A gaseous compound is composed of 80.0% carbon and 20.0% hydrogen by mass. Its density is 0.963 g/L. The density of oxygen gas at the same temperature and pressure is 1.032 g/L. Determine the molecular formula of the compound
Take a sample of 100 g. 80g C and 20g H.
mols C= 80/12 = 6.667
mols H= 20/1 = 20
Divide both by the smaller; i.e.,
6.6676/667 = 1
20/6.667 = 3
empirical formula is CH3 and empirical mass is 15.
p*M1 = d1RT and (pressure*molar mass = density*RT
p*M2 = d2RT
M1 is molar mass O2;M2 is molar mass compound.
Divide eqn 1 by eqn 2 to obtain
32/M2 = 1.032/0.963. P, R, T cancel
M2 = 32 x 0.963/1.032 = 29.98 = about 30
Then empirical mass x a number = molarmass
15 x ? = 30 and the number is 2 so
(CH3)2 or C2H6 is the molecular formula.
To determine the molecular formula of the compound, we need to calculate the molar mass and determine the empirical formula first.
1. Start by assuming we have 100 grams of the compound.
- This means we have 80 grams of carbon and 20 grams of hydrogen.
2. Calculate the number of moles for each element.
- Moles = Mass / Molar Mass
Carbon:
Molar mass of carbon = 12.01 g/mol
Moles of carbon = 80 g / 12.01 g/mol = 6.66 mol (approx.)
Hydrogen:
Molar mass of hydrogen = 1.008 g/mol
Moles of hydrogen = 20 g / 1.008 g/mol = 19.84 mol (approx.)
3. Find the ratio of the elements in the compound.
Divide both values by the smaller number of moles (in this case, hydrogen).
Carbon:
6.66 mol / 19.84 mol ≈ 1:2.98
Hydrogen:
19.84 mol / 19.84 mol ≈ 1:1
The ratio indicates that the empirical formula of the compound is CH3.
4. Determine the molar mass of the empirical formula.
Molar mass of CH3 = 12.01 g/mol (C) + 3 * 1.008 g/mol (H) = 15.04 g/mol
5. Now, we need to find the molecular formula of the compound.
To do this, we need to know the molar mass of the actual compound.
We can find it by dividing the density of the compound by the density of the empirical formula, and then multiply it by the molar mass of the empirical formula.
Molar mass of compound = (Density of compound / Density of CH3) * Molar mass of CH3
Density of compound = 0.963 g/L
Density of CH3 = 15.04 g/mol / 22.4 L/mol (molar volume at STP) = 0.671 g/L (approx.)
Molar mass of compound = (0.963 g/L / 0.671 g/L) * 15.04 g/mol ≈ 21.55 g/mol
6. Finally, determine the molecular formula.
Divide the molar mass of the compound by the molar mass of the empirical formula, and then multiply the empirical formula by this factor to get the molecular formula.
Molecular formula = (Molar mass of compound / Molar mass of CH3) * CH3
Molecular formula = (21.55 g/mol / 15.04 g/mol) * CH3
Molecular formula = 1.43 * CH3
Rounding to a whole number, we get the molecular formula C2H6.
Therefore, the molecular formula of the compound is C2H6 (ethane).