So I have a lab exam tomorrow and our instructor said that one of the questions would be as follows:
Prepare 100mL of 0.03 M Copper Sulfate, beginning from copper sulfate pentahydrate. Calculate the exact molarity to 3 sig figs.
Alright so I understand how to prepare 100mL of 0.003 CuSO4-5H2O by taking 0.03M X 0.1L to get .003 mol.
.003 Mol X the Molar Mass of CuSO4-5H2O which is 250G = .75G
Does this mean I add .75G to 99.25mL of water since the density of water is 1g/mL.
Is he asking us to maybe determine how much of copper sulfate pentahyrdate is actually copper sulfate. Then way out enough pentahyrdate to contain the amount of copper sulfate required to make 100mL of .03M copper sulfate?
So confused this should be basic Gen Chem 1 lab 101 stuff, but I can't seems to wrap my head around it can someone help me? I'd appreciate it thanks!
gms of CuSO4-5HOH = [Molarity needed][Volume needed (Liters)][formula Wt CuSO4] / [Purity fraction CuSO4 in pentahydrate]
=[(0.03M)(0.100L)(160g/mol)]/[0.64]=0.75gms CuSO4-5HOH + Solvent up to but not to exceed 100 ml total volume.
Preparing a solution of a given molarity involves calculating the amount of solute (in this case, copper sulfate) needed to achieve the desired concentration. Here's how you can approach the problem step by step:
1. Determine the molar mass of copper sulfate pentahydrate (CuSO4·5H2O). The molar mass can be calculated by adding up the atomic masses of its constituent elements. Copper (Cu) has a molar mass of 63.55 g/mol, sulfur (S) has a molar mass of 32.07 g/mol, oxygen (O) has a molar mass of 16.00 g/mol, and hydrogen (H) has a molar mass of 1.01 g/mol. Since there are 4 oxygens and 10 hydrogens in the pentahydrate, the molar mass is:
(1 * 63.55 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol) + (10 * 1.01 g/mol) = 249.71 g/mol
2. Calculate the amount of copper sulfate pentahydrate needed to make 100 mL of 0.03 M solution. Molarity (M) is defined as moles of solute per liter of solution. Since you want a 0.03 M solution, you need:
0.03 mol/L * 0.1 L = 0.003 mol
So, you need 0.003 moles of copper sulfate.
3. Now, we need to convert moles of copper sulfate pentahydrate to grams. We can do this by multiplying the number of moles by the molar mass of copper sulfate pentahydrate:
0.003 mol * 249.71 g/mol = 0.749 g
So, you need 0.749 grams of copper sulfate pentahydrate.
4. Dissolve the calculated mass of copper sulfate pentahydrate in enough water to make 100 mL of solution. Remember to take into account the density of water, which is approximately 1 g/mL. Therefore, the final volume of the solution will be slightly larger than 100 mL due to the added mass of copper sulfate pentahydrate. You can add this calculated mass of copper sulfate pentahydrate to just a small amount of water, dissolve it completely, and then dilute to the desired final volume (100 mL) using more water.
By following these steps, you can prepare 100 mL of a 0.03 M solution of copper sulfate pentahydrate with a molar mass rounded to 3 significant figures.
Hope this helps! Good luck with your lab exam!