Consider a mixture of the two solids BaCl2 * 2H2O (FW=244.26 g/mol) and KCl (FW=74.551 g/mol). When the mixture is heated to 160 C for 1 hour, the water of crystallization is driven off:
BaCl2 * 2 H2O (s) BaCl2(s) + 2 H2O(g)
A sample originally weighing 1.7839 g is found to weigh 1.5623 g after heating. Calculate the weight percent of Ba, K, and C1 in the original sample
1.7839g = mass BaCl2.2H2O + KCl
-1.5623g = mass after heating
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0.2216g = mass H2O driven off.
mass BaCl2.2H2O = 0.2216 x (molar mass BaCl2.2H2O/2*molar mass H2O) = ?
Then 1.7839-mass BaCl2.2H2O = mass KCl.
I will leave it for you to finish the problem. Post your work if you get stuck.
To calculate the weight percent of Ba, K, and Cl in the original sample, we need to determine the moles of BaCl2 * 2H2O and KCl in the sample and then calculate the weight percent using the molar masses.
Step 1: Calculate the moles of BaCl2 * 2H2O and KCl:
The molar mass of BaCl2 * 2H2O = 244.26 g/mol
The molar mass of KCl = 74.551 g/mol
Moles of BaCl2 * 2H2O = (mass of BaCl2 * 2H2O / molar mass of BaCl2 * 2H2O)
= (1.5623 g / 244.26 g/mol)
Moles of KCl = (mass of KCl / molar mass of KCl)
= (1.7839 g - 1.5623 g) / 74.551 g/mol
Step 2: Calculate the weight percent of Ba, K, and Cl in the original sample:
Weight percent of Ba = (moles of BaCl2 * 2H2O * molar mass of Ba) / (total moles * molar mass of the sample) * 100%
Weight percent of K = (moles of KCl * molar mass of K) / (total moles * molar mass of the sample) * 100%
Weight percent of Cl = ((2 * moles of BaCl2 * 2H2O + moles of KCl) * molar mass of Cl) / (total moles * molar mass of the sample) * 100%
Substitute the calculated values into the formulas to get the weight percent:
Weight percent of Ba = ((1.5623 g / 244.26 g/mol) * 137.327 g/mol) / ((1.5623 g / 244.26 g/mol) + ((1.7839 g - 1.5623 g) / 74.551 g/mol)) * 100%
Weight percent of K = (((1.7839 g - 1.5623 g) / 74.551 g/mol) * 39.0983 g/mol) / ((1.5623 g / 244.26 g/mol) + ((1.7839 g - 1.5623 g) / 74.551 g/mol)) * 100%
Weight percent of Cl = ((2 * (1.5623 g / 244.26 g/mol) + ((1.7839 g - 1.5623 g) / 74.551 g/mol)) * 35.453 g/mol) / ((1.5623 g / 244.26 g/mol) + ((1.7839 g - 1.5623 g) / 74.551 g/mol)) * 100%
To calculate the weight percent of Ba, K, and Cl in the original sample, we need to determine the weight of each element in the sample before and after heating.
1. Determine the weight of BaCl2 * 2H2O in the original sample:
- The molar mass of BaCl2 * 2H2O is 244.26 g/mol.
- Calculate the moles of BaCl2 * 2H2O in the original sample using the formula:
Moles = Mass / Molar mass
Moles = 1.7839 g / 244.26 g/mol = 0.007302 mol
- Calculate the weight of BaCl2 * 2H2O in the original sample:
Weight = Moles * Molar mass
Weight = 0.007302 mol * 244.26 g/mol = 1.7839 g
2. Determine the weight of BaCl2 after heating:
- Since the water of crystallization is driven off, the weight of BaCl2 will remain the same after heating.
- The weight of BaCl2 is 1.5623 g (as given in the question).
3. Determine the weight of H2O(g) after heating:
- The weight of H2O(g) can be calculated by subtracting the weight of BaCl2 after heating from the weight of the original sample:
Weight of H2O(g) = Weight of original sample - Weight of BaCl2 after heating
Weight of H2O(g) = 1.7839 g - 1.5623 g = 0.2216 g
4. Determine the weight of KCl after heating:
- Since KCl does not undergo any change during the heating process, its weight in the original sample remains the same.
- The weight of KCl is 1.7839 g (as given in the question).
5. Calculate the weight percent of each element:
- Weight percent of Ba = (Weight of Ba in the original sample / Weight of the original sample) * 100
Weight percent of Ba = (1.7839 g / 1.7839 g) * 100 = 100%
- Weight percent of K = (Weight of K in the original sample / Weight of the original sample) * 100
Weight percent of K = (0 g / 1.7839 g) * 100 = 0%
- Weight percent of Cl = (Weight of Cl in the original sample / Weight of the original sample) * 100
Weight percent of Cl = (1.5623 g / 1.7839 g) * 100 = 87.5%
Therefore, the weight percent of Ba, K, and Cl in the original sample is 100%, 0%, and 87.5%, respectively.