A hydrocarbon contains 85.7% carbon and the remainder

ishydroge. Its molar mass is 70g/mol. Determine the empirical

andmolecular formulas of the compound.

2NaBH4 (s) + I2 (s) - > B2H6 (g) + 2NaI (s) + H2 (g)

Suppose you use 1.203g NaBH4 and excess iodine, and youisolate

0.295g B2H6. What is the percent yield of B2H6?

CH₂; 67.1 %

Explanation:

To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation

Assume 100 grams of the compound.

# mol C = 85.7 g / 12.01 g/mol = 7.14 mol

# mol H = 14.3 g / 1.008 g/mol = 14.19 mol

The proportion is 14.9 mol H / 7.14 mol C = 2 mol H / 1 mol C

So the empirical formula is CH₂

For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.

We need to calculate the moles of NaBH₄ (M. W = 37.83 g/mol)

1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol

Theoretical yield from balanced chemical equation:

0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆

Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g / mol = 0.440 g

% yield = 0.295 g / 0.440 g x 100 = 67.1 %