Problem 116 - Chemistry and Chemical Reactivity

Question

Calcium and magnesium carbonates occur together in the mineral dolomite. Suppose you heat a sample of the mineral to obtain the oxides, \(\mathrm{CaO}\) and \(\mathrm{MgO},\) and then treat the oxide sample with hydrochloric acid. If \(7.695 \mathrm{g}\) of the oxide sample requires \(125 \mathrm{mL}\) of \(2.55 \mathrm{M} \mathrm{HCl},\) $$\begin{aligned} \mathrm{CaO}(\mathrm{s}) &+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \\\\\mathrm{MgO}(\mathrm{s}) &+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\end{aligned}$$ what is the weight percent of each oxide (CaO and \(\mathrm{MgO}\) ) in the sample?

Step-by-Step Solution

Answer

CaO is 58.8% and MgO is 41.2% of the sample.

Step 1: Determine Moles of HCl Used

First, calculate the number of moles of HCl used. Use the formula: \[ \text{Moles of HCl} = \text{Volume (L)} \times \text{Molarity (M)} \] Convert 125 mL to liters: \[ 125 \text{ mL} = 0.125 \text{ L} \] Now calculate the moles: \[ \text{Moles of HCl} = 0.125 \times 2.55 = 0.31875 \] moles.

Step 2: Relate Moles of HCl to Moles of CaO and MgO

Each reaction requires 2 moles of HCl per mole of oxide (CaO or MgO). Therefore, the total moles of oxide present is half the moles of HCl: \[ \text{Total moles of oxide} = \frac{0.31875}{2} = 0.159375 \] moles.

Step 3: Assume Variable Moles for CaO and MgO

Let \(x\) be the moles of CaO and \(y\) be the moles of MgO. Then, \(x + y = 0.159375\).

Step 4: Set Up Mass Equation for Oxides

Using the molar masses \(\text{CaO} = 56.08\) g/mol and \(\text{MgO} = 40.30\) g/mol, write the mass equation: \[ 56.08x + 40.30y = 7.695 \]

Step 5: Solve the System of Equations

We have two equations: \(x + y = 0.159375\) and \(56.08x + 40.30y = 7.695\). Solve these equations simultaneously: - From the first equation: \( y = 0.159375 - x \)- Substitute into the second equation: \[ 56.08x + 40.30(0.159375 - x) = 7.695 \]- Simplify and solve for \(x\): \[ 56.08x + 6.422125 - 40.30x = 7.695 \] \[ 15.78x = 1.272875 \] \[ x = \frac{1.272875}{15.78} \approx 0.0806 \]

Step 6: Calculate Moles and Mass of MgO

Substitute \(x\) back to calculate \(y\): \[ y = 0.159375 - 0.0806 = 0.078775 \]- Calculate the mass of MgO using \( y \): \[ \text{Mass of MgO} = 40.30 \times 0.078775 \approx 3.1735 \text{ g} \]

Step 7: Calculate Mass Percentages

Now calculate the weight percent of each oxide:- \(\text{Weight \% CaO} = \left( \frac{56.08 \times 0.0806}{7.695} \right) \times 100 \approx 58.8\% \)- \(\text{Weight \% MgO} = \left( \frac{40.30 \times 0.078775}{7.695} \right) \times 100 \approx 41.2\% \)

Key Concepts

Stoichiometry

Stoichiometry is essential in chemistry for calculating the amounts of reactants and products in a chemical reaction. This is done using a balanced chemical equation, which shows the stoichiometric coefficients—essentially the number of moles of each substance involved. By knowing the stoichiometry of a reaction, you can determine how much of each reactant is needed to produce a desired amount of product, or how much product can be formed from given quantities of reactants.
In the given exercise, stoichiometry allows us to understand the relationship between the moles of oxides (CaO and MgO) and hydrochloric acid (HCl). Both oxides require 2 moles of HCl to fully react, as depicted in the balanced reaction equations. This relationship was crucial in calculating the total moles of oxides based on the moles of HCl used during the experiment.

Molarity

Molarity (M) is a measurement of the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. This concept is key when dealing with reactions in a liquid phase, as it allows us to quantify the amount of dissolved substance.
The original exercise requires determining the molarity of HCl to calculate the moles of HCl used in the reaction. The formula for molarity is given by:

• M = \( \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

Here, the molarity of the hydrochloric acid was used to find out how many moles of HCl participated in the reaction with calcium oxide and magnesium oxide. This information was then used in stoichiometric calculations to find the weight percent of each oxide.

Gravimetric Analysis

Gravimetric analysis is a method of quantitative chemical analysis in which the amount of an analyte (substance of interest) is determined by measuring the mass of a solid. Typically, the steps include converting the analyte into a pure and stable compound, then weighing it.
In our exercise, gravimetric analysis involves the conversion of calcium and magnesium oxides into their chloride counterparts after reaction with HCl. The initial and final weights are used to find the mass percent composition of

• CaO
• MgO

in the sample. Considering the precise mass measurement, gravimetric analysis provides a reliable method to determine the exact amount of each component present.

Chemical Reactions

Chemical reactions are processes in which substances (reactants) are transformed into new substances (products). In stoichiometry, the balanced chemical equation represents how reactants convert to products in fixed ratios. This balance is crucial in ensuring that calculations related to reactants and products are accurate.
The given exercise features reactions between oxides (CaO and MgO) and hydrochloric acid (HCl). The equations:

• \( \text{CaO(s) + 2 HCl(aq)} \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O} \)
• \( \text{MgO(s) + 2 HCl(aq)} \rightarrow \text{MgCl}_2(aq) + \text{H}_2\text{O} \)

illustrate the conversion. They highlight that each mole of oxide reacts with two moles of HCl, forming chlorides and water. Understand this concept helps in solving for unknown concentrations in analytical chemistry tasks such as determining the weight percent of components in a mixture.