Question

3. An elemental analysis of an organic compound showed the following results: C = 82.66 %, H= 17.34 %. A 1.34 g of that compound was vaporized and was found to occupy 520 mL at -1 °C and 750 torr. What is the molecular weight and molecular formula of the compound? Compute the approximate density of ethane, CH3CH3, at 27°C and 3.00 atm. The molecular weight of ethane is 30.07. If the density of a gas is 1.7 g/L at 27°C and 1.5 atm, what is its molecular mass? PMM =D.RT

Answer 1

Empirical formula of hydrocarbon :

Let 100g of hydrocarbon is taken , moles of C = 82.66/12 = 6.888333

Moles of hydrogen. = 17.34/1 = 17.34

Simplest ratio of C: H = 1 : 2.517 = 2 : 5

Empirical formula = (C₂H₅)_n

The number of moles = PV/RT = (750/760)atm×0.520L/(0.0821×272) = 0.02298mol

Molecular mass = mass/number of mole = 1.34g/0.02298mol = 58.31 (Answer)

n = 58.31/29 = 2

The molecular formula of compound is C₄H₁₀ .

-------------

PM = d.R.T

Density = 3.00atm × 30.07g/mol /(0.0821atm-L/K.mol × 300K)

= 3.66 g/L

--------------------

PM = d.R.T

M = d.R.T /P = 1.7g/L × 0.0821atm-L/K.mol × 300K /1.5atm

= 27.914 g/mol

= 28 g/mol. (Answer)