0.0337g of a volatile liquid when vaporized completely at 100Pa and 300K occupied a volume of 20cm3. When this volume of vapor was completely burnt in an excess of oxygen, 40cm3 of carbon dioxide and 20cm3 of water vapor were formed. All volumes being measured at the same temperature and pressure. What is the formula of the compound? A: C2H2 B: C3H3 C: CH2CO D: CH3CHO
There are two or three ways of approaching this. The simplest way, but one that doesn't really cinch an answer is this.
Use PV = nRT. Solve for n = number of mols. Then n = g/molar mass. You have n and g, solve for molar mass. I get 42 for that. The molar mass of A is 26, B is 39, C is 42 and D is 44. So you assume C is the answer but it doesn't leave much room for rounding errors and that kind of thing.
If you wish to confirm that you can do this.
Use PV = nRT and convert 20 cc H2O to moles and convert 40 cc CO2 to moles.
Then grams = mol x molar mass; convert H2O and CO2 to grams H and grams C.
Add g H and g C and subtract from mass sample to find g oxygen (O, not O2).
Then convert g C, g H, g O to moles . moles = grams/molar mass
Finally, find the ratio of the elements to each other and that comes out CH2O.
If PV=nRT, (100)(2x10^-5)=(n)x(8.31)x(300), n is 8.02x10^-7. How did you manage to get the mr of 42?
To determine the formula of the compound, we need to analyze the stoichiometry of the reaction and calculate the number of moles of carbon, hydrogen, and oxygen in the compound.
Given:
Mass of the volatile liquid = 0.0337g
Volume of the vapor = 20cm^3
Volume of carbon dioxide = 40cm^3
Volume of water vapor = 20cm^3
Step 1: Calculate the number of moles of carbon dioxide and water vapor.
We know that 1 mole of any gas occupies 22.4L at standard temperature and pressure (STP). Since the volumes are given at the same temperature and pressure, we can apply this ratio to convert the volumes to moles.
Number of moles of carbon dioxide = (40cm^3 / 22.4L/mol) = 1.79 moles
Number of moles of water vapor = (20cm^3 / 22.4L/mol) = 0.89 moles
Step 2: Calculate the number of moles of carbon and hydrogen.
From the balanced chemical equation for the combustion of the compound, we can see that one mole of the compound produces one mole of carbon dioxide and one mole of water.
Since we have 1.79 moles of carbon dioxide and 0.89 moles of water, it means there must be the same number of moles of the compound. Therefore, we can conclude that the number of moles of carbon and hydrogen in the compound is also 1.79 and 0.89, respectively.
Step 3: Calculate the molar mass of carbon and hydrogen.
The molar mass of an element is the mass of one mole of that element. The molar mass of carbon (C) is approximately 12 g/mol, and the molar mass of hydrogen (H) is approximately 1 g/mol.
Step 4: Calculate the empirical formula.
To calculate the empirical formula, divide the number of moles of each element by the smallest number of moles. In this case, the smallest number of moles is 0.89.
Carbon: (1.79 moles / 0.89 moles) ≈ 2
Hydrogen: (0.89 moles / 0.89 moles) = 1
Therefore, the empirical formula is CH2.
Step 5: Determine the molecular formula.
To determine the molecular formula, we need to know the molar mass of the compound. The molar mass can be calculated by adding up the molar mass of each element.
Molar mass of carbon = (2 moles x 12 g/mol) = 24 g/mol
Molar mass of hydrogen = (2 moles x 1 g/mol) = 2 g/mol
The molar mass of the compound is approximately 26 g/mol.
Now, compare the molar mass of the compound with the empirical formula's molar mass (CH2 = 14 g/mol). If the molar mass is an integer multiple of the empirical formula's molar mass, we can determine the molecular formula.
The molecular mass is found to be approximately twice the empirical formula mass. Therefore, the molecular formula of the compound is C2H4.
Looking at the available options, none match the calculated molecular formula. Therefore, none of the options A, B, C, or D is the correct answer.
Note: The calculation assumes ideal behavior for gases and ignores any possible deviations.