Question

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

Answer 1

Answer:

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

Answer 2

Answer:

The molecular formula of the compound is COCl₂

Explanation:

Here we have

Mass of carbon in first sample = 0.305 g

Mass of oxygen in first sample = 0.407 g

Mass of chlorine in first sample = 1.805 g

At , Standard Temperature and pressure (STP) we have

P₂ = 760 mmHg = 101325 Pa

T₂ = 0 °C = 273 K

V₂ = 2 L = 0.002 m³

Therefore, from the universal gas equation, PV = nRT we have

n = PV/(RT)

Where:

R = Universal gas constant = 8.31451 J/gmol·K

Therefore,

Number of moles in the 8.84 g sample =  P₂V₂/(RT₂)

n = (101325 × 0.002)/(8.31451 × 273) = 8.923×10⁻² moles

Since 8.84 g = 8.923×10⁻² moles of the compound, we have

Number of moles = $\frac{Mass}{Molar \, Mass}$ ∴ Molar mass  is given by

Molar mass = $\frac{Mass}{Number \, of \, Moles}$ = $\frac{8.84}{8.923 \times 10^{-2}}$ = 99.07 grams

The mole ratio of the elements of the compound is given as;

Carbon, Molar mass = 12.017 g/mol

Number of moles in 0.305 g = 0.305/12.017 = 2.54×10⁻² moles

Similarly, oxygen O → 0.407/16 = 0.0254 moles

For chlorine Cl → 1.805/35.453 = 0.0509 moles  

Total number of moles = 0.1017 moles

Therefore

C = 0.0254 moles  

O = 0.0254 moles

Cl = 0.0509 moles

Divided by th smallest mole amount we have

C₁O₁Cl₂ or COCl₂

Checking Molar mass of C = 12.01, O = 16 and Cl₂ = 70.906

12.01 + 16 + 70.906 = 99

However, where the molar mass is calculated as 99.07 g we have the molecular formula as

COCl₂.