a gas has an empirical formula C6H7O5. When put in a 50mL bulb at 120 degrees celcius and 120kPa (pressure) it's mass is determined: the mass is 10.37 times as much as C6H6O. What is the molecular formula of the gas (C6H7O5)?
To find the molecular formula of the gas, which is represented by the empirical formula C6H7O5, we need to calculate the molar mass of this empirical formula. Then we can compare it to the molar mass determined experimentally.
1. Calculate the molar mass of the empirical formula:
- Carbon (C) has a molar mass of 12.01 g/mol.
- Hydrogen (H) has a molar mass of 1.008 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
Multiply the number of atoms for each element by their respective molar masses:
(6 C atoms * 12.01 g/mol) + (7 H atoms * 1.008 g/mol) + (5 O atoms * 16.00 g/mol) = 90.09 g/mol
The molar mass of the empirical formula C6H7O5 is 90.09 g/mol.
2. Determine the molar mass of the gas (C6H6O):
You mentioned that the mass of the gas is 10.37 times as much as C6H6O. Therefore, we need to divide the mass of the gas by 10.37 to find the molar mass of C6H6O.
3. Let's assume the mass of the gas is x grams. Dividing by 10.37:
x g / 10.37 = molar mass of C6H6O = 90.09 g/mol
Solving for x:
x g = 10.37 * 90.09 g/mol ≈ 934.0033 g/mol
4. Now we know that the molar mass of the gas is approximately 934.0033 g/mol.
5. To find the molecular formula, we need to compare the molar mass of the empirical formula (90.09 g/mol) to the molar mass of the gas (934.0033 g/mol).
Since 934.0033 g/mol is significantly greater than 90.09 g/mol, we can conclude that the empirical formula C6H7O5 is a submultiple of the molecular formula. This means we need to multiply the empirical formula by a whole number to get the molecular formula.
6. Calculate the ratio between the molar masses:
934.0033 g/mol / 90.09 g/mol ≈ 10.36
The ratio of the molar masses is approximately 10.36.
7. Notice that the ratio is very close to the mass ratio you provided, which is 10.37. This means the whole number multiplier is 10 or very close to it. Since we can't have fractional multipliers for the empirical formula, we will round it down to 10.
8. Multiply the empirical formula by the whole number multiplier (10):
10 * (C6H7O5) = C60H70O50
The molecular formula of the gas is C60H70O50.