Question

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with the carbonate and then "back-titrating" the remaining acid with \(\mathrm{NaOH}\). (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a \(0.1022-\mathrm{g}\) sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

Short answer

The molar mass of the carbonate is 197 g/mol and the metal M is Calcium.

Step-by-step solution

Write Reactions

(a) Reaction notes: When \(\mathrm{MCO}_{3}\) reacts with \(\mathrm{HCl}\), the reaction produces metal chloride (\(\mathrm{MCl}_2\)), water (\(\mathrm{H}_2\mathrm{O}\)) and carbon dioxide (\(\mathrm{CO}_2\)). When the excess \(\mathrm{HCl}\) is neutralized by \(\mathrm{NaOH}\), the products of the reaction are water (\(\mathrm{H}_2\mathrm{O}\)) and sodium chloride (\(\mathrm{NaCl}\)). Thus, the reactions are: \[\mathrm{MCO}_{3} + 2\mathrm{HCl}\rightarrow \mathrm{MCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\] \[\mathrm{HCl} + \mathrm{NaOH}\rightarrow \mathrm{H}_2\mathrm{O} + \mathrm{NaCl}\]

Calculation of moles of HCl

(b) First, calculate the total moles of \(\mathrm{HCl}\) added initially. Moles of \(\mathrm{HCl}\) can be calculated by multiplying the volume \(\mathrm{V}\) in liters by the molarity \(\mathrm{M}\), \(moles = M \times V\). Therefore: \[Total \ moles \ of \ \mathrm{HCl} = 0.0800 \ \mathrm{M} \times 0.02000\ \mathrm{L} = 0.00160 \ moles\] Now, calculate the moles of HCl that were neutralized by NaOH. From the equation of the reaction between NaOH and HCl, we see that 1 mole of NaOH neutralizes 1 mole of HCl. Moles of NaOH (which equal the moles of HCl neutralized) can be calculated in a similar way as before: \[Moles \ of \ neutralized \ \mathrm{HCl} = 0.1000 \ \mathrm{M} \times 0.00564 \ \mathrm{L} = 0.000564 \ moles\] By subtracting the moles of the neutralized HCl from total moles of HCl, we can find the moles of HCl which reacted with the carbonate: \[Moles \ of \ \mathrm{HCl} \ reacting \ with \ \mathrm{MCO}_{3} = 0.00160 \ moles - 0.000564 \ moles = 0.001036 \ moles\]

Calculate molar mass

Now, from the reaction between the metal carbonate and HCl, it is seen that 1 mole of the carbonate reacts with 2 moles of HCl. Hence, the moles of the metal carbonate can be calculated as the moles of HCl divided by 2: \[Moles \ of \ \mathrm{MCO}_{3} = 0.001036 \ moles \ / \ 2 = 0.000518 \ moles\] The molar mass of the metal carbonate is its given mass divided by the moles calculated. Therefore, the molar mass is: \[Molar \ mass \ of \ \mathrm{MCO}_{3} = 0.1022 \ \mathrm{g} \ / \ 0.000518 \ \mathrm{moles} = 197 \ \mathrm{g/mol}\] The molar mass of CO3 is 60 g/mol, so that of M is obtained by subtracting 60 from 197, which corresponds roughly to the molar mass of Calcium (40 g/mol), therefore M is Calcium \(M = Ca\).

Key concepts

Back-titration

In the realm of analytical chemistry, one particularly useful technique is back-titration. This process allows determination of the amount of an analyte in a sample by reacting it with a known excess of a standard reagent and then titrating this remaining reagent with a second standard solution. Unlike direct titration, where the end point directly indicates the amount of analyte present, back-titration works backwards from the excess.

Let's consider it through the perspective of the exercise: a known volume and concentration of hydrochloric acid, HCl, is reacted with a metal carbonate, MCO₃. Some of the HCl remains unreacted, and this excess is then neutralized ('back-titrated') with sodium hydroxide, NaOH. The moles of NaOH used give a direct measure of the moles of excess HCl, and when subtracted from the initial moles of HCl added, reveal the moles of HCl that reacted with the metal carbonate. This method is robust against errors caused by slow reactions or those that do not go to completion.

Stoichiometry

Stoichiometry is the mathematics behind chemistry. It is all about the quantitative relationship between reactants and products in a chemical reaction, based on the balanced chemical equation. For every stoichiometry problem, a balanced equation is the starting point, as it provides the mole ratios of reactants and products.

Within the given problem, stoichiometry is used to deduce the mole ratio between the metal carbonate and the hydrochloric acid. According to the balanced equation, one mole of MCO₃ reacts with two moles of HCl. By identifying the moles of HCl that reacted with the carbonate, we can calculate the moles of MCO₃ and ultimately its molar mass. This relationship is the key to translating between the mass of the substance and the moles we need for chemical equations, and it's especially useful for identifying unknown substances, much like our unknown metal M.

Chemical Reaction Equation

The chemical reaction equation provides a symbolic representation of a chemical reaction, showing the reactants transforming into products. Balancing this equation is vital, as it ensures that the law of conservation of mass is fulfilled, meaning the same number of each type of atom is present on both sides of the equation.

In our exercise, the reaction of the metal carbonate with HCl is represented as MCO₃ + 2HCl → MCl₂ + CO₂ + H₂O. This tells us precisely what is required and produced in the reaction. Additionally, it lays the foundation for all stoichiometric calculations by establishing the mole ratios. Understanding how to interpret and balance a chemical equation is crucial for anyone studying chemistry, as it provides the crucial link between reactants and products, allowing for detailed predictions about the outcomes of chemical reactions.