Question

35a

When 4.752 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 16.06 grams of CO₂ and 3.289 grams of H₂O were produced.
In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

35 part b

A 4.079 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 9.273 grams of CO₂ and 3.797 grams of H₂O are produced.
In a separate experiment, the molar mass is found to be 116.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

35 part c

A 10.64 gram sample of chromium is heated in the presence of excess chlorine. A metal chloride is formed with a mass of 32.40 g. Determine the empirical formula of the metal chloride.

Answer 1

A) % of C = 12 * 16.06 * 100 / 44 * 4.752 = 92.172 %

% of H = 2 * 3.289 * 100 / 18 * 4.752 = 7.69 %

Moles of C = 92.172/12 = 7.69

Moles of H = 7.69/1 = 7.69

Mole ratio of C and H

C : H = 7.69 : 7.69 = 1 : 1

So, empirical formula = C1H1 = CH

Empirical formula mass = 12*1 + 1*1 = 13 g/mol

n = 26.04 / 13 = 2

So, molecular formula = (CH)n = (CH)2 = C2H2 ... Answer

(B) % of C = 12 * 9.273 * 100 / 44 * 4.079 = 62 %

% of H = 2 * 3.797 * 100 / 18 * 4.079 = 10.343 %

% of O = 100 - (62 + 10.343) = 27.657 %

Moles of C = 62/12 = 5.167

Moles of H = 10.343/1 = 10.343

Moles of O = 27.657 / 16 = 1.7285

Mole ratio,

C : H : O = 5.167 : 10.343 : 1.7285

lowest number is 1.7285, so diving all moles by this number

C : H : O = (5.167/1.7285) : (10.343/1.7285) : (1.7285/1.7285)

C : H : O = 3 : 6 : 1

So, empirical formula = C3H6O

Empirical formula mass = 3*12 + 1*6 + 16 = 58 g/mol

n = 116.2/58 = 2

Molecular formula = (C3H6O)n = (C3H6O)2 = C6H12O2 .... Answer

C) Metal chloride will be Chromium chloride. All chromium is converted to Chromium chloride as Chlorine is in excess.

So, moles of Cr in sample = mass/molar mass = 10.64/52 = 0.205 moles

Mass of Cl in metal chloride = 32.4 - 10.64 = 21.76 g

Moles of Cl = 21.76/35.5 = 0.613

Mole ratio of Cr and Cl

Cr : Cl = 0.205 : 0.613 = 1 : 3

Hence, empirical formula = CrCl3 .... Answer