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'

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

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

the answer is na (sodium)