3. Another volatile liquid was used in a similar experiment. A student found that 0.32 g of vapor had a volume of 152ml at 100 celsius and 1 atm. What was the molar mass of this gas?
4. The empirical formula of the unknown gas in question 3 is CH4O. What is the molecular formula of this compound.
To determine the molar mass of the volatile liquid mentioned in question 3, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
1. Convert the volume from milliliters to liters: 152 mL = 0.152 L.
2. Convert the temperature from Celsius to Kelvin: 100 °C + 273.15 = 373.15 K.
3. Substitute the given values into the ideal gas law equation: (1 atm) * (0.152 L) = n * (0.0821 L∙atm/(mol∙K)) * (373.15 K).
4. Rearrange the equation to solve for the number of moles, n: n = (1 atm * 0.152 L) / (0.0821 L∙atm/(mol∙K) * 373.15 K).
5. Calculate the value of n: n ≈ 0.00667 moles.
6. Determine the molar mass by dividing the mass (0.32 g) by the number of moles:
Molar mass = mass / moles = 0.32 g / 0.00667 mol ≈ 48 g/mol.
Therefore, the molar mass of the volatile liquid is approximately 48 g/mol.
Now, moving on to question 4, to determine the molecular formula of the compound with the empirical formula CH4O:
1. Calculate the empirical formula mass by summing the atomic masses of the elements in the empirical formula:
CH4O = 12.01 g/mol (carbon) + 4 * 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen)
≈ 32.05 g/mol.
2. Determine the empirical formula mass ratio by dividing the molar mass of the compound (48 g/mol, as calculated in question 3) by the empirical formula mass:
48 g/mol / 32.05 g/mol ≈ 1.5.
3. Multiply the subscripts in the empirical formula by the empirical formula mass ratio obtained in step 2 to find the molecular formula:
C1H4O1 * 1.5 → C1.5H6O1.5.
The molecular formula of the compound is C1.5H6O1.5.