SME
Analytical Chemistry
Gravimetric analysis
Use this for “A 0.6020 g iron ore sample gives 0.3415 g of Fe₂O₃ (159.69 g/mol) after ignition. What percent of the ore is iron (55.85 g/mol)?”
Chart
This is an example. Type your own values to start a new problem.
GF
=
×
MA (g/mol)
MP (g/mol)
wA (%)
=
mP (g)
×
mₛ (g)
×
100
MA and MP are the molar masses of the analyte and the precipitate; GF turns grams of precipitate into grams of analyte, and wA is the analyte’s percent of the sample.
mA
Analyte mass
calculated · g
Formulas
GF
=
r
×
MA
MP
0.2473
=
1
×
35.45
143.32
mA = mP × GF
0.1773 = 0.7168 × 0.2473
wA
=
100
×
mA
mₛ
35.46
=
100
×
0.1773
0.5
Assumptions
•
The precipitate is pure, dried or ignited to constant mass, and has the known formula whose molar mass is MP.
•
All of the analyte ends up in the precipitate: none stays dissolved or is lost in washing.
•
r counts the analyte atoms in one formula of the precipitate: 1 for Cl in AgCl, 2 for Fe in Fe₂O₃.
Step-by-step
WE KNOW
mₛ = 0.5 g, mP = 0.7168 g, MA = 35.45 g/mol, MP = 143.32 g/mol, r = 1
FIND
gravimetric factor (GF), analyte mass (mA), analyte in the sample (wA)
Step 1 · Find gravimetric factor (GF)
Use GF = r × MA ÷ MP
Each
mole
of
precipitate
holds
r
moles
of
analyte,
so
the
grams
of
analyte
per
gram
of
precipitate
are
r
MA
over
MP.
GF
=
1
×
35.45
143.32
GF = 0.2473
Step 2 · Find analyte mass (mA)
Use mA = mP × GF
The factor turns grams of precipitate into grams of the analyte it holds.
mA = 0.7168 × 0.2473
mA = 0.1773 g
Step 3 · Find analyte in the sample (wA)
Use wA = 100 × mA ÷ mₛ
The analyte’s grams over the sample’s grams, as a percent.
wA
=
100
×
0.1773
0.5
wA
=
17.73
0.5
wA = 35.46%
Check
Put the numbers back in. Both sides should match.
0.2473
=
1
×
35.45
143.32
✓
0.1773 = 0.7168 × 0.2473 ✓
35.46
=
100
×
0.1773
0.5
✓
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