I did this experiment where I filled a graduated cylinder with water. The I put a stopper on it and put it under water and removed the stopper. The with a modified lighter (that wouldn't release sparks) I added butane gas. I got the following results:
mass of lighter = 15.18 g
final mass of lighter = 15.00 g
volume of butane = 80.0 mL
temperature = 25.0 degrees celsius
atmosperic pressure = 101.4 Kpa
1) What would the mass of butane be?
0.18 grams
2) Number of mols of butane?
0.003170846 mol.
3)Molar mass of butane?
56.8 g/mol
3) Calulate the volume of the number of moles found in questin 2 of butane at STP then use that value to find density. I don't know how to do this at all. ???
Ok, for 3) you need the volume of butane you measured, and the pressure of that butane. The pressure of the butane is atmospheric minus water vapor pressure (there is water vapor in there).
You get the water vapor pressure from a water vapor pressure table, knowing the temperature.
Then, you have volume, pressure, and number of moles. Now correct it to stp
PV=nRT and (letting small letters meaning your conditions, capital letters at stp)
pv=nRt
dividing the first equation by the second
PV/pv=T/t
and you are solving for V
V=pvT/Pt
density at stp
V/massbutane
Isn't the density the mass of butane/V
?
yes, you are correct, I was in too much of a hurry. Correct that.
To calculate the volume of the number of moles of butane at STP (Standard Temperature and Pressure), you can use the ideal gas law. The ideal gas law equation is given by:
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.
At STP conditions, the pressure is 1 atmosphere (1 atm) and the temperature is 0 degrees Celsius (273.15 Kelvin).
Assuming you have already calculated the number of moles of butane (0.003170846 mol) as in question 2, you can rearrange the ideal gas law equation to solve for the volume (V):
V = nRT / P
Substituting the values:
V = (0.003170846 mol) * (0.0821 L.atm/mol.K) * (273.15 K) / (1 atm)
Simplifying the equation:
V = 0.2227 L (rounded to four decimal places)
Now that you have the volume of the number of moles of butane at STP, you can use this value to calculate the density of butane.
Density is defined as mass divided by volume. In this case, the mass of the butane is 0.18 grams (as in question 1), and the volume is 0.2227 liters (as just calculated).
Density of butane = mass / volume
Density of butane = (0.18 g) / (0.2227 L)
Calculating the density:
Density of butane = 0.809 g/L (rounded to three decimal places)
So, the density of butane at STP is approximately 0.809 grams per liter.