Answer the following question about copper (II) sulfate pentahydrate:
a)how much water (in g) is there in 10.0 g of the compound listed above?
b)If you needed 2.50g of CuSO4 for a reaction but only had CuSO4 * 5H2O available, how many g would you need to use to get the same number of moles of CuSO4?
What percent H2O is in CuSO4.5H2O? If you calculate that, couldn't you do the others.
To answer these questions, we need to understand the concept of molar mass and the number of water molecules in copper (II) sulfate pentahydrate.
a) To determine the amount of water in 10.0 g of copper (II) sulfate pentahydrate, we can use the molar mass of the compound and the molar mass of water. The molar mass of copper (II) sulfate pentahydrate is the sum of the molar masses of its individual elements.
Step 1: Calculate the molar mass of copper (II) sulfate pentahydrate:
Cu: atomic mass = 63.55 g/mol
S: atomic mass = 32.07 g/mol
O: atomic mass = 16.00 g/mol (there are four oxygen atoms in the formula)
H: atomic mass = 1.01 g/mol (there are ten hydrogen atoms in the formula)
Multiply the atomic mass of each element by its respective subscript and sum these values:
Molar mass of CuSO4 * 5H2O = (63.55 g/mol) + (32.07 g/mol) + (16.00 g/mol × 4) + (1.01 g/mol × 10) = 249.69 g/mol
Step 2: Calculate the molar mass of water:
H: atomic mass = 1.01 g/mol
O: atomic mass = 16.00 g/mol
The molar mass of water is the sum of the atomic masses of hydrogen and oxygen:
Molar mass of H2O = (1.01 g/mol × 2) + (16.00 g/mol) = 18.02 g/mol
Step 3: Calculate the percentage of water in copper (II) sulfate pentahydrate:
The pentahydrate formula indicates that there are 5 water molecules for every molecule of copper (II) sulfate. The molar mass of these water molecules is:
Molar mass of H2O × 5 = 18.02 g/mol × 5 = 90.10 g/mol
Now, calculate the percentage of water in copper (II) sulfate pentahydrate:
(90.10 g/mol / 249.69 g/mol) × 100% = 36.1%
To find the amount of water in 10.0 g of copper (II) sulfate pentahydrate, we multiply the mass of the compound by the percentage of water:
Amount of water = 10.0 g × (36.1% / 100%) = 3.61 g
Therefore, there is 3.61 g of water in 10.0 g of copper (II) sulfate pentahydrate.
b) To find out how many grams of CuSO4 * 5H2O are needed to get the same number of moles of CuSO4 as 2.50 g, we need to consider their molar masses.
Step 1: Calculate the molar mass of CuSO4:
Cu: atomic mass = 63.55 g/mol
S: atomic mass = 32.07 g/mol
O: atomic mass = 16.00 g/mol (there are four oxygen atoms in the formula)
Molar mass of CuSO4 = (63.55 g/mol) + (32.07 g/mol) + (16.00 g/mol × 4) = 159.61 g/mol
Step 2: Calculate the molar mass of CuSO4 * 5H2O:
From the previous question, we know that the molar mass of CuSO4 * 5H2O is 249.69 g/mol.
Step 3: Calculate the number of moles of CuSO4 in 2.50 g:
The number of moles can be calculated using the formula:
Number of moles = Given mass / Molar mass
Number of moles of CuSO4 = 2.50 g / 159.61 g/mol = 0.0157 mol
Step 4: Convert moles of CuSO4 to moles of CuSO4 * 5H2O:
Since both compounds contain the same number of moles of CuSO4, we can set up the following ratio:
0.0157 mol CuSO4 = X mol CuSO4 * 5H2O
(X is the unknown number of moles of CuSO4 * 5H2O we need)
Step 5: Calculate the mass of CuSO4 * 5H2O:
To find the mass of CuSO4 * 5H2O, we need to know its molar mass:
Mass of CuSO4 * 5H2O = X mol × molar mass of CuSO4 * 5H2O
Solving for X, we can rearrange the equation:
X = (0.0157 mol CuSO4 × 249.69 g/mol) / 159.61 g/mol
X ≈ 0.0245 mol
Now, calculate the mass:
Mass of CuSO4 * 5H2O = 0.0245 mol × 249.69 g/mol ≈ 6.11 g
Therefore, you would need to use approximately 6.11 g of CuSO4 * 5H2O to get the same number of moles of CuSO4 as 2.50 g.