SME
Organic Chemistry I · Problem type
Rings and π bonds from the H₂ taken up
Use this for “0.750 g of C₁₀H₁₆ (136.24 g/mol) takes up 11.01 mmol of H₂ over Pd. How many rings and π bonds?”
Diagram
IHD
=
2 × 6 + 2 − 10
2
=
2.
Cyclohexene, C₆H₁₀: each π bond takes one H₂, so 1 π bond; rings = IHD − π bonds = 2 − 1 = 1.
Sample mass m = 0.8215 g
Molar mass M = 82.1 g/mol
Moles of compound n = 10 mmol
Moles of H₂ taken up nH₂ = 10 mmol
This is an example. Type your own values to start a new problem.
C
Carbon atoms
example
H
Hydrogen atoms
example
IHD
Index of hydrogen deficiency
calculated
m
Sample mass
example
M
Molar mass
calculated · g/mol
n
Moles of compound
calculated
nH₂
Moles of H₂ taken up
example
x
H₂ taken up per molecule
calculated
π
π bonds
calculated
rings
Rings
calculated
Formulas
IHD
=
2 ×
C
+ 2 −
H
2
2
=
2 × 6 + 2 − 10
2
M
=
12.011
×
C
+
1.008
×
H
82.1 = 12.011 × 6 + 1.008 × 10
n
=
m
M
0.01
=
0.8215
82.1
x
=
nH₂
n
1
=
0.01
0.01
π
=
x
rounded
to
the
ones
1 = 1 rounded to the ones
rings
=
IHD
−
π
1 = 2 − 1
Formulas are worked in mol. The step-by-step shows the conversions.
Assumptions
•
The compound is a hydrocarbon, CₓHᵧ: its IHD, (2C + 2 − H) ÷ 2, counts its rings and π bonds together.
•
H₂ with Pd adds to every C=C and C≡C, one H₂ for each π bond (two for C≡C); rings stay. A benzene ring takes up no H₂, so this page is for compounds without one.
•
The H₂ is measured, so the H₂ taken up per molecule is rounded to the nearest whole number of π bonds.
•
The picture draws one known structure that fits the counts, each π bond marked +H₂, when it has one.
Step-by-step
WE KNOW
C = 6, H = 10, m = 0.8215 g, nH₂ = 10 mmol
FIND
index of hydrogen deficiency (IHD), molar mass (M), moles of compound (n), H₂ taken up per molecule (x), π bonds (π), rings (rings)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to mol).
nH₂ = 10 mmol = 0.01 mol (1 mol = 1000 mmol)
Step 2 · Find index of hydrogen deficiency (IHD)
Use IHD = (2 × C + 2 − H) ÷ 2
An
open
chain
of
C
carbons
holds
2
C
+
2
H;
each
ring
or
π
bond
is
2
H
fewer,
so
halve
the
H
missing.
IHD
=
2 × 6 + 2 − 10
2
IHD
=
12 + 2 − 10
2
IHD
=
4
2
IHD = 2
Step 3 · Find molar mass (M)
Use M = 12.011 × C + 1.008 × H
Add
the
masses
of
the
atoms:
12.011
g/mol
for
each
C
and
1.008
g/mol
for
each
H
.
M
=
12.011
×
6
+
1.008
×
10
M
=
72.066
+
10.08
M
=
82.1
g/mol
Step 4 · Find moles of compound (n)
Use n = m ÷ M
Divide the mass by the molar mass.
n
=
0.8215
82.1
n
=
0.01
mol
Step 5 · Find H₂ taken up per molecule (x)
Use x = nH₂ ÷ n
Each
π
bond
takes
one
H₂,
so
the
moles
of
H₂
for
each
mole
of
compound
counts
the
π
bonds.
x
=
0.01
0.01
x
=
1
Step 6 · Find π bonds (π)
Use π = x rounded to the ones
A
molecule
holds
a
whole
number
of
π
bonds:
round
the
H₂
taken
up
per
molecule.
π
=
1
rounded
to
the
ones
π
=
1
Step 7 · Find rings (rings)
Use rings = IHD − π
The
IHD
counts
rings
and
π
bonds
together;
the
H₂
found
the
π
bonds,
so
the
rest
are
rings.
rings = 2 − 1
rings = 1
Step 8 · Convert the answers
Change each answer back to the unit you chose.
n = 0.01 mol = 10 mmol (1 mol = 1000 mmol)
Check
Put the numbers back in. Both sides should match.
2
=
2 × 6 + 2 − 10
2
✓
82.1 = 12.011 × 6 + 1.008 × 10 ✓
0.01
=
0.8215
82.1
✓
1
=
0.01
0.01
✓
1 = 1 rounded to the ones ✓
1 = 2 − 1 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.