Question

Both calcium chloride and sodium chloride are used to melt ice and snow on roads in the winter. A certain company was marketing a mixture of these two compounds for this purpose. A chemist, wanting to analyze the mixture, dissolved \(2.463 \mathrm{~g}\) of it in water and precipitated calcium oxalate by adding sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) The calcium oxalate was carefully filtered from the solution, dissolved in sulfuric acid, and titrated with 0.1000 \(M \mathrm{KMnO}_{4}\) solution. The reaction that occurred was \(6 \mathrm{H}^{+}+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{MnO}_{4}^{-} \longrightarrow\) $$ 10 \mathrm{CO}_{2}+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O} $$ The titration required \(21.62 \mathrm{~mL}\) of the \(\mathrm{KMnO}_{4}\) solution. (a) How many moles of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) were present in the calcium oxalate precipitate? (b) How many grams of calcium chloride were in the original \(2.463 \mathrm{~g}\) sample? (c) What was the percentage by mass of calcium chloride in the sample?

Short answer

The moles of \text{C}_2\text{O}_4^{2-} in the precipitate are calculated from the molarity and volume of \text{KMnO}_4 used, then converted to moles of \text{CaCl}_2, and finally to grams using the molar mass to find the mass and percentage of \text{CaCl}_2 in the sample.

Step-by-step solution

Calculate moles of KMnO4 used in titration

First, calculate the moles of \text{KMnO}_4 used in the titration. Since the concentration of \text{KMnO}_4 is given, use the equation: \[\text{moles} = \text{concentration} \times \text{volume in liters}\] In this case: \[\text{moles of KMnO}_4 = 0.1000 \text{ M} \times 0.02162 \text{ L}\]

Find moles of oxalate ion from moles of KMnO4

Use the stoichiometry of the balanced reaction to find the moles of \text{C}_2\text{O}_4^{2-}. According to the balanced reaction, 2 moles of \text{MnO}_4^- react with 5 moles of \text{C}_2\text{O}_4^{2-}. Accordingly: \[\text{moles of C}_2\text{O}_4^{2-} = (5/2) \times \text{moles of KMnO}_4\]

Calculate moles of calcium chloride present

In the calcium oxalate precipitate, there is a 1:1 molar ratio of calcium to oxalate. Thus, the moles of calcium chloride will be equal to the moles of oxalate ions. Therefore: \[\text{moles of CaCl}_2 = \text{moles of C}_2\text{O}_4^{2-}\]

Convert moles of calcium chloride to grams

To find the mass of calcium chloride, multiply the moles by the molar mass of calcium chloride (\text{CaCl}_2): \[\text{molar mass of CaCl}_2 = 40.08 + 2(35.45)\] \[\text{mass of CaCl}_2 = \text{moles of CaCl}_2 \times \text{molar mass of CaCl}_2\]

Calculate the percentage by mass of calcium chloride

Finally, to find the percentage by mass of calcium chloride in the sample: \[\text{percent by mass} = \left( \frac{\text{mass of CaCl}_2}{\text{mass of sample}} \right) \times 100\%\] Use the mass of the calcium chloride found in Step 4 and the original sample mass of 2.463 g.

Key concepts

Molar Mass Calculation

Understanding molar mass is fundamental to the field of chemistry, particularly when delving into stoichiometry and reaction equations. The molar mass of a substance is the weight of one mole (6.022 \(\times\) 10\(^{23}\) entities) of that substance, and it is expressed in grams per mole (g/mol). To calculate the molar mass, one must sum the atomic masses of all atoms in a formula unit of the compound. For example, in the case of calcium chloride (CaCl\(_2\)), the molar mass is found by adding the molar mass of one calcium atom (40.08 g/mol) with the molar mass of two chlorine atoms (2 \(\times\) 35.45 g/mol).

When combining these, we get:\[\text{Molar Mass of CaCl}_2 = 40.08 \text{g/mol} + 2(35.45 \text{g/mol}) = 110.98 \text{g/mol}\]
This calculation is vital for converting between mass and moles, which is a common step in solving many chemistry problems.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. It provides the basis for creating balanced equations, enabling chemists to predict the amounts of substances consumed and produced in a reaction. This is done using the mole ratio derived from the balanced equation.

In the problem regarding the calcium oxalate precipitate, the stoichiometric relationship between potassium permanganate (KMnO\(_4\)) and the oxalate ions (C\(_2\)O\(_4\)\(^{2-}\)), as outlined in the balanced reaction equation, is the key to finding the number of moles of oxalate ions present. Using mole ratios, we know that 2 moles of KMnO\(_4\) react with 5 moles of C\(_2\)O\(_4\)\(^{2-}\), establishing a 2:5 ratio. The information gleaned from stoichiometry is pivotal for informing subsequent calculations, like determining the moles of calcium chloride and, in turn, its mass.

Titration

Titration is an analytical method widely employed in chemistry to determine the concentration of an identified analyte. By gradually adding a reactant of known concentration, called the titrant, to a solution until the reaction reaches neutralization or equivalence, the quantity of the analyte can be measured. The point of neutralization or equivalence is often indicated by a color change with the aid of an indicator, or by reaching a particular value on a pH meter in acid-base titrations.

In the exercise provided, titration is employed to ascertain the amount of C\(_2\)O\(_4\)\(^{2-}\) in the calcium oxalate precipitate. The chemist uses the known concentration of the titrant, potassium permanganate (KMnO\(_4\)), to find out how many moles of C\(_2\)O\(_4\)\(^{2-}\) were present in the original sample. With the volume used in the titration (21.62 mL) and the molar concentration (.1000 M), the titration data feed into stoichiometric calculations to achieve accurate quantitative results about the sample's composition.