CuSO4+4NH3-Cu(NH3)4SO4
a. if you use 10.0g 0f CuSO4 and excess NH3, What is the theoretical yield of Cu(NH3)4SO4?
b. if you produce 12.6g of Cu(NH3)4SO4, What is the percent yield of the compund?
CuSO4+4NH3-Cu(NH3)4SO4
mols CuSO4 = g/molar mass.
Then ?mols CuSO4 = ? mol Cu(NH3)4SO4 where ? = the same number.
Then theoretical yield = grams x molar mass.
b.
%yield = (actual yield/theor yield)*100 = ?.
the answer of DrBob222 is almost all correct but the theoretical yield=mol*molar mass
a. Well, if you mix CuSO4 with excess NH3, you're definitely going to get some interesting chemistry going on. As for the theoretical yield of Cu(NH3)4SO4, that can be calculated by using stoichiometry. So let's get our calculations on like a math nerd at prom!
First, we need to find the molar mass of Cu(NH3)4SO4. Copper has a molar mass of 63.55 g/mol, nitrogen has a molar mass of 14.01 g/mol, hydrogen has a molar mass of 1.01 g/mol, sulfur has a molar mass of 32.07 g/mol, and oxygen has a molar mass of 16.00 g/mol. Adding all these up, we get a molar mass of Cu(NH3)4SO4 to be 245.78 g/mol.
Now, let's calculate the moles of CuSO4 by dividing the mass (10.0g) by the molar mass (159.61 g/mol). This gives us approximately 0.0627 mol.
According to the balanced equation, 1 mol of CuSO4 reacts to form 1 mol of Cu(NH3)4SO4. So our theoretical yield of Cu(NH3)4SO4 would be 0.0627 mol.
Finally, to find the theoretical yield in grams, we multiply the number of moles (0.0627 mol) by the molar mass (245.78 g/mol). The theoretical yield of Cu(NH3)4SO4 would be around 15.38 g.
b. Ah, percent yield, the bane of every chemist's existence. But fret not, my friend! We can tackle this puzzle together.
The percent yield is the actual yield (12.6g) divided by the theoretical yield (15.38g) multiplied by 100%. So, let's do the math: (12.6g / 15.38g) * 100% ≈ 82.0%.
Therefore, the percent yield of the compound is approximately 82.0%. Not too shabby, if you ask me!
To find the theoretical yield and percent yield for the reaction CuSO4 + 4NH3 → Cu(NH3)4SO4, we need to follow a step-by-step process.
Step 1: Calculate the molar mass of CuSO4.
The molar mass of CuSO4 is calculated by summing the atomic masses of its constituents: Cu (63.55 g/mol), S (32.07 g/mol), and 4*O (4*16.00 g/mol). Adding these values gives a molar mass of 159.61 g/mol.
Step 2: Convert the given mass of CuSO4 to moles.
According to the question, we have 10.0 g of CuSO4. To convert grams to moles, we divide the mass by the molar mass:
10.0 g / 159.61 g/mol = 0.0627 mol CuSO4
Step 3: Use the stoichiometric ratio to find moles of Cu(NH3)4SO4.
From the balanced equation, we can see that the stoichiometric ratio between CuSO4 and Cu(NH3)4SO4 is 1:1. This means that 1 mole of CuSO4 will produce 1 mole of Cu(NH3)4SO4.
Therefore, the moles of Cu(NH3)4SO4 produced will be the same as the moles of CuSO4:
0.0627 mol Cu(NH3)4SO4
Step 4: Convert moles of Cu(NH3)4SO4 to grams.
To convert moles to grams, we multiply the moles by the molar mass of Cu(NH3)4SO4. The molar mass can be calculated by adding the atomic masses of Cu, N, H, and S multiplied by their respective quantities in the compound:
Cu (63.55 g/mol)
N (14.01 g/mol)
H (1.01 g/mol)
S (32.07 g/mol)
Adding these values gives a molar mass of 187.68 g/mol.
Using the moles calculated in Step 3:
0.0627 mol * 187.68 g/mol = 11.749 g Cu(NH3)4SO4
Therefore, the theoretical yield of Cu(NH3)4SO4 is 11.749 g.
Now, let's move on to finding the percent yield.
Step 1: Calculate the percent yield.
The percent yield is calculated by dividing the actual yield (12.6 g) by the theoretical yield (11.749 g) and multiplying by 100:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Percent Yield = (12.6 g / 11.749 g) * 100 = 107.36%
Therefore, the percent yield of Cu(NH3)4SO4 is approximately 107.36%.