To calculate the percentage of MgCl2 in the given sample, we need to determine the amount of MgCl2 present in the sample and then calculate it as a percentage of the total sample mass.
Firstly, let's calculate the moles of AgNO3 used to precipitate chloride:
Moles of AgNO3 = molarity of AgNO3 × volume of AgNO3 solution
= 0.1030 M × 0.05000 L
= 0.00515 moles
AgNO3 reacts with chloride ions in a 1:2 ratio. So, moles of Cl- precipitated = 2 × moles of AgNO3 used
= 2 × 0.00515 moles
= 0.0103 moles
Next, let's calculate the moles of KSCN used to titrate the excess silver:
Moles of KSCN = molarity of KSCN × volume of KSCN solution
= 0.1260 M × 0.00865 L
= 0.0010899 moles
From the reaction, it is known that 1 mole of AgNO3 reacts with 1 mole of KSCN. Thus, the moles of AgNO3 remaining after titration = Moles of KSCN used
= 0.0010899 moles
To find the moles of AgNO3 reacted with MgCl2, subtract the remaining moles of AgNO3 from the total moles used:
Moles of AgNO3 reacted with MgCl2 = Total moles of AgNO3 used - Remaining moles of AgNO3
= 0.00515 moles - 0.0010899 moles
= 0.00406 moles
As per the reaction equation, 2 moles of AgNO3 react with 1 mole of MgCl2. So, the moles of MgCl2 present in the sample can be calculated as:
Moles of MgCl2 = (Moles of AgNO3 reacted with MgCl2) / 2
= 0.00406 moles / 2
= 0.00203 moles
Now, let's calculate the mass of MgCl2:
Mass of MgCl2 = Moles of MgCl2 × molar mass of MgCl2
= 0.00203 moles × (24.31 g/mol + 2 × 35.45 g/mol)
= 0.00203 moles × 95.21 g/mol
= 0.193 g
Finally, we can calculate the percentage of MgCl2 in the sample:
Percentage of MgCl2 = (Mass of MgCl2 / Mass of the sample) × 100
= (0.193 g / 0.272 g) × 100
= 71.03%
Therefore, the outcome of the analysis is that the sample contains approximately 71.03% MgCl2.