A student was asked to make some copper(II) sulfate-5-5 (CuSO4.5(H2O) BY reacting copper(II) oxide (CuO) with sulfuric acid.
a) Calculate the molar mass of the copper (II) sulfate-5-water
which is (64+32+64) + (10+80)=250 mol
b)Calculate the amount (in mol) of copper (II) sulfate-5-water in a 10,0g sample?
C) Calculate the mass of copper (II) oxide needed to make the 10,0g sample?
Can anyone help me with question b,c??
thanks you very much.
b) To calculate the amount (in mol) of copper (II) sulfate-5-water in a 10.0g sample, we need to use the molar mass calculated in part (a). The molar mass of copper(II) sulfate-5-water is 250 mol.
We can use the formula:
Amount (in mol) = Mass (in g) / Molar mass
Substituting the values, we have:
Amount (in mol) = 10.0g / 250 mol
Amount (in mol) = 0.04 mol
Therefore, the amount of copper (II) sulfate-5-water in a 10.0g sample is 0.04 mol.
c) To calculate the mass of copper (II) oxide needed to make the 10.0g sample, we need to find the molar mass of copper (II) oxide (CuO).
The molar mass of copper (II) oxide is 63.55g/mol + 16.00g/mol = 79.55g/mol.
We can use the formula:
Mass (in g) = Amount (in mol) * Molar mass
Substituting the values, we have:
Mass (in g) = 0.04 mol * 79.55 g/mol
Mass (in g) = 3.18g
Therefore, the mass of copper (II) oxide needed to make the 10.0g sample is 3.18g.
b) To calculate the amount (in mol) of copper (II) sulfate-5-water in a 10.0g sample, we first need to convert the mass of the sample into moles.
Step 1: Calculate the molar mass of copper (II) sulfate-5-water:
The molar mass of the copper (II) sulfate portion is (63.55 + 32.07 + 4*16.00) = 159.61 g/mol.
The molar mass of 5 water molecules is (5 * (2*1.01 + 16.00)) = 90.10 g/mol.
Therefore, the molar mass of copper (II) sulfate-5-water is 159.61 + 90.10 = 249.71 g/mol.
Step 2: Calculate the amount in moles:
Amount (in mol) = Mass (in g) / Molar mass (in g/mol)
Amount (in mol) = 10.0g / 249.71 g/mol
Now you can plug in the values and calculate the amount of copper (II) sulfate-5-water in a 10.0g sample.
c) To calculate the mass of copper (II) oxide needed to make the 10.0g sample, we can use stoichiometry.
Step 1: Write the balanced chemical equation for the reaction:
CuO + H2SO4 -> CuSO4 + H2O
Step 2: Determine the molar ratio between CuO and CuSO4:
From the balanced equation, we can see that 1 mole of CuO reacts to form 1 mole of CuSO4.
Step 3: Calculate the molar mass of CuO:
Using the atomic masses, we find:
Molar mass of CuO = 63.55 g/mol + 16.00 g/mol = 79.55 g/mol
Step 4: Determine the amount (in moles) of CuO needed:
From step 2, the molar ratio is 1:1. Therefore, the amount of CuO needed is equal to the amount of CuSO4-5H2O calculated in part b.
Step 5: Calculate the mass of CuO needed:
Mass = Amount (in mol) * Molar mass (in g/mol)
Now you can plug in the values and calculate the mass of CuO needed to make the 10.0g sample.
Amount of moles= mass/molar mass
No. of moles= 10.0g/250 g
No. of moles= 0.04 moles
c) CuO + H2SO4 ----> CuSO4 + H2O
mol 1 1 1 1
1 mole of CuO = 1 mole of CuSO4.5H2O
CuO= 64g + 16g= 80g
80 g of CuO = 250 g of CuSO4
x = 10g of CuSO4
10g x 80g/250g
800g/250
=3.2 grams of CuO