole ratio of the CO3^2- and [CaMg(CO₃)₂]

1 mol [CaMg(CO₃)₂] = 2 mol CO3^2-

therefore

0.019177 mol CO3^2- * 1 mol [CaMg(CO₃)₂] / 2 mol CO3^2-   = 0.009589 mol [CaMg(CO₃)₂]

now lets convert the moles of the [CaMg(CO₃)₂] to its mass

mass = moles * molar mass

molar mass of [CaMg(CO₃)₂] = 184.4 g per mol

mass of [CaMg(CO₃)₂] = 0.009589 mol * 184.4 g per mol

                                         = 1.77 g [CaMg(CO₃)₂]

now lets calculate the percent of the [CaMg(CO₃)₂] in the soil

% [CaMg(CO₃)₂] = ( mass of [CaMg(CO₃)₂] / mass of soil)*100%

                               = (1.77 g / 24.65 g)*100%

                              = 7.18 % [CaMg(CO₃)₂]

Therefore the percent of the [CaMg(CO₃)₂] in the soil is 7.18%