An identified covalent molecular compound contains only carbon, hydrogen, and oxygen. When 6.60 mg of this compound is burned, 19.67 mg of CO2 and 3.02 mg of H2O are produced. The freezing point I’m camphor is lowered by 10.4°C when 0.937 g of the compound is dissolved in 15.25 g of camphor.
What is the molecular formula of the unidentified compound ?
Choose appropriate coefficient in a molecular formula below
C H O
To find the molecular formula of the unidentified compound, we first need to determine the empirical formula.
We are given the masses of CO2 and H2O produced when the compound is burned:
- 19.67 mg of CO2
- 3.02 mg of H2O
To find the moles of carbon in CO2, we divide the mass of CO2 by its molar mass:
moles of carbon = 19.67 mg / (44.01 g/mol) = 0.446 mol
To find the moles of hydrogen in H2O, we divide the mass of H2O by its molar mass:
moles of hydrogen = 3.02 mg / (18.015 g/mol) = 0.167 mol
Next, we need to determine the amount of oxygen in the compound. Since the compound contains only carbon, hydrogen, and oxygen, we can calculate the difference:
moles of oxygen = Total moles of element - (moles of carbon + moles of hydrogen)
moles of oxygen = 0.446 mol + 0.167 mol - 0.937 mol/molecular coefficient
Now, let's calculate the molar mass of the compound:
molar mass of the compound = 0.937 g / (0.446 mol + 0.167 mol + moles of oxygen) = 0.937 g / (0.446 mol + 0.167 mol + moles of oxygen)
Now, we need to calculate the molality of the compound in the solution:
molality = moles of compound / kg of solvent
molality = (0.937 g / molar mass of the compound) / (15.25 g camphor / 1000 g)
Finally, we can calculate the molecular formula by dividing the molality by the empirical formula mass (mass of empirical formula - 12.01*g/mol C + 1.008 g/mol H + 16.00 g/mol O).
Let "a" be the coefficient for carbon, "b" be the coefficient for hydrogen, and "c" be the coefficient for oxygen.
The molecular formula is:
CₐHᵦOᶜ