What percent by mass does the water constitute in Na2SO4*10H2O?

Determine the formula for the hydrate of cobalt (II) acetate from the following data.
Mass of crucible and lid= 28.945g
Mass of crucible, lid, and hydrate= 29.986g
Mass of crucuble, lid, and anhydrous salt= 29.677g
Mass of crucible, lid, and anhydrous salt(2nd)= 29.673g

I think for the second one I need to to figure out how much the hydrate weighs by itself without the crucible then figure out how much weight the anhydrous salt lost during heating but I am not sure.

For the first question I don't know how to figure it out without knowing the grams so how do I do that?

For the first one, take a mole of the material. Na2SO4*10H2O has a molar mass of ??. And 10 moles H2O has a mass of ??. Then mass 10 moles H2O/mass 1 mole Na2SO4*10H2O and all that times 100 will be the percent water by mass.

For the second one, you are correct.
The mass of water lost = ??
The mass copper acetate converted to moles.
mass of water lost converted to moles.
Then look at the ratio of the moles and determine the number of moles of water per 1 mole copper acetate. Post your work if you get stuck.

you must have a formula to solve any mole because anything in solving a numbers like mole have formula you must search the formula to be easy sor you to finish your work....

To calculate the percent by mass of water in a hydrate, you need to use the following formula:

Percent by mass of water = (mass of water / mass of hydrate) x 100

For the first question, you are given the formula Na2SO4 * 10H2O. This means that there are 10 water molecules associated with each molecule of Na2SO4. To find the percent by mass of water, you need to know the molar mass of Na2SO4 and the molar mass of water.

The molar mass of Na2SO4 is:
(2 x atomic mass of Na) + atomic mass of S + (4 x atomic mass of O)

The atomic mass of Na is 22.99 g/mol, the atomic mass of S is 32.06 g/mol, and the atomic mass of O is 16.00 g/mol. Plugging in the values, you get:
(2 x 22.99) + 32.06 + (4 x 16.00) = 142.04 g/mol

Now, calculate the molar mass of water, which is 18.015 g/mol.

Next, calculate the mass of water in the hydrate by subtracting the mass of the anhydrous salt from the mass of the hydrate.

Mass of water = (mass of hydrate - mass of anhydrous salt)
= 29.986 g - 29.677 g
= 0.309 g

Finally, calculate the percent by mass of water using the formula mentioned earlier:

Percent by mass of water = (mass of water / mass of hydrate) x 100
= (0.309 g / 29.986 g) x 100
= 1.03%

Therefore, water constitutes approximately 1.03% of the mass of Na2SO4*10H2O.

For the second question, you are given the masses of the crucible, lid, hydrate, and anhydrous salt in two separate measurements. To find the formula for the hydrate, you need to calculate the number of moles of water and anhydrous salt.

First, calculate the mass of the hydrate by subtracting the mass of the crucible and lid from the mass of the crucible, lid, and hydrate:

Mass of hydrate = mass of crucible, lid, and hydrate - mass of crucible and lid
= 29.986 g - 28.945 g
= 1.041 g

Calculate the mass of the anhydrous salt for the first measurement:

Mass of anhydrous salt (measurement 1) = mass of crucible, lid, and anhydrous salt - mass of crucible and lid
= 29.677 g - 28.945 g
= 0.732 g

Calculate the mass of the anhydrous salt for the second measurement:

Mass of anhydrous salt (measurement 2) = mass of crucible, lid, and anhydrous salt (2nd) - mass of crucible and lid
= 29.673 g - 28.945 g
= 0.728 g

To find the moles of water, divide the mass of water in the hydrate by its molar mass:

Moles of water = (mass of water / molar mass of water)
= (0.309 g / 18.015 g/mol)
= 0.0171 mol

To find the moles of anhydrous salt, divide the mass of anhydrous salt by its molar mass:

Moles of anhydrous salt = (mass of anhydrous salt / molar mass of anhydrous salt)
= (0.732 g / molar mass of anhydrous salt) (using measurement 1)
= (0.728 g / molar mass of anhydrous salt) (using measurement 2)

To determine the formula for the hydrate, compare the ratio of moles of water to moles of anhydrous salt. The simplest whole number ratio will give you the formula of the hydrate.

In this case, the ratio is approximately:
Moles of water : Moles of anhydrous salt = 0.0171 : 0.0289 (using measurement 1)
= 0.0171 : 0.0288 (using measurement 2)

To simplify the ratio, you can divide both values by 0.0171 to get approximately:
Moles of water : Moles of anhydrous salt = 1 : 1.69 (using measurement 1)
= 1 : 1.68 (using measurement 2)

From the ratio, it can be deduced that the formula of the hydrate is Co(CH3COO)2 * 1.68H2O. The number of water molecules is approximate and can be rounded to the nearest whole number to get the final formula, which would be Co(CH3COO)2 * 2H2O.

Surely you know something about how this should b done. Tell us what you don't understand.