two hydrates were weighed, heated to drive off waters of hydration, and then cooled. The residues were then rewieghed. Based on the follwing results, what are the formulas of the gydrates?
Compound. Initial mass. M after cooling
NiSO4*xH2O 2.08 1.22
CoCl*xH2O 1.62 .88
#1. mass water evolved = 2.08-1.22 = 0.86 grams.
mass anhydrous NiSO4 = 1.22 grams
moles H2O = 0.86/18 =0.0478
moles NiSO4 = 1.22/154.76 = 0.00788.
Divide by the smaller to find the ratio to 1 mole NiSO4.
0.00788/0.00788 = 1 NiSO4
0.0478/0.00788 = 6.06
round these to 1 NiSO4 to 6 H2O and the formula is
NiSO4.6H2O.
Check my work.
The Co one is done the same way. Check to make sure that Co formula isn't CoCl2.xH2O.
Sorry that might be a little unlclear..I hope this helps
Compound: NiSO4*xH2O
Initial mass: 2.08
Mass after cooling: 1.22
Compound: CoCl*xH2O
Initial mass: 1.62
Mass after cooling: .88
I understand now...THANK YOU!
To determine the formulas of the hydrates, we need to compare the initial masses with the masses after cooling. The difference in mass corresponds to the mass of water that was driven off during heating.
Let's start with the compound NiSO4*xH2O. The initial mass of the compound was 2.08 g, and after cooling, the mass reduced to 1.22 g. The difference in mass is 2.08 g - 1.22 g = 0.86 g.
The mass of water lost during heating is 0.86 g. To calculate the number of moles of water lost, we need to divide the mass by the molar mass of water (18.015 g/mol).
0.86 g / 18.015 g/mol ≈ 0.0477 mol
Now, let's consider the compound CoCl*xH2O. The initial mass of the compound was 1.62 g, and after cooling, the mass reduced to 0.88 g. The difference in mass is 1.62 g - 0.88 g = 0.74 g.
The mass of water lost during heating is 0.74 g. Dividing this by the molar mass of water gives:
0.74 g / 18.015 g/mol ≈ 0.041 moles
To find the value of "x" in each formula, we need to calculate the ratio of moles of water lost to the moles of anhydrous salt remaining after heating.
For NiSO4*xH2O:
Moles of water lost = 0.0477 mol
Moles of anhydrous salt remaining = 1.22 g / (molar mass of NiSO4 ≈ 154.76 g/mol) = 0.00788 mol
The ratio of moles of water lost to moles of anhydrous salt remaining is:
0.0477 mol / 0.00788 mol ≈ 6
Therefore, the formula for NiSO4*xH2O is NiSO4*6H2O.
For CoCl*xH2O:
Moles of water lost = 0.041 mol
Moles of anhydrous salt remaining = 0.88 g / (molar mass of CoCl ≈ 129.84 g/mol) = 0.00677 mol
The ratio of moles of water lost to moles of anhydrous salt remaining is:
0.041 mol / 0.00677 mol ≈ 6
Therefore, the formula for CoCl*xH2O is CoCl*6H2O.
In conclusion, the formulas of the hydrates are NiSO4*6H2O and CoCl*6H2O.