you are given samples of two pure compounds, one of which is a gas and one of which is a liquid at room temperature. by elemental analysis, both compounds are found to have the following composition (by mass):
52% C
13% H
35% O
(a) Determine the empirical formula of both compounds.
(b) The molar mass of both compounds is determined to be 46.0g/mol. Name one method for calculating the molar mass of a pure gas and one method for calculating the molar mass of a pure liquid.
Take a 100 g sample which will give you
52 g C
13 g H
25 g O
Convert each to moles.
52/12.01 = ??
13/1.008 = ??
35/16 = ??
Now you want to find the whole number ratios of these values. The easiest way to do that is to divide the smallest number by itself. That assures you of getting 1.00000 for that number. Then divide the other numbers by the same small number and round all to whole numbers. That will give you the empirical formula.
You can estimate the molar mass of a liquid by using the ebullioscopic method (boiling point elevation method). For a gas one can use the Victor Meyer method.
To determine the empirical formula for both compounds, we need to calculate the simplest ratio of atoms present in each compound. Let's call the gas compound X and the liquid compound Y.
(a) To determine the empirical formula:
1. Assume we have 100g of each compound. This allows us to work with mass percentages directly.
2. Convert the mass percentages to grams.
- Compound X:
- C: 52g (52% of 100g)
- H: 13g (13% of 100g)
- O: 35g (35% of 100g)
- Compound Y:
- C: 52g (52% of 100g)
- H: 13g (13% of 100g)
- O: 35g (35% of 100g)
3. Convert the grams of each element to moles.
- Compound X:
- C: 52g / molar mass of C (12.01 g/mol) = 4.33 moles of C
- H: 13g / molar mass of H (1.01 g/mol) = 12.87 moles of H
- O: 35g / molar mass of O (16.00 g/mol) = 2.19 moles of O
- Compound Y:
- C: 52g / molar mass of C (12.01 g/mol) = 4.33 moles of C
- H: 13g / molar mass of H (1.01 g/mol) = 12.87 moles of H
- O: 35g / molar mass of O (16.00 g/mol) = 2.19 moles of O
4. Divide each mole value by the smallest mole value obtained.
- Compound X:
- C: 4.33 moles of C / 2.19 moles of O ≈ 1.98
- H: 12.87 moles of H / 2.19 moles of O ≈ 5.88
- O: 2.19 moles of O / 2.19 moles of O = 1.00
- Compound Y:
- C: 4.33 moles of C / 2.19 moles of O ≈ 1.98
- H: 12.87 moles of H / 2.19 moles of O ≈ 5.88
- O: 2.19 moles of O / 2.19 moles of O = 1.00
The empirical formula for both compounds, based on the simplified ratio of atoms, is approximately:
Compound X: C₂H₆O
Compound Y: C₂H₆O
(b) Method for calculating the molar mass of a pure gas:
One method is to measure the gas's density and use the ideal gas law. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (K). By measuring the pressure, volume, and temperature of the gas, we can solve for the number of moles (n). Then, by dividing the mass of the gas by the number of moles, we can calculate the molar mass of the gas.
Method for calculating the molar mass of a pure liquid:
One method is to measure the boiling point elevation or freezing point depression of the liquid. Both of these properties depend on the concentration of solute particles in the liquid. By measuring the change in boiling or freezing point and using the equation ΔT = K·m·i, where ΔT is the change in temperature, K is the cryoscopic/molal constant, m is the molality, and i is the van't Hoff factor, we can calculate the molality (m) of the liquid solution. From there, we can use the equation m = (moles of solute) / (mass of solvent in kg) to calculate the number of moles of solute. Finally, by dividing the mass of solute by the number of moles, we can determine the molar mass of the pure liquid compound.