A solution contains 11.70 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -7.01 ∘C. The mass percent composition of the compound is 38.70% C, 9.74% H, and the rest is O.
What is the molecular formula of the compound?
Express your answer as a molecular formula
So far I got CH3O as my empirical, but I can't seem to figure out how to find the molecular formula
Let the molality of the solution be m
Given,
Freezing point = -7.01 degree C
=> Depression in freezing point = delta Tf = 7.01 degree C
Kf for water = 1.86
We know that,
delta Tf = Kf x m
=> 7.01 = 1.86 x m
=> m = 3.77
The molality of the solution is 3.77 m
Molality = Moles of solute / Mass of solvent (in kg)
Mass of water = 50 x 1 = 50 g = 0.05 kg
=> 3.77 = Moles of unknown compound / 0.05
=> Moles of unknown compound = 3.77 x 0.05 = 0.1884 moles
Mass of unknown compound = 11.7 g
=> Molar Mass of unknown compound = 11.7 / 0.1884 = 62.1 g / mol
Given,
The mass percent composition of the compound is 38.70% C, 9.74% H, and the rest is O.
Assuming 100 g of compound
Mass of C = 38.7 g
=> Moles of C = 38.7 / 12 = 3.225
Mass of H = 9.74
=> Moles of H = 9.74 / 1 = 9.74 moles
Mass of O = 51.56 g
=> Moles of O = 51.56 / 16 = 3.2225
Dividing each moles by 3.225, we get
Moles of C = 1
Moles of H = 3
Moles of O = 1
=> The emperical formula of the compound is CH3O
Emperical Mass = 12 + 3 + 16 = 31
n = Molar Mass / Emperical Mass = 62.1 / 31 = 2
=> Molecular Formula = C2H6O2
A solution contains 11.70 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume...
A solution contains 10.35 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.26 ∘C. The mass percent composition of the compound is 60.98% C, 11.94% H, and the rest is O. What is the molecular formula of the compound?
When 0.60 grams of a non-electrolyte unknown compound is dissolved in 10 grams of water, the freezing point of the solution is -1.86o C. Given that the freezing point of pure water is 0o C and that the freezing-point depression constant for water is 1.86 o C/mol/kg, calculate the molecular mass (g/mol) of the unknown compound.
The freezing point of a solution that contains 1.00 g of an unknown compound, (A), dissolved in 10.0 g of benzene is found to be 2.17 oC. The freezing point of pure benzene is 5.48 oC. The molal freezing point depression constant of benzene is 5.12 oC/molal. What is the molecular weight of the unknown compound?
A solution containing 1.00 g of an unknown non-electrolyte liquid and 9.00 g water has a freezing point of -3.33 oC. The Kf = 1.86 oC/m for water. Calculate the molar mass of the unknown liquid, in g/mol.
A solution is prepared by dissolving 0.107 g of an unknown non-electrolyte in 88.1 g of camphor. The freezing point of the solution was measured to be 178.0�C. What is the molecular weight of the compound? (The freezing point of pure camphor is 179.5�C and kf= 40�C/m)?
39.636 g of an unknown compound X is dissolved in 170.86 mL. What is the molarity of X? Here is some information to determine the identity of X. When 1.00g of X is dissolved in water and allowed to react wtih AgNO3, all the chlorine in X precipitates and 1.95g of AgCl is collected. When 1.00g of X undergoes complete combustion, 0.900 g of CO2 and 0.735 g of H2O were collected. The empirical and molecular formula are identical. Compound...
0.2650 g of a compound of unknown molecular mass were dissolved in 18.00 mL of a non-ionizing solvent with specific gravity of 0.7480. The pure solvent was determined to have a freezing point of 6.80°C. The freezing point of the solution was determined graphically. Trial one yielded a freezing point of 5.31°C for the solution. Trial two indicated the freezing point to be 5.23°C. Kf (solvent) = 12.8 C°/m a. Calculate the molecular mass of the unknown. b. If 5.50...
Calculate the freezing point of a solution that contains 5.42 g of glucose, a non-electrolyte, dissolved in 102 g of water.
When 9.31 g of an unknown non-electrolyte is dissolved in 50.0 g of benzene, the boiling point increased by 3.16 degrees C. If the Kbp of the solvent is 2.53 K/m, calculate the molar mass of the unknown solute. The answer is 149 ± 2% looking for explanation how to work this problem
3a. Calculate the molar mass (in g/mol) of an unknown 1:1 electrolyte if 0.482 g dissolved in 223.1 mL of water at 74.75 °C has an osmotic pressure of 54.4 mmHg. R = 0.082058 L⋅atm⋅mol−1⋅K−1. 1.00 atm = 760 mmHg. Report your answer to THREE significant figures. 3b. Calculate the required mass of an unknown nonelectrolyte (ℳ = 131.5599 g/mol) dissolved in 140.1 g of solvent that gives a solution that boils at 36.04 °C. The boiling point of the...