Problem 116 - Chemistry and Chemical Reactivity

Question

A 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 mL. of 2.55 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: 58.52%, MgO: 41.48%

Step 1: Convert Molarity to Moles of HCl

First, calculate the number of moles of HCl used in the reaction. Use the formula \( ext{moles} = ext{molarity} \times ext{volume in liters} \). Here, the volume is 125 mL, which is 0.125 liters. Thus,\[ ext{moles of HCl} = 2.55 ext{ M} \times 0.125 ext{ L} = 0.31875 ext{ moles of HCl} \]

Step 2: Determine Moles of Oxides Reacted

Each mole of CaO or MgO reacts with 2 moles of HCl according to the given chemical equations. Therefore, the total moles of oxides \(( ext{CaO} + ext{MgO})\) is half of the moles of HCl used because both oxides use '2 moles of HCl per mole' as a factor.\[ ext{Total moles of Oxides} = \frac{0.31875}{2} = 0.159375 \text{ moles} \]

Step 3: Set Up System of Equations for CaO and MgO

Let \( x \) be the moles of CaO and \( y \) be the moles of MgO. We have:1. \( x + y = 0.159375 \)2. Using their respective molar masses, we also know the total mass of the oxide mixture: - Molar mass of \( ext{CaO} \) is 56.08 g/mol. - Molar mass of \( ext{MgO} \) is 40.3 g/mol.Thus, \( 56.08x + 40.3y = 7.695 \text{ g} \).

Step 4: Solve Simultaneous Equations

Solve the two equations:\[ x + y = 0.159375 \]\[ 56.08x + 40.3y = 7.695 \]First, solve for \( x \) in terms of \( y \) from the first equation:\[ x = 0.159375 - y \]Substitute into the second equation:\[ 56.08(0.159375 - y) + 40.3y = 7.695 \]Simplify and solve for \( y \):\[ 8.9439 - 56.08y + 40.3y = 7.695 \] \[ -15.78y = -1.2489 \] \[ y = 0.07914 \text{ moles of MgO} \]Substitute \( y \) back to find \( x \):\[ x = 0.159375 - 0.07914 = 0.080235 \text{ moles of CaO} \]

Step 5: Calculate Mass of Each Oxide

Calculate the mass from the moles for each oxide:- Mass of \( ext{CaO} \): \[ 0.080235 \times 56.08 = 4.50185 \text{ g} \]- Mass of \( ext{MgO} \): \[ 0.07914 \times 40.3 = 3.19315 \text{ g} \]

Step 6: Find Weight Percent of Each Oxide

Calculate the weight percent of each oxide in the mixture:- Weight percent of \( ext{CaO} \): \[ \frac{4.50185}{7.695} \times 100 = 58.52\% \]- Weight percent of \( ext{MgO} \): \[ \frac{3.19315}{7.695} \times 100 = 41.48\% \]

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of one or more substances into different substances. Here, we heat a sample of dolomite, resulting in the formation of calcium oxide (CaO) and magnesium oxide (MgO).
These oxides react with hydrochloric acid ( HCl), a common acid in chemistry lab reactions. Two separate reactions occur:

• CaO(s) + 2HCl(aq) → CaCl e2(aq) + H 2O(l)
• MgO(s) + 2HCl(aq) → MgCl e2(aq) + H 2O(l)

Both oxides consume acids in a 1:2 ratio; meaning one mole of oxide reacts with two moles of HCl. This stoichiometric ratio is essential for calculating how much of each reactant is used or leftover after the reaction. Remember, a balanced equation ensures that atoms are conserved, and this is crucial when dealing with stoichiometry.

Molarity

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of a solute per liter of solution (mol/L). In this problem, the molarity of HCl is given as 2.55 M. To find the total number of moles of HCl:

• Use the formula: number of moles = molarity × volume in liters.

Given a 125 mL volume, convert it to liters by dividing by 1000, resulting in 0.125 L. Then, calculate:\[\text{moles of HCl} = 2.55 \, \text{M} \times 0.125 \, \text{L} = 0.31875 \, \text{moles of HCl}\]
This calculation is crucial because it allows us to determine how many oxides reacted with the hydrochloric acid, based on the stoichiometry of the reaction.

Simultaneous Equations

Simultaneous equations are a system of equations used to find multiple unknowns. In our exercise, we have two unknowns: the moles of CaO ( x) and the moles of MgO ( y). By setting up a system of equations based on stoichiometry and total mass, we can solve for these values.

• Equation 1: Total moles of oxides: x + y = 0.159375
• Equation 2: Total mass of oxides: 56.08x + 40.3y = 7.695 g

To solve these, express x as x = 0.159375 - y, and substitute into the second equation. This simplifies to finding y, as the moles of MgO. Once y is calculated, substitute it back to find x, the moles of CaO. Systematically working through these equations ensures accurate separation of the different substances in the mixture.

Weight Percent Calculation

Weight percent is a way to express the concentration of a component in a mixture. It is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100.For our task, after finding the mass of each oxide:

• Mass of CaO is calculated from its moles: \(0.080235 \, \text{mol} \times 56.08 \, \text{g/mol} = 4.50185 \, \text{g}\).
• Mass of MgO is similarly found: \(0.07914 \, \text{mol} \times 40.3 \, \text{g/mol} = 3.19315 \, \text{g}\).

Calculate the weight percent:

• CaO: \(\frac{4.50185}{7.695} \times 100 = 58.52\%\)
• MgO: \(\frac{3.19315}{7.695} \times 100 = 41.48\%\)

These percentages give insight into the composition of the sample, providing a clear understanding of how much each component contributes to the total mass of the sample.