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:
C) C₁₄H₁₁O₆
Explanation:
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)
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
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).
