Question

Compounds \(X\) and \(Y\) have the same molecular formula and contain \(18.25 \%\) titanium, \(40.54 \%\) chlorine, \(36.59 \%\) oxygen, and \(4.61 \%\) hydrogen by mass. When \(0.5000 \mathrm{g}\) of \(\mathrm{X}\) was dissolved in water and treated with silver nitrate solution, \(0.2731 \mathrm{g}\) of silver chloride were precipitated. \(0.5000 \mathrm{g}\) of \(\mathrm{Y}\) were dissolved in water and titrated against \(0.2000 \mathrm{moldm}^{-3}\) silver nitrate solution, of which \(28.2 \mathrm{cm}^{3}\) were required. Identify \(X\) and \(Y\), and determine any possible isomers. (Section 28.3 )

Short answer

Compounds "X" and "Y" are isomers of Titanium trichloride hexahydrate following TiCl₃(OH)₃ with varied hydration or arrangement.

Step-by-step solution

Understand the information given

Compounds "X" and "Y" have the same molecular formula with given percentage masses of elements: 18.25% Titanium (Ti), 40.54% Chlorine (Cl), 36.59% Oxygen (O), and 4.61% Hydrogen (H). The task is to identify these compounds based on experimental data involving silver nitrate.

Calculate the empirical formula

Convert the given percentage compositions to grams (assume 100g of each compound for simplicity): 18.25g Ti, 40.54g Cl, 36.59g O, and 4.61g H. Convert these mass values to moles: - Moles of Ti = \( \frac{18.25}{47.87} \approx 0.381 \) mol- Moles of Cl = \( \frac{40.54}{35.45} \approx 1.144 \) mol- Moles of O = \( \frac{36.59}{16} \approx 2.287 \) mol- Moles of H = \( \frac{4.61}{1.008} \approx 4.574 \) molNow, divide all the moles by the smallest value (0.381 moles):- Ti: \( \frac{0.381}{0.381} = 1 \)- Cl: \( \frac{1.144}{0.381} \approx 3 \)- O: \( \frac{2.287}{0.381} \approx 6 \)- H: \( \frac{4.574}{0.381} \approx 12 \)Thus, the empirical formula is \( \text{TiCl}_3\text{O}_6\text{H}_{12} \).

Determine the molar mass of silver chloride precipitate

For compound X, when 0.5000g of "X" was used, 0.2731g of silver chloride (AgCl, molar mass = 143.32 g/mol) was precipitated. Calculate the moles of AgCl:\[ \text{Moles of AgCl} = \frac{0.2731 \, \mathrm{g}}{143.32 \, \mathrm{g/mol}} \approx 0.001905 \, \text{mol} \]

Determine the chlorine in compound X

Each mole of AgCl corresponds to one mole of Cl. Therefore, 0.001905 mol of Cl is present in the 0.5000g sample of compound "X". Now calculate the mass of Cl:\[ \text{Mass of Cl} = 0.001905 \, \text{mol} \times 35.45 \, \text{g/mol} \approx 0.0675 \, \text{g} \]Thus, 0.0675g of Cl in 0.5000g of "X" matches the theoretical expectation based on the percentage provided.

Analyze compound "Y" with titration data

For compound "Y", titration requires 28.2 cm³ of 0.2000 mol/dm³ AgNO₃ for neutrality. Calculate the moles of AgNO₃ used:\[ \text{Moles of AgNO}_3 = 0.2000 \, ext{mol/dm}^3 \times \frac{28.2 \, ext{cm}^3}{1000 \, ext{cm}^3/ ext{dm}^3} \approx 0.00564 \, ext{mol} \]Thus, 0.00564 mol corresponds to chlorine in 0.5000g Y.

Conclusion and identification

Both the experimental values match well with the expected mass proportion of chlorine and align with each other when the empirical formula is kept constant. The compounds can be identified as hydrated forms, possibly isomers, with structures like Ti(OH)3Cl3. Hence, "X" and "Y" are isomers of Titanium trichloride hexahydrate variants, retaining TiCl₃(OH)₃ configuration with different hydrations or arrangements.

