SME
Grade 10 Science · Problem type
Empirical formula from percents
Use this to find the simplest whole-number formula from the percent of each element.
Chart
40.00% + 6.70% + 53.30% = 100% of the whole.
This is an example. Type your own values to start a new problem.
%C
Percent carbon
example · %
%H
Percent hydrogen
example · %
%O
Percent oxygen
calculated · %
nC
Moles of C in 100 g
worked out (not typed) · mol
nH
Moles of H in 100 g
worked out (not typed) · mol
nO
Moles of O in 100 g
worked out (not typed) · mol
s
Smallest amount
worked out (not typed) · mol
C
C per smallest
worked out (not typed)
H
H per smallest
worked out (not typed)
O
O per smallest
worked out (not typed)
Formulas
%
C
+
%
H
+
%
O
=
100
40.00 + 6.70 + 53.30 = 100
nC
=
%
C
12.01
3.331
=
40.00
12.01
nH
=
%
H
1.008
6.647
=
6.70
1.008
nO
=
%
O
16.00
3.331
=
53.30
16.00
s
=
smallest
of
nC,
nH
and
nO
3.331 = smallest of 3.331, 6.647 and 3.331
C
=
nC/
s
1 = 3.331/3.331
H
=
nH/
s
1.996 = 6.647/3.331
O
=
nO/
s
1 = 3.331/3.331
Assumptions
•
Take 100 g of the compound, so each percent is that many grams.
•
Change grams to moles, then divide every amount by the smallest.
•
Round ratios near a whole number; if one ends in 0.5 or 0.33, multiply them all by 2 or 3 first.
Step-by-step
WE KNOW
%C = 40.00%, %H = 6.70%
FIND
percent oxygen (%O), moles of C in 100 g (nC), moles of H in 100 g (nH), moles of O in 100 g (nO), smallest amount (s), C per smallest (C), H per smallest (H), O per smallest (O)
Step 1 · Find percent oxygen (%O)
Use %C + %H + %O = 100
The three elements make up the whole.
%
O
=
100
−
%
C
−
%
H
%
O
=
100
−
40.00
−
6.70
%
O
=
53.30%
Step 2 · Find moles of C in 100 g (nC)
Use nC = %C ÷ 12.01
In
100
g
the
percent
is
grams:
divide
by
12.01
g
per
mole
of
C
.
nC
=
40.00
12.01
40.00%
of
100
g
is
40.00
g
of
C
40.00
g
12.01
g/mol
=
3.331
mol
nC = 3.331 mol
Step 3 · Find moles of H in 100 g (nH)
Use nH = %H ÷ 1.008
In
100
g
the
percent
is
grams:
divide
by
1.008
g
per
mole
of
H
.
nH
=
6.70
1.008
6.70%
of
100
g
is
6.70
g
of
H
6.70
g
1.008
g/mol
=
6.647
mol
nH = 6.647 mol
Step 4 · Find moles of O in 100 g (nO)
Use nO = %O ÷ 16.00
In
100
g
the
percent
is
grams:
divide
by
16.00
g
per
mole
of
O
.
nO
=
53.30
16.00
53.30%
of
100
g
is
53.30
g
of
O
53.30
g
16.00
g/mol
=
3.331
mol
nO = 3.331 mol
Step 5 · Find smallest amount (s)
Use s = smallest of nC, nH and nO
The scarcest element sets the unit: one of its atoms.
s
=
smallest
of
3.331,
6.647
and
3.331
s
=
3.331
mol
Step 6 · Find C per smallest (C)
Use C = nC/s
Divide
by
the
smallest:
C
atoms
for
each
atom
of
the
scarcest
element.
C
=
3.331/3.331
C
=
1
(about
1)
Step 7 · Find H per smallest (H)
Use H = nH/s
Divide
by
the
smallest:
H
atoms
for
each
atom
of
the
scarcest
element.
H
=
6.6468254/3.3305579
H
=
1.996
(about
2)
Step 8 · Find O per smallest (O)
Use O = nO/s
Divide
by
the
smallest:
O
atoms
for
each
atom
of
the
scarcest
element.
O
=
3.331/3.331
O
=
1
(about
1;
the
empirical
formula
is
CH₂
O
)
Check
Put the numbers back in. Both sides should match.
40.00 + 6.70 + 53.30 = 100 ✓
3.331
≈
40.00
12.01
✓
6.647
≈
6.70
1.008
✓
3.331
≈
53.30
16.00
✓
3.331 = smallest of 3.331, 6.647 and 3.331 ✓
1 = 3.331/3.331 ✓
1.996 ≈ 6.6468254/3.3305579 ✓
1 = 3.331/3.331 ✓
Topic check
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