# how do I solve for:

How many grams of water would be lost form 100.0grams of CuCl2 x 5H2O when the hydrated salt was heated at 1000degreesC for one hour in a crucible?

What is the percent composition of water in the compound:

Percentwater= 5*(2+16)/(5*18 + atomic mass Cu + 2*atomicmassCl) * 100

That will tell you how much water is there.

Bob Pursley is right. Essentially all of the hydrated water would be lost at that extreme temperature, and you would be left with the anhydrous form, CuCl2.

## To solve for the grams of water lost, you need to calculate the percent composition of water in the compound. The formula provided can help you determine the amount of water present.

Percentwater = 5 * (2 + 16) / (5 * 18 + atomic mass Cu + 2 * atomic mass Cl) * 100

In this formula, 5 represents the coefficient of water (5H2O), and (2 + 16) represents the molecular weight of water. The denominator of the fraction accounts for the atomic masses of all the elements present in the compound.

To calculate the percent composition of water, you will need the atomic mass of copper (Cu) and chlorine (Cl). Look up their atomic masses on the periodic table.

Once you have determined the percent composition of water, you can calculate the amount of water lost from 100.0 grams of CuCl2 x 5H2O. Multiply the percent composition by the total mass of the compound (100.0 grams) to find the grams of water lost.

It's important to note that at the extreme temperature of 1000 degrees Celsius, essentially all of the hydrated water would be lost, leaving behind the anhydrous form of CuCl2.