The correct answer would be C. a precipitate
Hope this helped!
I can’t see it can you write the problem ?
Answer:
![B_3H_6N_3](https://tex.z-dn.net/?f=B_3H_6N_3)
Explanation:
Hello there!
In this case, since the mass percentages in a compound which is wanted to know the molecular formula, can be assumed to be the masses, we first need to compute the moles they have in the formula unit:
![n_B=40.29gB*\frac{1molB}{10.811gB} =3.73molB\\\\n_H=7.51gH*\frac{1molH}{1.01gH} =7.44molH\\\\n_N=52.20gN*\frac{1molN}{14.01gN} =3.73molN](https://tex.z-dn.net/?f=n_B%3D40.29gB%2A%5Cfrac%7B1molB%7D%7B10.811gB%7D%20%3D3.73molB%5C%5C%5C%5Cn_H%3D7.51gH%2A%5Cfrac%7B1molH%7D%7B1.01gH%7D%20%3D7.44molH%5C%5C%5C%5Cn_N%3D52.20gN%2A%5Cfrac%7B1molN%7D%7B14.01gN%7D%20%3D3.73molN)
Next, we divide each moles by the fewest ones (3.73 mol) in order to find the subscript in the empirical formula first:
![B:\frac{3.73}{3.73}=1 \\\\H:\frac{7.44}{3.73}=2\\\\N:\frac{3.73}{3.73}=1](https://tex.z-dn.net/?f=B%3A%5Cfrac%7B3.73%7D%7B3.73%7D%3D1%20%5C%5C%5C%5CH%3A%5Cfrac%7B7.44%7D%7B3.73%7D%3D2%5C%5C%5C%5CN%3A%5Cfrac%7B3.73%7D%7B3.73%7D%3D1)
Then, the empirical formula is BH2N whose molar mass is 26.83 g/mol, so the ratio of molecular to empirical is 80.50/26.83=3; therefore, the molecular formula is three times the empirical one:
![B_3H_6N_3](https://tex.z-dn.net/?f=B_3H_6N_3)
Best regards!
I assume C₃H₃O.
Hope I helped! Tell me if I'm wrong!
B , your products are on the right side of the reaction. The reactants are on the left side