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

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ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.