Spring 2024 AS Sessions:
Session 1 – Radioactivity and Ionisation Energies (NS)
Session 2 – Gravimetric Practical (PB)
Session 3 – Instrumental Analysis (NS)
Session 4 – Making up a Standard Solution and Titrations (PB)
Gravimetric Analysis
Silver Nitrate
Mr
?
Gravimetric method – to find moles XCl
XCl(aq) + AgNO3(aq) 🡪 AgCl(s) + XNO3(aq)
Cl-(aq) + Ag+(aq) 🡪 AgCl(s)
XCl (aq)
EXCESS
Remember
nAgCl = nXCl
XCl(aq) + AgNO3(aq) 🡪 AgCl(s) + XNO3(aq)
XCl(aq) + AgNO3(aq) 🡪 AgCl(s) + XNO3(aq)
n(XCl) = 0.00407 mol
0.305
Molar mass = 0.305
0.00407
XCl =74.94
X = 74.94 – 35.5 = 39.43
Summary
Mr of silver chloride
moles of XCl
Applying to a similar situation
Questions
What is the charge of the carbonate ion?
What does this tell us about the group of M?
In practical tems how will gravimetric analysis be done to establish M’s identity?
Accurately weigh the carbonate M2CO3
Form a ppt of CaCO3 from the M2CO3
Filter, dry to constant mass and weigh ppt
of CaCO3
Find moles of CaCO3 ppt formed
Use ppt equation to relate to moles of CaCO3 to moles of M2CO3
Making the precipitate
Add CaCl2 (aq)
M2CO3(aq) + CaCl2(aq)
CaCO3(s) + 2 MCl(aq)
Filter, dry to constant mass and weigh ppt
Find moles of CaCO3 ppt formed
Moles of CaCO3 = 1.838/100.1
= 0.01836
Mass of CaCO3 formed = 1.838g
Use eqn to find moles of M2CO3
Moles of M2CO3 = 0.01836
Moles of M2CO3 = 0.01836
Mr (M2CO3) = mass/moles
Calculation to find Mr
Mr = 1.972/0.01836
= 107.4
Identity of M in M2CO3
M= 107 – 60 = 23
2
Mass of boat and contents
= 1.4614g
Mass of emptied boat
= 0.9662g
Mass added = 1.4614-0.9662
= 0.4952g
Mass added = 1.4614-0.9662
= 0.4952g
Error in weighing = 0.5246-0.4952
= 0.0294g
% error = 0.0294/0.4952 x 100
= 5.94% !!!
Spring 2024 AS Sessions:
Session 1 – Radioactivity and Ionisation Energies (NS)
Session 2 – Gravimetric Practical (PB)
Session 3 – Instrumental Analysis (NS)
Session 4 – Making up a Standard Solution and Titrations (PB)