A certain non-volatile, non-electrolyte contains 40% carbon, 6.7% hydrogen and 53.3% oxygen. Aqueous solution of this solute containing 5.00% by mass of the solute boils at 100.1 5 ∘ C . The ratio of

Question

A certain non-volatile, non-electrolyte contains 40% carbon, 6.7% hydrogen and 53.3% oxygen. Aqueous solution of this solute containing 5.00% by mass of the solute boils at 100.1 5 ∘ C . The ratio of molecular weight to empirical weight of the compound is ( K b ​ = 0.512 )

TestHub · Accepted Answer

Determination of empirical formula:

Relative number of C atoms $= \frac{40}{12} = 3.33$

Relative number of H atoms $= \frac{6.7}{1} = 6.7$

Relative number of O atoms $= \frac{53.3}{16} = 3.33$

Simplest whole ratio of C $= \frac{3.33}{3.33} = 1$

Simplest whole ratio of H $= \frac{6.7}{3.33} = 2$

Simplest whole ratio of O $= \frac{3.33}{3.33} = 1$

Therefore, the empirical formula is CH₂O.

Empirical formula weight $= 30$ .

Determination of molecular formula:

$m = \frac{Δ T _{b}}{K _{b}} = \frac{0.1 5 ^{\circ} C}{0.51 2 ^{\circ} C / m} = 0.29 m$

Now we know that $n = \frac{m}{M}$ .

$0.29 m o l / k g = \frac{50 g /0.950 k g}{M}$

$M = \frac{50 g /0.950 k g}{0.29 m o l / k g} = 180 g / m o l$

Therefore, the value of $n = \frac{M o lwt .}{E m p . f or m u l a wt .} = \frac{180}{30} = 6$ .

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