Two liquids are analyzed and found to both be 85.7% carbon and 14.3% hydrogen. At 750 mmHg and 150 C, both are gases. At these conditions, an 800.0 mL bottle will hold 1.60 g of Compound A, while the same bottle will hold 2.22 g of Compound B. Find the molecular formulas of both compounds A and B. (Hint: use percent composition to find the empirical formula, and then use the Ideal gas law to find moles of each.) Enter your answer for both molecular formulas like this: XY, X2Y

Molecular formula A: C₅H₁₀

Molecular formula B: C₇H₁₄

Explanation:

It is possible to obtain empirical formula of compounds using percent composition, thus:

C: 85.7% × (1mol / 12.01g) = 7.136 moles C

H: 14.3% × (1mol / 1.01g) = 14.158 moles H

Mole ratio of H:C is:

14.158mol / 7.136mol = 2

That means in compounds A and B you have 2 hydrogens per atom of carbon and empirical formila is:

CH₂

Using PV = nRT, moles of A and B are:

Where P is pressure (750mmHg / 760 = 0.987atm), V is volume (0.8000L), R is gas constant (0.082atmL/molK), and T is temperature (150°C +273.15 = 423.15K)

Moles A and B: n = PV / RT

n = 0.987atm×0.8000L / 0.082atmL/molK×423.15K

n = 0.0228 moles of A and B.

Using the mass of A and B it is possible to find molar mass of each compound:

A = 1.60g / 0.0228mol = 70.31g/mol

B = 2.22g / 0.0228mol = 97.56g/mol

As empirical formula of both compounds is CH₂, (molar mass = 14.03g/mol). Molecular formula of compounds is:

A = 70.31g/mol / 14.03g/mol = 5 → Molecular formula: 5×CH₂ = C₅H₁₀

B = 97.56g/mol / 14.03g/mol = 7 → Molecular formula: 7×CH₂ = C₇H₁₄

56 moles of   lieter in chemistry

un recipient conține 6,4 moli de gaz. gazul hidrogen reprezintă 25% din totalul kilometrilor din container. în cazul în care presiunea totală este de 1.24 atm ceea ce este