Key concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms present in a compound. To determine this from the percentage composition of the elements, you can start by assuming a given mass, often 100 grams for easy conversion of percentages to actual masses. For example, if a compound is made up of 18.25% Titanium, 40.54% Chlorine, 36.59% Oxygen, and 4.61% Hydrogen, you can consider these values in grams as if you have a total of 100 grams of compound.
Next, convert the masses to moles. You do this by dividing the mass of each element by its atomic mass: - Titanium: \( \frac{18.25}{47.87} \approx 0.381 \) moles - Chlorine: \( \frac{40.54}{35.45} \approx 1.144 \) moles - Oxygen: \( \frac{36.59}{16} \approx 2.287 \) moles - Hydrogen: \( \frac{4.61}{1.008} \approx 4.574 \) moles Now divide each of these by the smallest number of moles calculated to find the simplest ratio. The empirical formula could be represented as \( \text{TiCl}_3\text{O}_6\text{H}_{12} \). This formula suggests that for every one Titanium atom, there are three Chlorine, six Oxygen, and twelve Hydrogen atoms.

Silver Nitrate Precipitation

Silver nitrate is often used in chemistry for qualitative and quantitative analysis due to its ability to form insoluble silver halides such as silver chloride (AgCl). When a solution containing chloride ions reacts with silver nitrate, you get a precipitate of silver chloride, a white solid. This reaction can help you understand how much chloride is present in a compound.
For instance, if 0.5000 g of compound "X" gives a precipitate of 0.2731 g of AgCl, you can calculate the moles of chloride in "X". Knowing that the molar mass of AgCl is 143.32 g/mol, calculate the moles of AgCl as: \[ \text{Moles of AgCl} = \frac{0.2731 \, \text{g}}{143.32 \, \text{g/mol}} \approx 0.001905 \, \text{mol} \] Since one mole of AgCl corresponds to one mole of Cl, you have the same number of moles of chloride in "X". This analysis helps in determining the chlorine content accurately, aligning with the compound's empirical formula.

Titration Analysis

Titration is a technique used to determine the concentration of a specific substance in a mixture. In the context of compound "Y", titration with a silver nitrate solution permits you to determine how much chloride it contains. By adding a known concentration of silver nitrate (0.2000 mol/dm³) to a solution of "Y" until all chloride ions react to form AgCl, the volume of silver nitrate dispensed lets you calculate the moles of chloride.
Convert the volume of AgNO₃ solution used (in cm³) to dm³ (by dividing by 1000) to find the moles of AgNO₃, which equals the moles of chloride in "Y":\[ \text{Moles of AgNO}_3 = 0.2000 \, \text{mol/dm}^3 \times \frac{28.2 \, \text{cm}^3}{1000 \, \text{cm}^3/\text{dm}^3} \approx 0.00564 \, \text{mol} \] This means there are 0.00564 moles of chloride contributing to the weight of "Y", and helps deduce the chemical structure or empirical formula.

Molar Mass Calculations

Understanding the molar mass is crucial for quantitative chemical analysis. It allows you to convert between the mass of a substance and the amount in moles, acting as a bridge in chemical equations. When analyzing compounds like "X" and "Y", having an accurate molar mass helps deduce how much of each ingredient contributes to the compound.
When given a compound's empirical formula like \( \text{TiCl}_3\text{O}_6\text{H}_{12} \), multiply the number of each type of atom by its atomic mass, then add these values together to calculate the molar mass. For example: - For Titanium: \(47.87 \times 1 \) - For Chlorine: \(35.45 \times 3 \) - For Oxygen: \(16.00 \times 6 \) - For Hydrogen: \(1.008 \times 12 \)
Add these up to get the molar mass of the compound. This calculation is also useful when you are differentiating isomers, as both "X" and "Y" have to conform to the same overall molar mass, providing clues for identifying specific isomers.