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 = 
∴ Molar mass is given by
Molar mass = 
= 
= 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.
