A cylinder was found in a storeroom of a manufacturing company. The cylinder was so old that the label had fallen off, and no one remembered what the cylinder held. A 5 mg sample was found to occupy 4.13 mL at 23 Celsius and 745 torr. The sample was also found to be composed of only carbon and hydrogen. Identify the gas.
n = PV/RT 5mg = 0.005g
0.98 atm * 0.00413/0.0821*296K = 1.665E-4 mol(what? since comp.not known
M.W.= 0.005/1.665E-4 = 30.03 g/mol
Molecular formula: C2H6 by manipulating the numbers of hydrogen. What is the right logical way to proceed though. Don't I have to follow pattern to find molecular formula?
No, I think this is logical enough. CH4 would be 16, CH3- would be 15 and 30/15 = 2 so two units of CH3 would do it and CH3CH3 (or C2H6) does the trick. You know anything above C3 won't work because 3 x 12 = 36 and that leaves no rooom for H atoms. Thus, C1 and C2 are the only possibilties.
Well, isn't it convenient that the only possibilities for the mystery gas are C1 and C2? It's like the gas is playing a little game with us, keeping us on our toes. But don't worry, I have the solution for you – it's like a jigsaw puzzle, but with atoms!
Based on your calculations, the molecular formula of the gas is indeed C2H6, or ethane for those who are familiar with chemical names. So it seems our mystery gas was just a little bit of ethane hiding away in that cylinder. Maybe it was feeling a bit shy, but now it can finally come out and join the party!
Yes, your logical approach is correct. By analyzing the given information, you can determine the molecular formula of the gas.
Since the sample is composed only of carbon and hydrogen, you can narrow down the possible molecular formulas based on the number of carbon atoms and the maximum number of hydrogen atoms allowed.
The sample weighs 5 mg, which is equivalent to 0.005 g. Using the ideal gas law equation, n = PV/RT, you can calculate the number of moles of the gas in the sample.
The volume is given as 4.13 mL, which is equivalent to 0.00413 L. The temperature is 23 degrees Celsius, which can be converted to 296 K. The pressure is given as 745 torr, which is equivalent to 0.98 atm. Plugging these values into the equation, you can calculate the number of moles.
n = (0.98 atm * 0.00413 L) / (0.0821 * 296 K) = 1.665E-4 mol
To find the molecular weight (MW), you divide the mass by the number of moles.
MW = 0.005 g / 1.665E-4 mol = 30.03 g/mol
Now, you need to determine the best molecular formula given this molecular weight. You start with CH4, which has a molecular weight of 16 g/mol. Since the calculated molecular weight is higher than that, you move on to CH3, which has a molecular weight of 15 g/mol.
To obtain a molecular weight of 30.03 g/mol, you can have two units of CH3, resulting in the molecular formula CH3CH3, which is also known as C2H6.
Therefore, the gas in the cylinder is likely ethane (C2H6).
To identify the gas in the cylinder, we can use the ideal gas law equation n = PV/RT.
First, we need to convert the mass of the sample from milligrams to grams. Since 1 gram is equal to 1000 milligrams, we have 5 mg = 0.005 g.
Next, we need to convert the volume from milliliters to liters. Since 1 liter is equal to 1000 milliliters, we have 4.13 mL = 0.00413 L.
Now, we can substitute the values into the ideal gas law equation:
n = (P * V) / (R * T)
Where:
n = number of moles
P = pressure in atmospheres (torr can be converted to atm by dividing by 760)
V = volume in liters
R = gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (23 Celsius = 23 + 273.15 Kelvin)
Plugging in the values, we get:
n = (0.745 atm * 0.00413 L) / (0.0821 L·atm/mol·K * 296 K)
n = 1.665E-4 mol
Now, to find the molar mass of the gas, we can use the formula:
Molar mass = mass of sample (in grams) / number of moles
Molar mass = 0.005 g / 1.665E-4 mol
Molar mass ≈ 30.03 g/mol
Since the sample is composed of only carbon and hydrogen, we can determine the molecular formula by examining the molar mass. Looking at the possibilities, if we have a molecular formula of CH4 (methane), the molar mass would be 16 g/mol, which is lower than the observed molar mass of 30.03 g/mol. So, we need to consider a higher molecular formula.
Now, let's analyze the molecular formula of CH3-. The molar mass is 15 g/mol. To obtain a molar mass of 30.03 g/mol, we would need two units of CH3- (2 * 15 g/mol = 30 g/mol). Therefore, the molecular formula would be CH3CH3, which represents ethane (C2H6).
To verify that C2H6 is the correct molecular formula, we can compare it to other possibilities. For example, any molecular formula containing more than two carbon atoms (such as C3H8) would exceed the observed molar mass of 30.03 g/mol, as carbon atoms have a molar mass of 12 g/mol. Therefore, considering the limitations of the molar mass, the logical conclusion is that the gas in the cylinder is ethane (C2H6).