Question

One method for chemical analysis involves finding some reagent that will precipitate the species of interest. The mass of the precipitate is then used to determine what mass of the species of interest was present in the original sample. For example, calcium ion can be precipitated from solution by addition of sodium oxalate. The balanced equation is $$\mathrm{Ca}^{2+}(a q)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \rightarrow \mathrm{CaC}_{2} \mathrm{O}_{4}(s)+2 \mathrm{Na}^{+}(a q)$$ Suppose a solution is known to contain approximately 15 g of calcium ion. Show by calculation whether the addition of a solution containing \(15 \mathrm{g}\) of sodium oxalate will precipitate all of the calcium from the sample.

Short answer

Based on the calculations, the addition of a solution containing 15 g of sodium oxalate will not be enough to precipitate all of the calcium ions from the sample, as sodium oxalate is the limiting reactant. More sodium oxalate would be required to react with approximately 15 g of calcium ions.

Step-by-step solution

Identify the molar mass of the participating substances

In order to convert the mass of the substances involved to moles, we need to know their molar mass. Using the periodic table, we can find the molar mass of calcium (Ca), sodium (Na), carbon (C), and oxygen (O) which are the components of the substances involved: - Ca: 40.08 g/mol - Na: 22.99 g/mol - C: 12.01 g/mol - O: 16.00 g/mol Now we can calculate the molar masses of calcium ion (Ca²⁺), sodium oxalate (Na₂C₂O₄), and calcium oxalate (CaC₂O₄): For Ca²⁺, the molar mass is equal to the molar mass of calcium: M(Ca²⁺) = 40.08 g/mol For Na₂C₂O₄, M(Na₂C₂O₄) = 2 * M(Na) + 2 * M(C) + 4 * M(O) = 2 * 22.99 + 2 * 12.01 + 4 * 16.00 = 133.99 g/mol

Calculate the moles of calcium ions and sodium oxalate

Now, we will convert the given mass of calcium ions and sodium oxalate to moles using their molar masses: moles of Ca²⁺ = 15 g / 40.08 g/mol = 0.374 moles moles of Na₂C₂O₄ = 15 g / 133.99 g/mol = 0.112 moles

Determine the limiting reactant using stoichiometry

According to the balanced reaction equation: \(Ca^{2+}(aq) + Na_{2}C_{2}O_{4}(aq) \rightarrow CaC_{2}O_{4}(s) + 2Na^{+}(aq)\) We can observe that 1 mole of calcium ions reacts with 1 mole of sodium oxalate to form 1 mole of calcium oxalate. Therefore, the stoichiometric ratio of Ca²⁺ to Na₂C₂O₄ is 1:1. Now let's compare the mole ratio of Ca²⁺ and Na₂C₂O₄ to the stoichiometric ratio: moles of Ca²⁺ / moles of Na₂C₂O₄ = 0.374 moles / 0.112 moles = 3.34 Since the mole ratio of Ca²⁺ to Na₂C₂O₄ (3.34) is greater than the stoichiometric ratio (1:1), it means that there are not enough moles of sodium oxalate to react with all the calcium ions present. Hence, sodium oxalate is the limiting reactant.

Conclusion

Based on the calculations, the addition of a solution containing 15 g of sodium oxalate will not be enough to precipitate all of the calcium ions, as sodium oxalate is the limiting reactant. More sodium oxalate would be required to react with approximately 15 g of calcium ions.

Key concepts

Stoichiometry

Stoichiometry is the mathematical relationship between the amounts of reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. Consequently, the amount of each element must be the same before and after the reaction takes place.

Stoichiometry calculations often begin with a balanced chemical equation. From this equation, we can derive molar ratios which tell us how the reactants combine and how much of each product they yield. For example, in the equation \( \mathrm{Ca}^{2+}(aq) + \mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}(aq) \rightarrow \mathrm{CaC}_{2}\mathrm{O}_{4}(s) + 2\mathrm{Na}^{+}(aq) \), we see a 1:1 molar ratio between calcium ions and sodium oxalate. We also see that one mole of calcium ions will produce one mole of calcium oxalate.

To convert between the mass of a substance and the number of moles, we use the substance's molar mass. The steps for stoichiometry typically follow identifying the molar mass, converting mass to moles, using the molar ratio to calculate the mass of the desired product or reactant, and identifying the limiting reactant if necessary.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two solutions containing soluble salts are combined to form one or more insoluble products, known as precipitates. Such reactions can be used to separate or remove ions from a solution, as seen in the formation of calcium oxalate in our example.

When sodium oxalate is added to a solution containing calcium ions, the two react to form calcium oxalate, which is insoluble in water and precipitates out of the solution. This process is represented by the equation \( \mathrm{Ca}^{2+}(aq) + \mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}(aq) \rightarrow \mathrm{CaC}_{2}\mathrm{O}_{4}(s) + 2\mathrm{Na}^{+}(aq) \). The solid \( \mathrm{CaC}_{2}\mathrm{O}_{4} \) can be filtered from the mixture, allowing for the quantitative analysis of the calcium content.

The extent of a precipitation reaction depends on the solubility rules for the salts involved. For analysis and synthesis purposes, knowing these rules helps predict the outcome of mixing various ionic solutions and designing methods for separating compounds.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, stopping the reaction from proceeding because there is no more of that substance left to react. Identifying the limiting reactant is crucial because it dictates the maximum amount of product that can be formed.

To find the limiting reactant, we compare the mole ratio of the reactants with the stoichiometric ratio from the balanced chemical equation. In the reaction between calcium ions and sodium oxalate, if the mole ratio does not match the stoichiometric 1:1 ratio, then the reactant with fewer moles becomes the limiting reactant.

In our example, due to having lower moles compared to calcium ions, sodium oxalate is the limiting reactant. Therefore, even though there might be an excess of calcium ions, the production of calcium oxalate will be limited to the amount of sodium oxalate present. Any chemical analysis or synthesis aiming for maximum yields needs to consider the limiting reactant to adjust the quantities of reactants and ensure all of the desired product can be collected.