The composition of a propane clathrate was found to contain 40.1 mL of water.
Assuming that all the cavities of the clathrate was filled with propane, how many grams of propane did the clathrate contain?
To find the number of grams of propane in the clathrate, we need to convert the volume of water to moles using the molar volume of water and then convert moles of water to moles of propane using the stoichiometry of the clathrate.
Here's how to calculate it step by step:
1. Calculate the moles of water: First, convert the volume of water (40.1 mL) to liters by dividing it by 1000:
Volume of water (in liters) = 40.1 mL / 1000 = 0.0401 L
The molar volume of water at standard temperature and pressure (STP) is approximately 22.4 L/mol.
Moles of water = Volume of water / Molar volume of water
= 0.0401 L / 22.4 L/mol
2. Determine the stoichiometry of the clathrate: The clathrate is assumed to be completely filled with propane. The chemical formula of propane is C3H8, which means it contains 3 carbon atoms and 8 hydrogen atoms.
Since all cavities are filled with propane, we can assume that the ratio of propane (C3H8) to water (H2O) is 1:1 in the clathrate.
3. Convert moles of water to moles of propane: Since the stoichiometry ratio between propane and water is 1:1, the moles of propane in the clathrate will be the same as the moles of water.
4. Calculate the molar mass of propane: The molar mass of propane (C3H8) can be calculated by summing the atomic masses from the periodic table:
Molar mass of propane = (3 * atomic mass of carbon) + (8 * atomic mass of hydrogen)
5. Calculate the mass of propane in grams: Finally, multiply the moles of propane by the molar mass to get the mass of propane in grams.
Now you have all the steps to solve the problem. Plug in the values into the respective equations to find the answer.