Note: The weird name you have used for FeSO4 is not an IUPAC approved name.
12.3 g = mass FeSO4 + H2O
-06.0 g = mass FeSO4
--------------------
6.3 g = mass water of crystallization.
%H2O = (mass H2O/mass sample)*100 = ?
12.3 g = mass FeSO4 + H2O
-06.0 g = mass FeSO4
--------------------
6.3 g = mass water of crystallization.
%H2O = (mass H2O/mass sample)*100 = ?
1. Determine the mass of the water lost:
Mass of hydrated salt (before heating) = 12.3g
Mass of anhydrous salt (after heating) = 6g
Mass of water lost = Mass of hydrated salt - Mass of anhydrous salt
= 12.3g - 6g
= 6.3g
2. Calculate the percentage by mass of water:
% mass of water lost = (Mass of water lost / Mass of hydrated salt) x 100
= (6.3g / 12.3g) x 100
≈ 51.22%
Therefore, the % by mass of water of crystallization lost during heating is approximately 51.22%.
1. Start by calculating the initial mass of the compound (iron(II) tetraoxosulphate(VI)):
Initial mass = mass of anhydrous salt + mass of water of crystallization
= 6g (given) + mass of water of crystallization
2. Next, subtract the mass of the anhydrous salt from the initial mass to obtain the mass of water lost during heating:
Mass of water lost = Initial mass - mass of anhydrous salt
= Initial mass - 6g
3. Now, we need to calculate the percentage by mass of water of crystallization. This can be done using the following formula:
% by mass = (Mass of water lost / Initial mass) x 100
Let's substitute the values into the formula:
% by mass = (Mass of water lost / Initial mass) x 100
= [(Initial mass - 6g) / Initial mass] x 100
However, we do not have the value for the initial mass. To find this, we need additional information. Please provide the mass of the iron(II) tetraoxosulphate(VI) crystals mentioned in the question.