Butane is trapped as described in the Procedure for this experiment in a graduated cylinder at an atmospheric pressure of 1.007 atm. The water temperature is recorded to be 20.0 °C, and the volume of gas trapped is 26.7 mL with a mass of 0.066 grams. What is the molar mass (in g/mol) of butane based on these calculations?
Question
Butane is trapped as described in the Procedure for this experiment in a graduated cylinder at an atmospheric pressure of 1.007 atm. The water temperature is recorded to be 20.0 °C, and the volume of gas trapped is 26.7 mL with a mass of 0.066 grams. What is the molar mass (in g/mol) of butane based on these calculations?
Solution
To calculate the molar mass of butane, 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.
Step 1: Convert the given values to the appropriate units.
- Pressure (P) = 1.007 atm
- Volume (V) = 26.7 mL = 0.0267 L (since 1 mL = 0.001 L)
- Temperature (T) = 20.0 °C = 293.15 K (since 273.15 K = 0 °C)
Step 2: Rearrange the ideal gas law equation to solve for n (number of moles), n = PV/RT.
Step 3: Substitute the given values into the equation.
- For this calculation, we'll use the value of R that has the units L·atm/(K·mol), which is 0.0821 L·atm/(K·mol).
- n = (1.007 atm * 0.0267 L) / (0.0821 L·atm/(K·mol) * 293.15 K) = 0.0011 mol
Step 4: Calculate the molar mass of butane.
- Molar mass = mass of the sample / number of moles = 0.066 g / 0.0011 mol = 60 g/mol
So, the molar mass of butane is approximately 60 g/mol.
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