A hydrate of zinc nitrate has the formula Zn(NO3)2 . xH2O. If the mass of 1 mol of anhydrous zinc nitrate is 63.67% of the mass of 1 mol of the hydrate, what is the value of x?
So the %H2O = 100 - 63.67 = 36.33
a 100 g sample contains 64.67 g Zn(NO3)2 and 36.33 g H2O.
mols H2O = 36.33/18.015
mols Zn(NO3)2 = 63.67/molar mass zinc nitrate (don't include the water).
Find the ratio of mols H2O to 1 mol Zn(NO3)2. That will be x in the formula.
x=6
To find the value of x in the formula Zn(NO3)2 . xH2O, we need to use the information given about the mass of 1 mol of anhydrous zinc nitrate and the hydrate.
Let's start by understanding what a hydrate is. A hydrate is a compound that has water molecules trapped within its crystal structure. In this case, the hydrate of zinc nitrate has water molecules represented by xH2O.
According to the given information, the mass of 1 mol of anhydrous zinc nitrate is 63.67% of the mass of 1 mol of the hydrate. We can set up the following equation to solve for x:
Mass of anhydrous zinc nitrate = 0.6367 * Mass of hydrate
Now, we need to determine the molar mass of anhydrous zinc nitrate and the hydrate. The molar mass of anhydrous zinc nitrate is the molar mass of zinc (Zn) plus twice the molar mass of nitrate (NO3):
Molar mass of anhydrous zinc nitrate = Molar mass of Zn + 2 * Molar mass of NO3
Let's calculate the molar mass of Zn and NO3:
Molar mass of Zn = 65.38 g/mol (atomic mass of Zn)
Molar mass of NO3 = 62.00 g/mol (atomic mass of N) + 3 * 16.00 g/mol (atomic mass of O)
Molar mass of NO3 = 62.00 g/mol + 48.00 g/mol = 110.00 g/mol
Now we can calculate the molar mass of anhydrous zinc nitrate:
Molar mass of anhydrous zinc nitrate = 65.38 g/mol + 2 * 110.00 g/mol
Molar mass of anhydrous zinc nitrate = 285.38 g/mol
Next, we need to determine the molar mass of the hydrate. The molar mass of the hydrate can be calculated by adding the molar mass of anhydrous zinc nitrate and the molar mass of xH2O:
Molar mass of hydrate = Molar mass of anhydrous zinc nitrate + Molar mass of xH2O
The molar mass of water (H2O) is 18.02 g/mol.
Molar mass of hydrate = 285.38 g/mol + 18.02 g/mol * x
Now we can substitute the given information into our equation:
285.38 g/mol = 0.6367 * (285.38 g/mol + 18.02 g/mol * x)
Simplifying the equation:
285.38 = 0.6367 * (285.38 + 18.02 * x)
Divide both sides of the equation by 0.6367:
285.38 / 0.6367 = 285.38 + 18.02 * x
447.91 = 285.38 + 18.02 * x
Subtract 285.38 from both sides of the equation:
447.91 - 285.38 = 285.38 - 285.38 + 18.02 * x
162.53 = 18.02 * x
Now divide both sides of the equation by 18.02 to solve for x:
162.53 / 18.02 = 18.02 * x / 18.02
x = 9
Therefore, the value of x is 9.