Problem 90 - Chemistry: The Central Science

Question

A \(1.248-g\) sample of limestone rock is pulverized and then treated with \(30.00 \mathrm{~mL}\) of \(1.035 \mathrm{M} \mathrm{HCl}\) solution. The excess acid then requires \(11.56 \mathrm{~mL}\) of \(1.010 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percent by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

Step-by-Step Solution

Answer

The percent by mass of calcium carbonate in the limestone rock can be calculated by first finding the moles of HCl that reacted with CaCO3, then the moles of CaCO3 that reacted, and the mass of CaCO3. Finally, divide the mass of CaCO3 by the initial mass of the limestone rock and multiply by 100. Following these steps, we get: Percent by mass of CaCO3 = \(\dfrac{\mathrm{Mass\,of\,CaCO_{3}}}{1.248\,\mathrm{g}} \times 100\)

Step 1: Calculate the moles of excess HCl

Using the second chemical reaction, find the moles of excess \(\mathrm{HCl}\) based on the volume of \(\mathrm{NaOH}\) and its molarity. Moles of excess \(\mathrm{HCl} = \mathrm{Volume\,of\, NaOH\,(L)} \times \mathrm{M\,of\,NaOH} \) Moles of excess \(\mathrm{HCl} = 0.01156\,\mathrm{L} \times 1.010\,\mathrm{M} \)

Step 2: Calculate the moles of HCl that reacted with CaCO3

Now that we have the moles of excess \(\mathrm{HCl}\), we can find the moles of \(\mathrm{HCl}\) that reacted with \(\mathrm{CaCO_{3}}\). To do this, subtract the moles of excess \(\mathrm{HCl}\) from the total moles of \(\mathrm{HCl}\) initially present in the solution. Total moles of \(\mathrm{HCl} = \mathrm{Volume\,of\, HCl\,(L)} \times \mathrm{M\,of\,HCl} \) Total moles of \(\mathrm{HCl} = 0.03000\,\mathrm{L} \times 1.035\,\mathrm{M} \) Moles of \(\mathrm{HCl}\) that reacted with \(\mathrm{CaCO_{3}} = (\text{Total moles of}\, \mathrm{HCl}) - (\text{Moles of excess}\, \mathrm{HCl})\)

Step 3: Calculate the moles of CaCO3 that reacted with HCl

Using the balanced chemical equation for the reaction between \(\mathrm{CaCO_{3}}\) and \(\mathrm{HCl}\), we can find the moles of \(\mathrm{CaCO_{3}}\) that reacted with \(\mathrm{HCl}\). From the equation, there is a 1:2 mole ratio between \(\mathrm{CaCO_{3}}\) and \(\mathrm{HCl}\). Moles of \(\mathrm{CaCO_{3}} = \text{Moles of}\, ＼mathrm{HCl\,that\,reacted\,with\,CaCO_{3}} / 2\)

Step 4: Calculate the mass of CaCO3 that reacted with HCl

Now that we have found the moles of \(\mathrm{CaCO_{3}}\) that reacted with \(\mathrm{HCl}\), we can calculate the mass of \(\mathrm{CaCO_{3}}\) by multiplying the moles with the molar mass of \(\mathrm{CaCO_{3}}\). Mass of \(\mathrm{CaCO_{3}} = \mathrm{Moles\,of\,CaCO_{3}} \times \mathrm{Molar\,mass\,of\,CaCO_{3}} \) Molar mass of \(\mathrm{CaCO_{3}} = 40.08\,\mathrm{g/mol\,Ca} + 12.01\,\mathrm{g/mol\,C} + 3 \times 16.00\,\mathrm{g/mol\,O} = 100.09\,\mathrm{g/mol}\)

Step 5: Calculate the percent by mass of CaCO3 in the limestone

To find the percent by mass of \(\mathrm{CaCO_{3}}\) in the limestone rock, divide the mass of \(\mathrm{CaCO_{3}}\) found in step 4 by the initial mass of the limestone rock and multiply by 100. Percent by mass of \(\mathrm{CaCO_{3}} = \dfrac{\mathrm{Mass\,of\,CaCO_{3}}}{\mathrm{Mass\,of\,limestone\,rock}} \times 100\) Percent by mass of \(\mathrm{CaCO_{3}} = \dfrac{\mathrm{Mass\,of\,CaCO_{3}}}{1.248\,\mathrm{g}} \times 100\) By following these steps, you will be able to calculate the percent by mass of calcium carbonate in the limestone rock.

Key Concepts

Stoichiometry

Stoichiometry is the mathematical relationship between reactants and products in a chemical reaction. It's a crucial concept for understanding how to translate the amounts of one substance into the corresponding amounts of another substance involved in a chemical equation.
For instance, in the case of calcium carbonate reacting with hydrochloric acid (\(\mathrm{CaCO}_3 + 2\mathrm{HCl} \rightarrow \mathrm{CaCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\)), stoichiometry tells us that 1 mole of \(\mathrm{CaCO}_3\) reacts with 2 moles of \(\mathrm{HCl}\). To solve stoichiometric problems, we often start by balancing the chemical equation and then use mole ratios to calculate the amounts of reactants or products.

Molarity and Titration

Molarity (\(M\)) is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution. In titrations, it's a key factor that allows us to determine the amount of substance in a given volume of liquid.
In the example problem, titration helps to find the excess amount of \(\mathrm{HCl}\) that did not react with the limestone. This is achieved by neutralizing the excess acid with a known concentration and volume of sodium hydroxide (\(\mathrm{NaOH}\)). Knowing the molarity and volume of the \(\mathrm{NaOH}\) solution used allows us to calculate the moles of \(\mathrm{HCl}\) left over, thereby determining the amount that reacted with the limestone.

Molar Mass Calculation

The molar mass is the weight of one mole of a substance, usually expressed in grams per mole (\(g/mol\)). It's the sum of the atomic masses of all the atoms in a molecule. For calcium carbonate (\(\mathrm{CaCO}_3\)), the molar mass is calculated by adding the atomic masses of calcium, carbon, and oxygen.
Knowing the molar mass is essential for converting between moles and grams. In the context of our textbook exercise, we need the molar mass of calcium carbonate to determine the mass that reacted with hydrochloric acid. This step is critical for figuring out the percentage of \(\mathrm{CaCO}_3\) in the limestone.

Chemical Reaction Quantification

Chemical reaction quantification involves calculating the amount of reactants or products in a chemical reaction. It uses stoichiometry, molar masses, and volumes of solutions to determine the quantity of substances involved.
In the provided exercise, after finding the moles of excess \(\mathrm{HCl}\) using titration, you can quantify the chemical reaction by using the stoichiometry of the balanced equation to find the moles of calcium carbonate that reacted. This is followed by finding the mass using the molar mass. Finally, the chemical reaction is quantified by determining the percentage of \(\mathrm{CaCO}_3\) in the original limestone sample. This encompasses all the steps of chemical quantification, from reacting moles to determining mass percentages.