A1g sample of enriched water, a mixture of H2O AND D2O, reacted completely with Cl2 to give a mixture of HCl and DCl. The HCl and DCl were then dissolved in pure H2O to make a 1.00 L solution. A 25.00 mL, sample of the 1 L solution was reacted with excess AgNO3 and 0.3800g of an AgCl precipitate formed. What was the mass % of D2O in the original sample of enriched water?
[H|D]2O + Cl2 = [H|D]Cl
[H|D]Cl + AgNO3 = AgCl + [H|D]NO3
Molar mass H2O: W1 = 18.01532 g/mol
Molar mass D2O: W2 = 20.02763 g/mol
Molar mass AgCl: W3 = 143.3214 g/mol
Sample mass: Ms = 1g
Precipitate: Mp = 0.3800g
Assay Volume: Vs = 1.00L
Aliquot Volume: Va = 25.00mL
Let n1 and n2 be the amounts of hydrogen and deuterium in the 1g mixed-hydrogen sample.
Ms = n1 W1 + n2 W2
or
1g = 18.01532 n1 + 20.02763 n2
The mole amount of silver chloride precipitate equals the mole amount of mixed-hydrogen in the aliquot, which is a representative fraction of that in the original sample.
(W3/Mp) = (n1+n2)(Va/Vs).
or
(143.3214/0.3800) = (n1+n2)(0.02500/1.00)
Solve these simultaneous equations for n1 and n2.
Mass % of D2O is then: (100% n2 W2 / Ms)
46.0%
To calculate the mass % of D2O in the original sample of enriched water, we need to determine the number of moles of D2O and H2O present in the sample.
Let's break down the information given step by step:
1. Enriched Water Reaction:
- A 1g sample of enriched water, which is a mixture of H2O and D2O, reacted completely with Cl2 to give a mixture of HCl and DCl.
2. Dissolving HCl and DCl in Pure H2O:
- The resulting HCl and DCl mixture were then dissolved in pure H2O to make a 1.00 L solution.
3. Reacting the 1 L Solution with AgNO3:
- A 25.00 mL sample of the 1 L solution was reacted with excess AgNO3.
- 0.3800g of AgCl precipitate formed.
Now, let's determine the number of moles of D2O and H2O involved in the reaction:
1. Moles of AgCl Precipitate:
- We have 0.3800g of AgCl precipitate formed from the reaction between AgNO3 and the HCl and DCl in the 25.00 mL sample.
- The molar mass of AgCl is 143.32 g/mol.
- We can calculate the number of moles of AgCl using the following formula:
Moles of AgCl = Mass of AgCl / Molar Mass of AgCl
Moles of AgCl = 0.3800g / 143.32 g/mol
2. Moles of Chloride Ions:
- Since AgCl precipitate formed from the reaction between AgNO3 and the chloride ions from HCl and DCl, the moles of AgCl directly represent the moles of chloride ions in the solution.
- Therefore, the number of moles of chloride ions present in the 25.00 mL sample is also the same as the moles of AgCl.
3. Moles of DCl and HCl:
- We know that HCl and DCl are both present in the 1.00 L solution.
- However, the mass % of D2O in the original sample is NOT affected by the Cl2 reaction, as the reaction only changes the D2O and H2O components, not the Cl2.
- Therefore, the number of moles of DCl and HCl in the 1.00 L solution is the same as the number of moles of chloride ions determined earlier.
4. Moles of D2O and H2O:
- Since the enriched water sample only contains D2O and H2O, the total moles of D2O and H2O can be calculated using the following equation:
Moles of D2O + Moles of H2O = Moles of DCl + Moles of HCl
- The moles of DCl and HCl were determined to be the same as the moles of chloride ions.
5. Mass % of D2O:
- The mass % of D2O in the original sample can be calculated as follows:
Mass % of D2O = (Moles of D2O / Total Moles of D2O and H2O) * 100
By following these steps and performing the necessary calculations, we can determine the mass % of D2O in the original sample of enriched water.