SME
Analytical Chemistry · Problem type
Propagating uncertainty into a concentration
Use this for “0.4000 ± 0.0003 g of KCl (74.55 g/mol) is made up to 0.5000 ± 0.0002 L. Give the molarity with its uncertainty.”
This is an example. Type your own values to start a new problem.
m
Mass of solute
example · g
sₘ
Uncertainty in the mass
example · g
V
Volume of solution
example · L
sV
Uncertainty in the volume
example · L
M
Molar mass
example · g/mol
c
Concentration
calculated · M
rel
Relative uncertainty
calculated
sc
Uncertainty in the concentration
calculated · M
Formulas
c
=
m
M
×
V
0.01711
=
0.25
58.44 × 0.25
rel
=
√(
(
sₘ
m
)
2
+
(
sV
V
)
2
)
8.9443 ×
10
-4
=
√(
(
0.0002
0.25
)
2
+
(
0.0001
0.25
)
2
)
sc
=
c
×
rel
1.5305 ×
10
-5
=
0.01711
×
(8.9443
×
10
-4
)
Assumptions
•
The errors in the mass and the volume are random and independent of each other.
•
In a product or quotient the relative uncertainties add in quadrature: square, add, take the root.
•
The molar mass is exact, so it adds no uncertainty.
Step-by-step
WE KNOW
m = 0.25 g, sₘ = 0.0002 g, V = 0.25 L, sV = 0.0001 L, M = 58.44 g/mol
FIND
concentration (c), relative uncertainty (rel), uncertainty in the concentration (sc)
Step 1 · Find concentration (c)
Use c = m ÷ (M × V)
Moles
of
solute
are
m
M
;
spread
them
through
the
volume
in
liters.
c
=
0.25
58.44 × 0.25
c
=
0.25
14.61
c
=
0.017112
M
Step 2 · Find relative uncertainty (rel)
Use rel = √((sₘ ÷ m)² + (sV ÷ V)²)
c
is
a
quotient
of
m
and
V
,
so
their
relative
uncertainties
add
in
quadrature.
rel
=
√(
(
0.0002
0.25
)
2
+
(
0.0001
0.25
)
2
)
rel
=
√
0.0008
2
+
0.0004
2
rel
=
√((6.400
×
10
-7
)
+
(1.600
×
10
-7
))
rel
=
√
8.000 ×
10
-7
rel
=
8.94427 ×
10
-4
Step 3 · Find uncertainty in the concentration (sc)
Use sc = c × rel
The
relative
uncertainty
is
sc
c
,
so
scale
it
by
the
concentration.
sc
=
0.017112
×
(8.94427
×
10
-4
)
sc
=
1.5305 ×
10
-5
M
Check
Put the numbers back in. Both sides should match.
0.017112
=
0.25
58.44 × 0.25
✓
8.94427 ×
10
-4
=
√(
(
0.0002
0.25
)
2
+
(
0.0001
0.25
)
2
)
✓
1.5305 ×
10
-5
=
0.017112
×
(8.94427
×
10
-4
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Error analysis and statistics
Main lesson
Rejecting a suspect reading (Q test)
Use this for “Four titrations take 24.18, 24.21, 24.22 and 24.39 mL. Qcrit is 0.829 for four readings at 95%. May 24.39 mL be rejected?”
Comparing a mean with a known value (t-test)
Use this for “A standard certified at 5.00 mg/L lead reads 5.21 mg/L on average over 6 runs, s = 0.18 mg/L. Is the method biased at 95%?”