A compound containing sulfur,nitrogen,and hydrogen has a composition of 86.6% sulfur,12.6% nitrogen,and 0.91% hydrogen by mass.A solution prepared by dissolving 3.130 g of this compound in 25.00 g of bromoform,CHBr₃,produces a solution that freezes at -0.31°C.What is the molecular formula of the compound? (K_f for bromoform is 14.4°C/m,and pure bromoform has a freezing point of 7.8°C.) A)SNH B)S₃N₂H₂ C)S₆N₂H₂ D)S₃NH E)S₂N₂H₂

Question

Quizplus · Accepted Answer

1. Calculate the molar mass of the compound:

- % mass of sulfur = 86.6%, so assume 86.6 g of sulfur in a 100 g sample of the compound 
- % mass of nitrogen = 12.6%, so assume 12.6 g of nitrogen in a 100 g sample of the compound 
- % mass of hydrogen = 0.91%, so assume 0.91 g of hydrogen in a 100 g sample of the compound

- Then, calculate the number of moles for each element:

- Moles of Sulfur = 86.6 g / 32.06 g/mol = 2.698 mol 
- Moles of Nitrogen = 12.6 g / 14.01 g/mol = 0.899 mol 
- Moles of Hydrogen = 0.91 g / 1.01 g/mol = 0.902 mol

2. Find the empirical formula:

- Divide each of the mole values by the smallest mole value (which is the number of moles of nitrogen, 0.899):

- Moles of Sulfur = 2.698 mol / 0.899 mol = 3.00 
- Moles of Nitrogen = 0.899 mol / 0.899 mol = 1.00 
- Moles of Hydrogen = 0.902 mol / 0.899 mol = 1.00

- The empirical formula is therefore SNH3.

3. Determine the molecular formula:

- To find the molecular formula, we need to know the molecular mass of the compound. We can calculate this by adding the molar masses of the elements in the empirical formula:

- Molar mass of Sulfur = 32.06 g/mol 
- Molar mass of Nitrogen = 14.01 g/mol 
- Molar mass of Hydrogen = 1.01 g/mol

- Molecular mass = (3 x 32.06 g/mol)+(1 x 14.01 g/mol)+(3 x 1.01 g/mol) = 128.23 g/mol

- Now, we can find the multiple that we need to multiply the empirical formula by, in order to get the molecular formula:

- Molecular mass of compound / empirical formula mass = multiple 
- Multiple = 128.23 g/mol / 31.03 g/mol = 4.13

- Molecular formula = empirical formula x multiple 
- Molecular formula = SNH12

4. Calculate the new freezing point:

- ΔT = Kf x molality of solution 
- Molality of solution = moles of solute / kg of solvent

- We are given that the compound was dissolved in 25.00 g of bromoform, which has a Kf value of 14.4 °C/m.

- First, calculate the moles of solute:

- Moles of compound = 3.130 g / 128.23 g/mol = 0.0244 mol

- Then, calculate the molality of the solution:

- Molality = 0.0244 mol / 0.025 kg = 0.976 m

- Finally, calculate the freezing point depression:

- ΔT = 14.4 °C/m x 0.976 m = 14.05 °C

- This means that the freezing point of the solution is lowered by 14.05°C.

- The freezing point of pure bromoform is 7.8°C, so the new freezing point of the solution can be calculated as:

- New freezing point = 7.8°C - 14.05°C = -6.25°C

5. Compare the empirical formula mass to the molar mass of the compound:

- Empirical formula mass: 32.06 + 14.01 + 3(1.01) = 31.03 g/mol 
- Molecular formula mass: 128.23 g/mol

- The molecular formula mass is more than 4 times the empirical formula mass, which matches our multiple calculated above.

- Therefore, the molecular formula is SS16N12H12, which simplifies to SS4N3H3.

- The correct answer is C.

A Compound Containing Sulfur,nitrogen,and Hydrogen Has a Composition of 86