Plaster of Paris is the compound calcium sulfate hemihydrate, CaSO4*1/2H2O
What mass of water is liberated when a 2.763 gram sample is heated to constant mass?
The sample is 2.763 grams
How much of the 2.763 gram sample is water? I need an amount in grams.
Amount of anhyrdate + amount of water = amount of sample
Please help
To find the mass of water liberated when heating the sample, we need to determine the amount of water in the sample.
Step 1: Calculate the molar mass of the compound.
- The molar mass of calcium sulfate hemihydrate (CaSO4*1/2H2O) can be calculated by adding the molar masses of its individual elements.
- Calcium (Ca): 40.08 g/mol
- Sulfur (S): 32.06 g/mol
- Oxygen (O): 16.00 g/mol (x4 for 4 atoms) = 64.00 g/mol
- Hydrogen (H): 1.01 g/mol (x2 for 2 atoms) = 2.02 g/mol
- Molar mass of CaSO4 = 40.08 + 32.06 + 64.00 = 136.14 g/mol
- Molar mass of H2O = 2.02 x 2 = 4.04 g/mol
Step 2: Determine the moles of the sample.
- Divide the mass of the sample by its molar mass to find the number of moles.
- Moles of sample = mass of sample / molar mass of sample = 2.763 g / 136.14 g/mol ≈ 0.0203 mol
Step 3: Calculate the moles of water.
- Since there is one mole of water (H2O) for every half mole of calcium sulfate hemihydrate (CaSO4*1/2H2O), we need to divide the moles of the sample by 0.5 to get the moles of water.
- Moles of water = moles of sample / 0.5 = 0.0203 mol / 0.5 ≈ 0.0407 mol
Step 4: Convert the moles of water to grams.
- Multiply the moles of water by the molar mass of water to find the mass of water.
- Mass of water = moles of water x molar mass of water = 0.0407 mol x 4.04 g/mol ≈ 0.1649 g
Therefore, approximately 0.1649 grams of water will be liberated when the 2.763 grams sample is heated to constant mass.
To determine the mass of water liberated when a 2.763 gram sample of plaster of Paris is heated, we need to set up an equation using the principle of conservation of mass.
First, let's determine the amount of anhydrous calcium sulfate (CaSO4) in the sample. We know the molar mass of CaSO4 is 136.14 g/mol, so we can calculate the moles of anhydrous calcium sulfate present.
Moles of CaSO4 = mass of CaSO4 / molar mass of CaSO4
Moles of CaSO4 = 2.763 g / 136.14 g/mol
Moles of CaSO4 = 0.0203 mol
Since the compound CaSO4*1/2H2O has a 1:1 stoichiometric ratio of calcium sulfate to water, the moles of water liberated will be equal to the moles of anhydrous calcium sulfate. Therefore, the moles of water liberated is also 0.0203 mol.
Next, we need to convert the moles of water to grams using its molar mass. The molar mass of water (H2O) is 18.015 g/mol.
Mass of water liberated = moles of water * molar mass of water
Mass of water liberated = 0.0203 mol * 18.015 g/mol
Mass of water liberated = 0.3668 g
Therefore, when a 2.763 gram sample of plaster of Paris is heated to constant mass, approximately 0.3668 grams of water will be liberated.
You can do this the long way or a short way or one that's in between. The short way isn't taught anymore I don't think. The long way is to write an equation, balance it, and solve as if it were a stoichiometry problem. Here is a worked example that will solve all of your stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
The short way is
2.763g x (molar mass 0.5H2O/molar mass CaSO4.0.5H2O) = ? g H2O.
The in between way is to calculate the %H2O in CaSO4.1/2H2O.
That is (1/2 molar mass H2O/molar mass CaSO4.1/2H2O) x 100 = %H2O
Then (%H2O/100)*2.763 = g H2O (note the short way and the in between way are really the same thing but the approach is different)