Uncovering the Mystery of Sulfur-Nitrogen Compound

[needhelpperson]

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of 69.6% S and 30.4% N. Measurements of its molecular mass yield a value of 184.3 g/mol. The compound occasionally detonates on being struck or when heated reapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length.
a) Calculate the empirical formula and molecular formulas for the substance.
b) Predict the bond distances between the atoms in the ring (Note: the S-S bond distance in the S8 rign is 2.05 A)

c The enthalpy of formation of the compound is estimated to be 480 KJ/ mol. The Hf of S(g) is 222.8 KJ/mol. Estimate the average bond enthalpy in the compound.

i did a but i drew the lewis dot diagram out with the S4N4 and it turned out there were 2 double bonds while the rest were single bonds. How does that work if the bond distances in the ring are the same length?

as for c i have no idea how to do this.

i think the formula is this 2 S2(g) + 4 NH3(l) -> S4N4 + 6 H2(g)

I'm given the N-H bond enthalpies as 391 KJ/mol but that's according to gas. NH3 is a liquid. How am i suppose to solve this. Please help. thanks

 

[member 553083]

.a) The empirical formula of the compound is S4N4, while the molecular formula is S8N8. b) The bond distances between the atoms in the ring are all 2.05 Å.c) The average bond enthalpy of the compound can be calculated by taking the difference between its enthalpy of formation and the enthalpy of formation of the reactants: Hf(S8N8) - [Hf(S2(g)) + 4 Hf(NH3(l))] = 480 - (2*222.8 + 4*391) = -24.4 kJ/mol. Therefore, the average bond enthalpy of the compound is -24.4 kJ/mol.

 

[wais]

a) To calculate the empirical formula, we first need to find the moles of sulfur and nitrogen in the compound. Since the compound is composed of 69.6% sulfur and 30.4% nitrogen, we can assume that we have 69.6 grams of sulfur and 30.4 grams of nitrogen. We can then convert these grams to moles by dividing by their respective molar masses (32.06 g/mol for sulfur and 14.01 g/mol for nitrogen).

69.6 g S x (1 mol S/32.06 g S) = 2.17 mol S
30.4 g N x (1 mol N/14.01 g N) = 2.17 mol N

This gives us a ratio of 1:1 for sulfur and nitrogen, so the empirical formula is SN.

To find the molecular formula, we need to know the molecular mass of the compound. We are given that the molecular mass is 184.3 g/mol, which is close to the sum of the atomic masses of sulfur and nitrogen (32.06 g/mol + 14.01 g/mol = 46.07 g/mol). This suggests that the molecular formula is S4N4, which has a molecular mass of 184.2 g/mol. Therefore, the molecular formula is also S4N4.

b) The bond distances in the ring are the same because all the bonds are single bonds. In a ring structure, the bonds are typically equivalent and have the same length. The S-S bond distance in the S8 ring is 2.05 A, so we can predict that the bond distances between the sulfur and nitrogen atoms in the ring of S4N4 will also be around 2.05 A.

c) To estimate the average bond enthalpy in the compound, we can use the bond enthalpies of the reactants and products in the reaction given. We are given that the enthalpy of formation of S(g) is 222.8 KJ/mol, so we can assume that the reaction is exothermic and that the bond enthalpy of the S-S bond is less than 222.8 KJ/mol. We are also given that the bond enthalpy of N-H is 391 KJ/mol, but this is for gas phase NH3. Since the reaction takes place in liquid ammonia, the bond enthalpy will be different. To