Problem 156 - General Chemistry

Question

The active ingredients in an antacid tablet contained only calcium carbonate and magnesium carbonate. Complete reaction of a sample of the active ingredients required \(41.33 \mathrm{~mL}\) of \(0.08750 M\) hydrochloric acid. The chloride salts from the reaction were obtained by evaporation of the filtrate from this titration; they weighed \(0.1900 \mathrm{~g}\). What was the percentage by mass of the calcium carbonate in the active ingredients of the antacid tablet?

Step-by-Step Solution

Answer

51.2%

Step 1: Write Balanced Chemical Reactions

Calcium carbonate and magnesium carbonate react with hydrochloric acid as follows:\[\begin{align*}\text{CaCO}_3 + 2\text{HCl} & \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \\text{MgCO}_3 + 2\text{HCl} & \rightarrow \text{MgCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\end{align*}\]

Step 2: Calculate Moles of HCl Used

The moles of hydrochloric acid can be determined using the formula \[ n = M \times V \]where \(M\) is molarity and \(V\) is volume in liters.\[n = 0.08750 \text{ M} \times 0.04133 \text{ L} = 0.003619 \text{ mol HCl}\]

Step 3: Establish the Stoichiometry

From the balanced equations in Step 1, we see that:- Each mole of \(\text{CaCO}_3\) requires 2 moles of \(\text{HCl}\).- Each mole of \(\text{MgCO}_3\) requires 2 moles of \(\text{HCl}\).Thus, total \(0.003619\) moles of \(\text{HCl}\) would mean half this amount in moles of carbonate salt is reacted:\[ n = \frac{0.003619}{2} = 0.0018095 \text{ mol (CaCO}_3 + \text{MgCO}_3) \]

Step 4: Calculate Molar Mass of Chloride Salts

The molar masses are:\- CaCl_2 = 110.98 \text{ g/mol} \\- MgCl_2 = 95.21 \text{ g/mol} \[ \]Let \(x\) be the moles of \(\text{CaCO}_3\) and \(y\) be the moles of \(\text{MgCO}_3\).From Step 3, \[ x + y = 0.0018095 \].

Step 5: Relate Total Mass of Chlorides

The chloride salt mass is the sum of the mass of calcium chloride and magnesium chloride obtained:\[ 110.98x + 95.21y = 0.1900 \text{ g} \].

Step 6: Solve System of Equations

Using the equations from Steps 4 and 5:\[\begin{align*}x + y &= 0.0018095 \110.98x + 95.21y &= 0.1900\end{align*}\]Substitute \(y = 0.0018095 - x\) into the second equation and solve for \(x\):\[110.98x + 95.21(0.0018095 - x) = 0.1900\]\[110.98x + 0.1722 - 95.21x = 0.1900\]\[15.77x = 0.0178 \x = \frac{0.0178}{15.77} = 0.00113\]

Step 7: Calculate Mass and Percentage of Calcium Carbonate

The mass of \(\text{CaCO}_3\) corresponding to \(x = 0.00113\) moles is:\[ 0.00113 \times 100.09 \text{ g/mol} = 0.113 g \]The percentage by mass of \(\text{CaCO}_3\) in the total weight of chloride (\text{active ingredients}) is:\[\frac{0.113}{0.113 + 0.077} \times 100\% = 51.2\%\]

Step 8: Final Answer: Percentage of Calcium Carbonate

The percentage by mass of calcium carbonate in the active ingredient of the antacid tablet is approximately 51.2%.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It helps us understand the quantities needed and produced in a balanced chemical equation.
In stoichiometry, the coefficients of a balanced chemical equation are used to calculate the ratio of moles between the reactants and products.
This ensures the law of conservation of mass is satisfied, meaning matter is neither created nor destroyed during a chemical reaction.

• For calcium carbonate reacting with hydrochloric acid, the balanced equation is: \( \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \)
• This tells us that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid to form 1 mole of calcium chloride, alongside carbon dioxide and water as by-products.

This stoichiometric calculation is crucial for determining how much calcium carbonate was present based on the moles of hydrochloric acid used in the reaction.

Moles Calculation

Understanding how to calculate moles is critical in quantitative chemistry. The mole is a unit used to express the amount of a chemical substance.
The number of moles can be calculated using the formula \( n = M \times V \), where \( n \) represents the number of moles, \( M \) is the molarity, and \( V \) is the volume in liters.
For our exercise, we calculated the moles of hydrochloric acid used as follows:

• The molarity, \( M \), was given as 0.08750 M.
• The volume, \( V \), converted to liters from milliliters (41.33 mL = 0.04133 L).
• The resulting moles of \( \text{HCl} \) was found to be \( 0.003619 \text{ mol} \).

This represents the total amount of hydrochloric acid that reacted with the carbonates in the antacid tablet.

Molar Mass

Molar mass is the mass of one mole of a given substance and is often expressed in grams per mole (g/mol).
It is crucial for converting between mass and moles in stoichiometry calculations.
For instance, in our exercise, the molar masses of the products needed were:

• \( \text{CaCl}_2 \) (calcium chloride) = 110.98 g/mol
• \( \text{MgCl}_2 \) (magnesium chloride) = 95.21 g/mol

These values are essential for calculating the total mass of chloride salts formed in the reaction. By knowing these molar masses, we can accurately determine the contribution of each component to the total mass of salts (0.1900 g in this particular reaction). This, in turn, helps us figure out the percentage of each compound in the original mixture.

Chemical Reaction

In chemistry, understanding the chemical reaction is pivotal. A chemical reaction involves the transformation of reactants into products.
Each reaction can be detailed with a balanced chemical equation, showing the conversion process.
For calcium carbonate reacting with hydrochloric acid, the equation is:

• \( \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \)
• For magnesium carbonate, the equation is similar: \( \text{MgCO}_3 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \)

The production of calcium chloride and magnesium chloride, alongside carbon dioxide and water, implies that the carbonate components are fully consumed in the reaction with hydrochloric acid. Recognizing the complete consumption of products helps compute the amount of original reactants.

Percentage Composition

Percentage composition in chemistry refers to the proportion by mass of each component in a compound or mixture.
It gives insight into the distribution of substances within a sample, often expressed as a percentage.
To find the percentage by mass of calcium carbonate in the antacid tablet, we had:

• First, solve for the moles of \( \text{CaCO}_3 \) using a system of equations based on the stoichiometry of the reaction.
• The mass of \( \text{CaCO}_3 \) was found to be 0.113 g.
• The mass of \( \text{MgCO}_3 \) was complementary to the total resulting from the total chloride salt mass.

Using these masses, we calculate the percentage:
\( \frac{0.113}{0.113 + 0.077} \times 100\% \approx 51.2\% \). This percentage represents the share of \( \text{CaCO}_3 \) in the overall compound of the active ingredients.