Question

A sample of impure tin of mass 0.535 g is dissolved in strong acid to give a solution of \(S n^{2+} .\) The solution is then titrated with a 0.0448 \(\mathrm{M}\) solution of \(\mathrm{NO}_{3}^{-},\) which is reduced to \(\mathrm{NO}(g) .\) The equivalence point is reached upon the addition of 0.0344 L of the \(\mathrm{NO}_{3}^{-}\) solution. Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents.

Short answer

The percent by mass of tin in the original sample is calculated using the moles of NO3- from titration and the stoichiometry of the reaction.

Step-by-step solution

- Calculate moles of NO3- used in titration

To find the moles of nitrate ion (\text{NO}_3^-) used in the titration, use the concentration of the nitrate solution and the volume used at the equivalence point. The formula to calculate moles is: \[ \text{moles} = \text{concentration (M)} \times \text{volume (L)} \]

- Determine moles of Sn2+ reacting with NO3-

Since we're at the equivalence point, the amount of nitrate reacted is stoichiometrically equivalent to the amount of Sn2+ in the solution. Thus, the moles of Sn2+ is equal to the moles of NO3- used.

- Calculate mass of Sn from moles of Sn2+

The molar mass of Sn is required to find the mass of tin from the moles of Sn2+. Use the formula: \[ \text{mass of Sn} = \text{moles of Sn2+} \times \text{molar mass of Sn} \]

- Calculate percent by mass of tin in the original sample

To find the percent by mass of tin in the impure sample, divide the mass of Sn calculated in Step 3 by the original mass of the impure sample and then multiply by 100 to get the percentage. The formula is: \[ \text{Percent by mass} = \left(\frac{\text{mass of Sn}}{\text{mass of impure sample}}\right) \times 100\% \]

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It's like a recipe for chemistry; just like baking a cake requires a certain ratio of ingredients, chemical reactions require substances in precise ratios. By understanding these ratios, we can predict how much of each substance is needed or produced during a reaction.

For instance, in a titration involving a strong acid and a base, the stoichiometry of the reaction allows us to calculate the concentration of an unknown solution by using a solution with a known concentration. Titration is a typical application, where the concept of equivalents or molarity works hand in hand with stoichiometry to determine concentration and purity, as in the textbook exercise involving the titration of tin with nitrate ions.

Molar Mass Calculation

Calculating molar mass is crucial in chemistry for converting between grams and moles. The molar mass of an element is the weight of 6.022 x 10^23 (Avogadro's number) of its atoms, usually expressed in grams per mole (g/mol).

For a compound, the molar mass is the sum of the molar masses of its constituent atoms. For example, to find the molar mass of Sn (tin), we would look at the periodic table, where the atomic weight of tin is approximately 118.71 g/mol. Knowing the molar mass allows chemists to measure out quantities of a substance in moles, which are easier to relate to reactions on the molecular level, as highlighted in the steps for solving the exercise.

Percent by Mass

Percent by mass (also known as weight percent or mass percent) is a way of expressing the concentration of a component in a mixture or element in a compound. It is defined as the mass of the component divided by the total mass of the mixture or compound, multiplied by 100%.

In the context of the textbook exercise, percent by mass helps determine the purity of a sample. Once the mass of tin recovered from the reaction is known, we can compare it to the original sample's mass to find out how much of the sample was tin. This percentage tells us the purity of the sample, which is essential for many practical applications, such as manufacturing or substance verification.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two substances. The substance that gives away electrons is oxidized, while the substance that accepts electrons is reduced. These reactions are fundamental to various processes, such as metabolism in biological systems and corrosion of metals.

In the exercise we're discussing, NO3- acts as the oxidizing agent and is reduced to NO, while Sn2+ is oxidized. Titration in this redox reaction helps to pinpoint the exact moment when the reacting substances have combined in the correct stoichiometric proportions, which is known as the equivalence point. Understanding redox reactions is integral to scientific fields such as energy generation, material science, and environmental monitoring.