A sample of a gas contains 0.305 g of carbon, 0.407 g of oxygen and 1.805 g of chlorine. A different sample of the same gas having a mass of 8.84 g occupies 2 L at STP. What is the molecular formula for the compound

The molecular formula of the compound containing 0.305 g of carbon, 0.407 g of oxygen and 1.805 g of chlorine is COCl₂

We'll begin by calculating the empirical formula of the compound. This can be obtained as follow:

C = 0.305 g

O = 0.407 g

Cl = 1.805 g

Divide by their molar mass

C = 0.305 / 12 = 0.0254

O = 0.407 / 16 = 0.0254

Cl = 1.805 / 35.5 = 0.0508

Divide by the smallest

C = 0.0254 / 0.0254 = 1

O = 0.0254 / 0.0254 = 1

Cl = 0.0508 / 0.0254 = 2

Thus, the empirical formula of the compound is COCl₂

Next, we shall determine the number of mole of the gas that occupies 2 L at STP.

At standard temperature and pressure (STP),

22.4 L = 1 mole

Therefore,

2 L = 2 / 22.4

2 L = 0.089 mole

Thus, the number of mole of the gas that occupied 2 L at STP is 0.089 mole.

Next, we shall determine the molar mass of the gas.

Mass of gas = 8.84 g

Mole of gas = 0.089 mole.

Molar mass = mass / mole

Molar mass of gas = 8.84 / 0.089

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

Empirical formula = COCl₂

Molar mass = 99.3 g/mol

Molecular formula = Empirical × n = molar mass

[COCl₂]n = 99.3

[12 + 16 + (2×35.5)]n = 99.3

99n = 99.3

Divide both side by 99

n = 99.3 / 99

Molecular formula = [COCl₂]n  

Molecular formula = [COCl₂] × 1

Therefore, the molecular formula of the compound is COCl₂

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Molecular formula = empirical formula  = COCl2

Explanation:

Step 1: Data given

MAss of carbon = 0.305 grams

MAss of oxygen = 0.407 grams

Mass of chlorine = 1.805 grams

A different sample of the same gas having a mass of 8.84 g occupies 2 L at STP

Atomic mass of C = 12.01 g/mol

Molar mass of O2 = 16.0 g/mol

Molar mass of Cl2 = 35.45 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = 0.305 grams / 12.01 g/mol

Moles C = 0.0254 moles

Moles O = 0.407 grams / 16.0 g/mol

Moles O = 0.0254 moles

Moles Cl = 1.805 grams / 35.45 g/mol

Moles Cl = 0.0509 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.0254 moles / 0.0254 moles = 1

O: 0.0254 moles / 0.0254 moles = 1

Cl: 0.0509 / 0.0254 = 2

The empirical formula is COCl2

Step 4: Calculate moles

At STP we have 22.4 L for 1 mol

For 2 L we have 1 / 11.2 = 0.0893 moles

Step 5: Calculate the molar mass

Molar mass = mass / moles

Molar mass = 8.84 grams / 0.0893 moles

Molar mass = 98.99 g/mol

Step 6: Calculate the molar mass of the empirical formula

COCl2 = 12.01 + 16.0 + 70.9 ≠ 98.91 g/mol

Step 7: Calculate the molecular formula

Molecular formula = empirical formula  = COCl2

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml