Answer:
[tex]B_3H_6N_3[/tex]
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:
[tex]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[/tex]
Next, we divide each moles by the fewest ones (3.73 mol) in order to find the subscript in the empirical formula first:
[tex]B:\frac{3.73}{3.73}=1 \\\\H:\frac{7.44}{3.73}=2\\\\N:\frac{3.73}{3.73}=1[/tex]
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:
[tex]B_3H_6N_3[/tex]
Best regards!
