Chemistry, 27.09.2019 02:30 tateandvioletAHS14AY

Analysis of a gaseous chlorofluorocarbon, cclxfy, shows that it contains 11.79% c and 69.57% cl. in another experiment, you find that 0.107 g of the compound fills a 458-ml flask at 25 °c with a pressure of 21.3 mm hg. what is the molecular formula of the compound?

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Answer from: ehaynie

The molecular formula = $C_2Cl_{4}F_2$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 11.79

Molar mass of C = 12.0107 g/mol

% moles of C = 11.79 / 12.0107 = 0.9816

% of Cl = 69.57

Molar mass of Cl = 35.453 g/mol

% moles of Cl = 69.57 / 35.453 = 1.9623

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

% moles of F = 18.64 / 18.998 = 0.9812

Taking the simplest ratio for C, Cl and F as:

0.9816 : 1.9623 : 0.9812

= 1 : 2 : 1

The empirical formula is = CCl_2F

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25 + 273.15) K = 298.15 K

Volume = 458 mL = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 0.00052445 moles

Given that :

Amount = 0.107 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.00052445= \frac{0.107\ g}{Molar\ mass}$

$Molar\ mass= 204.0233\ g/mol$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

The molecular formula = $C_2Cl_{4}F_2$

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Analysis of a gaseous chlorofluorocarbon, cclxfy, shows that it contains 11.79% c and 69.57% cl. in...

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