Question

An unknown compound is composed of 65.44% C, 29.07% O, and 5.49% H. A sample weighing 5.34 g, when dissolved in 60.00 g H2O, lowers the freezing point to -0.600°C. What is the molecular formula of the compound? (Kf for water = 1.86°C/m; C = 12.0, O = 16.0, H = 1.0 g/mol). A) C7H5O3 B) CHO C) C14H11O6 D) C3H3O E) C5H5O5'

Answer 1

Answer:

The molecular formula of the compound is C14H11O6

Explanation:

To lower the freezing point we have to apply this formula

ΔT = Kf . molality

ΔT = 0,6°C

So ΔT / Kf = molality ( moles of solute in 1kg of solvent)

0,6° / 1,86 m/°C = 0,322 moles

This moles are in 1kg of water, but I dissolved the solute in 60 g so the rule of three will be

1000 g water _____ 0,322 moles

60 g water _______ (60 . 0,322)/ 1000 = 0,01932 moles

This moles are the mass of the weigh I dissolved, 5,34 g

So let's find out the molar mass

0,01932 moles are ____ 5.34 g

1 mol is _____________ (5,34 . 1 )/ 0,01932 = 276.39 g/m

Option C is the answer

C14H11O6 = 12.14 + 11.1 + 16.6 = ≅ 276

Answer 2

Answer:

C) C₁₄H₁₁O₆

Explanation:

• With the percent composition we can determine the empirical formula of the compound:

Assuming there's 100 g of compound, the moles of each element would be:

65.44 g C ÷ 12 g/mol = 5.45 mol C

29.07 g O ÷ 16 g/mol = 1.82 mol O

5.49 g H ÷ 1 g/mol = 5.49 mol H

We divide those values by the lowest among them:

C⇒ 5.45 / 1.82 = 2.99 ≅3

O⇒ 1.82 / 1.82 = 1

H ⇒5.49 / 1.82 = 3.01 ≅3

So the empirical formula is C₃H₃O (55 g/mol)

• With the freezing point decrease we can calculate the compound's  molecular weight

The freezing point decrease can be expressed by the formula:

ΔT = Kf * molality

And using the data given by the problem we can calculate the molality

0.6°C = 1.86 °C/m * molality

molality = 0.322 m

Using the definition of molality we can calculate the moles of the compound present in 5.34 g:

0.322 m = moles / 0.06 kgH₂O

moles = 0.01932 moles Compound

With that we can calculate the molar mass of the compound:

5.34 g / 0.01932 mol = 276.40 g/mol

• Now we divide the molar mass of the compound by the molar mass of the empirical formula:

276.40 / 55 = 5.02 ≅  5

We multiply by a factor of 5 each component in the empirical formula:

C₁₅H₁₅O₅

The closest option is C).