Q85E - Chemistry (OpenStax)

Question

A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B₂O₃. What are the empirical and molecular formulas of the compound?

 

Step-by-Step Solution

Answer

Empirical formula = B₁H₃

Molecular formula =  B₂H₆

Information given

We can write the equation

 

B_x H_y   + O₂→B₂O₃ +   H₂O

0.025 g                     0.063 g

 

0.063 g of B₂O₃  x    1 mol of B₂O₃   x       2 mol B  x    10.81g of B

                                 69.62g of B₂O₃    1 mol B₂O₃      1 mol B    

 

=   0.01956 g of B

 

B_x H_y   =    0.025 g

 

So g of H =   0.025 g – 0.01956 g =   0.00544 g of H

 

Required formula

%    Grams \( \to \)      mols   \( \to \)     mol ratio   \( \to \)    empirical formula   \( \to \)      molecular formula

0.01956 g of B x (1 mol B/10.81 g mol B) = (0.0081 mol B/0.0081) = 1 mol B

 

 

0.00544 g of H   x (1 mol H/1g mol H) = (0.00540 mol H/0.0081)   =    3 mol H

Empirical formula = B₁H₃

Molecular mass/ Empirical mass = 28/ 13.83 = 2

Molecular formula = B₂H₆