The mass of the CuSO4 x 5H2O sample is 1.664 g. The mass of the recovered copper is 1.198 g.

Calculate the percent copper in pure copper sulfate pentahydrate(this is the theoretical percent copper). One mole of copper has a mass of 63.55 g, and by addition, one mole of CuSO4 x 5H2O has a mass of 249.5 g.

Answered above.

Why did the copper go to therapy? Because it had pentahydrate issues!

To calculate the percent copper in pure copper sulfate pentahydrate, we can use the formula:

Percent Copper = (Mass of Recovered Copper / Mass of CuSO4 x 5H2O) x 100%

Let's plug in the given values:

Mass of Recovered Copper = 1.198 g
Mass of CuSO4 x 5H2O = 1.664 g

Percent Copper = (1.198 g / 1.664 g) x 100%

Calculating this, we get:

Percent Copper = 71.99%

So, the theoretical percent copper in pure copper sulfate pentahydrate is approximately 72%. Keep in mind, this is just a theoretical value and may differ in actual experimental results.

To calculate the percent copper in pure copper sulfate pentahydrate (CuSO4 x 5H2O), we can use the equation:

Percent copper = (mass of recovered copper / mass of CuSO4 x 5H2O sample) x 100

Given the following values:
- Mass of CuSO4 x 5H2O sample = 1.664 g
- Mass of recovered copper = 1.198 g
- One mole of copper (Cu) = 63.55 g
- One mole of CuSO4 x 5H2O = 249.5 g

First, we need to convert the mass of the CuSO4 x 5H2O sample to moles. To do this, divide the mass by the molar mass:

Moles of CuSO4 x 5H2O = (mass of CuSO4 x 5H2O sample) / (molar mass of CuSO4 x 5H2O)

Molar mass of CuSO4 x 5H2O = molar mass of Cu + molar mass of S + 4 * (molar mass of O) + 5 * (molar mass of H2O)

Molar mass of CuSO4 x 5H2O = (63.55 g/mol) + (32.07 g/mol) + 4 * (16.00 g/mol) + 5 * (18.02 g/mol)

Once you calculate the molar mass, substitute it into the equation:

Moles of CuSO4 x 5H2O = 1.664 g / (molar mass of CuSO4 x 5H2O)

Now, let's calculate the moles of copper (Cu) present in the recovered copper. This can be done using the molar mass of copper:

Moles of copper = (mass of recovered copper) / (molar mass of Cu)

Molar mass of Cu = 63.55 g/mol

Next, we can calculate the percent copper in pure copper sulfate pentahydrate by dividing the moles of copper by the moles of CuSO4 x 5H2O and multiplying by 100:

Percent copper = (moles of copper / moles of CuSO4 x 5H2O) x 100

Finally, substitute the calculated values into the equation to find the percent copper in pure copper sulfate pentahydrate.

So what's the question? Percent Cu in the sample?

(1.198/1.664)*100 = ??

Theoretical yield =
(mass Cu/mass hydrate)*100 = (63.55/249.5)*100 = ??

The question is asking for percent of copper in the sample( so the first step shown above), but when I had to calculate the experimental percent I did the same thing, and i got 72%( is that the correct number of significant figures)?

I also wanted to know when I have to calculate the error of the % of copper what number do I use as true value and actual value?

The mass of the CuSO4 x 5H2O sample is 1.664 g. The mass of the recovered copper is 1.198 g.

1. Compute the experimental percent by mass of copper in the sample:
1.1664/1.998x 100=71.81%

2. Calculate the percent copper in pure copper sulfate pentahydrate(this is the theoretical percent copper). One mole of copper has a mass of 63.55 g, and by addition, one mole of CuSO4 x 5H2O has a mass of 249.5 g:

63.55/249.5x100= 25.47%

3. In order to calculate the error of the % Cu and the percent error of the % Cu I know I am suppose to use 1.198 as the experimental value, but I am not sure what to use for the true value.