An unknown gaseous hydrocarbon consists of 85.63% carbon by mass. A 0.335 g sample of the gas occupies a volume of 0.107 L at STP. What is the identity of the gas?
Thanks so much, I would of never figured out that out myself past the empirical formula since I forgot about the 22.4L factor.
The molar mass turns out to be 70, and since the only answer I was given that adds up to 70 is C5H10, it's an easy pick. Pentane!
Nope. Pentane is C5H12. If C5H10 is the choice you are given, that is pentene.
Well, if this were a guessing game I'd say the gas is probably a carbonated soda, because that's about 85% water, right? But, alas, this is a chemistry question so we have to be a bit more scientific.
To find the identity of the gas, we can start by calculating the moles of carbon in the sample. We know that the mass of carbon is 85.63% of the total mass, so we can calculate:
Mass of carbon = 0.85 * 0.335 g = 0.28475 g
Next, we need to convert the mass of carbon to moles. Since the atomic mass of carbon is 12.01 g/mol, we can calculate:
Moles of carbon = 0.28475 g / 12.01 g/mol ≈ 0.0237 mol
Now, let's use the ideal gas law to find the molar volume of the gas at STP. At STP (standard temperature and pressure), 1 mole of gas occupies 22.4 L. Therefore, we can calculate the molar volume of our gas:
Molar volume of gas = 0.107 L / 0.0237 mol ≈ 4.51 L/mol
Now comes the tricky part - figuring out the molar mass of the gas. We know that the gas is a hydrocarbon, so it consists of carbon and hydrogen. The molar mass of the gas can be calculated using the following formula:
Molar mass = (Mass of carbon + Mass of hydrogen) / Moles of gas
Since we know the mass of carbon (0.28475 g), and we can assume that the mass of hydrogen is the remaining mass, we can calculate:
Mass of hydrogen = Mass of gas - Mass of carbon = 0.335 g - 0.28475 g = 0.05025 g
Now, let's plug these values into the formula:
Molar mass ≈ (0.28475 g + 0.05025 g) / 0.0237 mol ≈ 14.17 g/mol
Now, we need to find the molecular formula of the gas. The molecular formula of the gas is CH4 (methane). Congratulations, you've found the identity of the gas! Just remember, don't try making any jokes with this gas - it's highly flammable!
To determine the identity of the unknown gaseous hydrocarbon, we need to follow a series of steps:
Step 1: Calculate the molar mass of the unknown hydrocarbon:
We know that the hydrocarbon is 85.63% carbon by mass. This means that if we assume a 100 g sample of the hydrocarbon, 85.63 g of it will be carbon. To calculate the mass of hydrogen, we subtract the mass of the carbon from the total mass:
100 g - 85.63 g = 14.37 g
Next, we need to convert these masses into moles. The molar mass of carbon is approximately 12.01 g/mol, and the molar mass of hydrogen is approximately 1.01 g/mol.
For carbon:
85.63 g ÷ 12.01 g/mol = 7.13 mol
For hydrogen:
14.37 g ÷ 1.01 g/mol = 14.23 mol
Step 2: Calculate the empirical formula of the hydrocarbon:
To find the empirical formula, we need to find the simplest whole-number ratio between carbon and hydrogen atoms. To do this, we divide the number of moles of each element by the smaller value. In this case, we divide both values by 7.13 (the number of moles of carbon):
Carbon: 7.13 mol ÷ 7.13 mol = 1 mol
Hydrogen: 14.23 mol ÷ 7.13 mol = 1.99 mol
After rounding to the nearest whole number, we find that the empirical formula is CH2.
Step 3: Determine the molecular formula using the molar mass of the empirical formula:
To find the molecular formula, we need to know the molar mass of the empirical formula CH2. The molar mass is approximately 14.03 g/mol (12.01 g/mol for carbon + 2 * 1.01 g/mol for hydrogen).
To calculate the molecular formula, we divide the experimental molar mass of the compound by the molar mass of the empirical formula:
Experimental molar mass = 0.335 g ÷ 0.107 L × (22.4 L/mol)
Experimental molar mass = 0.335 g ÷ 0.107 L × (22.4 L/mol) = 0.74 g/mol
Finally, we divide the experimental molar mass by the molar mass of the empirical formula to find the whole-number multiple:
0.74 g/mol ÷ 14.03 g/mol = 0.05
Since the result is approximately 0.05, we can assume that the molecular formula is the same as the empirical formula, CH2.
Therefore, the identity of the gas is ethylene (C2H4).
Take a 100 g sample which gives you
85.63 g C
14.14.37 g H.
Convert to mols.
85.63/12 = about 7
14.37/1 = about 14
Empirical formula is CH2.
At STP a mole of gas occupies 22.4L; therefore, 0.107/22.4 = about 0.05 mols but you need to be more exact than that.
mols = grams/molar mass. You know mols and you know grams, solve for molar mass. I get approximately 72.
Since this is a hydrocarbon we know the formula will be
CH3(CH2)nCH3
72-30(for the two CH3 groups) = 40 left for the CH2. Then 40/14 = about 2.8 or so, round to 3.0 so the formula must be
CH3(CH2)3CH3 or C5H12. That is a gas. Do you know the name?