When 15.5 g of an organic compound known to be 70.58 % c , 5.9 % h , and 23.50 % o by mass is dissolved in 718.1 g of cyclohexane, the freezing point is 3.29 ∘ c . the normal freezing point of cyclohexane is 6.59 ∘ c . what is the molecular formula for the organic compound? assume that the organic compound is a molecular solid and does not ionize in water. kf values for various solvents are given here.

The molecular formula for the organic compound is C₈H₈O₂

Explanation:

70.58 % C , 5.9 % H , and 23.50 % O

This is a percent by mass, so:

in 100 g of compound I have 70.58 g of C, 5.9 g of H and 23.5 g of O.

We need to apply the formula for freezing point depression, the colligative property.

ΔT = Kf . m

Where ΔT means T°F (pure solvent) − T°F (solution)

Kf, the cryoscopic constant and m is molality (mol of solute in 1kg of solvent); Kf for cyclohexane is 20.8 °C/m

6.59°C - 3.29°C = 20.8 °C/m . m

3.3°C = 20.8 °C/m . m

3.3°C /20.8 m/°C = molal

0.158 m/kg

In 1kg of solvent I have 0.158 moles, but I have 718.1 g of cyclohexane

1000 g 0.158 moles of solute

718.1 g 0.114 moles of solute

To find out the molar mass of my compound I have to divide

mass/mol → 15.5 g /0.114 m = 136.04 g/m

Now I have to work with the percent and think this rules of three:

100 g of compound has 70.58 g of C __5.9 g of H __23.5 g of O

136.04 g of compound has ___  96.02 g of C __8.02 g of H __ 32 g of O

Molar mass of C = 12 g/m

Molar mass of H = 1 g/m

Molar mass of O = 16 g/m

g / molar mass = moles

C: 96 g / 12 g/m = 8

H: 8 g / 1 g/m =8

O: 32 g/ 16 g/m = 2

The molecular formula for the organic compound is C₈H₈O₂

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

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