Question

A compound has the empirical formula . a 256-ml flask, at 373 k and 750. torr, contains 0.527 g of the gaseous compound. give the molecular formula.

Answer 1

The empirical formula of the compound is BH3, and the molecular formula is B₂H₆.

Step-by-step explanation:

The empirical formula of a compound is obtained by determining the simplest whole-number ratio of the atoms present. The given compound has the empirical formula BH3. To find the molecular formula, we need to compare the molar mass of the empirical formula to the experimental molar mass.

To calculate the molar mass of BH3, we add up the atomic masses of the elements: B(10.81 g/mol) + H(1.01 g/mol) × 3 = 12.84 g/mol. The experimental molar mass is calculated using the ideal gas law: PV = nRT. Rearranging the formula to isolate n and substituting the given values, we have n = PV / RT. Plugging in the corresponding values, we find n = (0.750 atm × 0.256 L) / (0.0821 L atm mol^-1 K^-1 × 373 K) = 0.0217 mol.

Dividing the experimental molar mass (0.527 g) by the molar mass of the empirical formula (12.84 g/mol) gives us the number of empirical formula units present in one molecule of the compound: 0.527 g / 12.84 g/mol ≈ 0.041 moles. Since the result is close to 2, we multiply the empirical formula by 2 to give the molecular formula: B₂H₆.

Answer 2

Answer: From the ideal gas law, MM=mRTPV; where MM = molecular mass; m = mass; P = pressure in atmospheres; V= volume in litres; R = gas constant with appropriate units. So, 0.800â‹…gĂ—0.0821â‹…Lâ‹…atmâ‹…Kâ’1â‹…molâ’1Ă—373â‹…K0.256â‹…LĂ—0.987â‹…atm = 97.0 gâ‹…molâ’1. nĂ—(12.01+1.01+2Ă—35.45)â‹…gâ‹…molâ’1 = 97.0â‹…gâ‹…molâ’1. Clearly, n = 1. And molecular formula = C2H2Cl2. I seem to recall (but can't be bothered to look up) that vinylidene chloride, H2C=C(Cl)2 is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required.