(a)calculate the mass percent of carbon in the hydrated form of the solid that has the formula BeC2O4 X 3 H2O (s).

I divided 24.02 g by 151.08 g and got .16, or 16%. is that correct?

Also, I'm having trouble with this follow up question.

When heated to 220 degrees C, BeC2O4 X 3 H2O dehydrates completely as respresented: BeC2O4 X 3 H2O (s)--> BeC2O4 + 3H20 (g). If 3.21 g of BeC2O4 X 3 H2O is heated to 220 degrees C, calculate the:

(i) mass of BeC2O4 (s) formed
(ii) volume of H2O (g) released, measured at 220 degrees C and 735 mm Hg

Yes, the answer comes out to be 15.9%. As for the second part, it is a simple problem.

3.21g BeC2O4 3H2O x (1 mol BeC2O4 3H2O/151.03g) x (1 mol BeC2O4/1 mol BeC2O4) x (97.01/1 mol BeC2O4) = 2.06 g

ii. Since it is measured at STP, we can assume that 22.4 L of the substance = 1 mol.

3 mol H2Ox (22.4 L H2O/1 mol H2O) = 67.2 L H20

The measurements are not at STP, they are at 220 celsius and 735 mm Hg. so you use PV=nRT in this instance

I didn't check the formula weight of the BeC2O4.3H2O for the first, but the procedure is right.

On the second, calculate the percent composition of 3H2O in the BeC2O4.2H2O, then you can do i easily.

for ii, change it to moles, then V= nRT/P

To calculate the mass percent of carbon in the hydrated form of the solid BeC2O4 X 3 H2O (s), we need to determine the molar mass of carbon in BeC2O4 X 3 H2O and the molar mass of the compound as a whole. Then we can calculate the mass percent.

1. Calculate the molar mass of carbon (C):
The molar mass of carbon is 12.01 g/mol.

2. Calculate the molar mass of BeC2O4 X 3 H2O:
To calculate the molar mass of BeC2O4 X 3 H2O, we need to add up the molar masses of the elements in the formula. Be has a molar mass of 9.01 g/mol, C has a molar mass of 12.01 g/mol, O has a molar mass of 16.00 g/mol, and H has a molar mass of 1.01 g/mol. Since there are three water molecules (H2O), we need to multiply the molar mass of H2O by 3:
Molar mass of BeC2O4 X 3 H2O = (9.01 g/mol) + (2 * 12.01 g/mol) + (4 * 16.00 g/mol) + (3 * (2 * 1.01 g/mol)) = 151.08 g/mol.

3. Calculate the mass percent of carbon:
Mass percent of carbon = (molar mass of carbon / molar mass of BeC2O4 X 3 H2O) * 100%
= (12.01 g/mol / 151.08 g/mol) * 100%
= 7.95%

Therefore, the mass percent of carbon in the hydrated form of BeC2O4 X 3 H2O (s) is approximately 7.95%, not 16%.

Now let's move on to the follow-up question.

(i) To calculate the mass of BeC2O4 (s) formed when 3.21 g of BeC2O4 X 3 H2O is heated to 220 degrees C, we need to find the molar mass of BeC2O4.

1. Calculate the molar mass of BeC2O4:
The molar mass of BeC2O4 is (9.01 g/mol) + (2 * 12.01 g/mol) + (4 * 16.00 g/mol) = 105.04 g/mol.

2. Calculate the moles of BeC2O4 X 3 H2O:
Moles = mass / molar mass
= 3.21 g / 151.08 g/mol
= 0.0213 mol

3. Since the reaction shows that BeC2O4 X 3 H2O decomposes into BeC2O4, the number of moles of BeC2O4 formed will be the same as the number of moles of BeC2O4 X 3 H2O used. Therefore, the mass of BeC2O4 (s) formed will be:
Mass of BeC2O4 (s) = moles of BeC2O4 X molar mass of BeC2O4
= 0.0213 mol * 105.04 g/mol
= 2.23 g

(ii) To calculate the volume of H2O (g) released at 220 degrees C and 735 mm Hg, we can use the ideal gas law.

1. Calculate the number of moles of H2O released:
Moles = mass / molar mass
= 3.21 g / (3 * 18.02 g/mol)
= 0.0593 mol

2. Use the ideal gas law (PV = nRT) to calculate the volume of H2O (g):
V = (n * R * T) / P
= (0.0593 mol * 0.0821 L·atm/mol·K * (220 + 273.15) K) / (735 mm Hg * 1 atm / 760 mm Hg)
= 2.28 L

Therefore, the volume of H2O (g) released, measured at 220 degrees C and 735 mm Hg, is approximately 2.28 L.