SME
General Chemistry I
Stoichiometry
Use this for “A 0.2500 g sample of a compound of C, H and O burns to 0.3664 g of CO₂ and 0.1500 g of H₂O. Find its empirical formula.”
Diagram
n(C)
=
0.3664
44.01
=
8.33 ×
10
-3
mol;
n(H)
=
2
×
0.15
18.02
=
0.0166
mol.
m(O) = 0.25 − 12.01 × 8.33 × 10⁻³ − 1.008 × 0.0166 = 0.133 g, so n(O) = 8.33 × 10⁻³ mol.
Ratio C : H : O = 1 : 2 : 1, so CH₂O.
All C ends in CO₂ and all H in H₂O; O is the mass left over.
This is an example. Type your own values to start a new problem.
m
Sample mass
example · g
mCO₂
CO₂ mass
example · g
mH2O
H₂O mass
example · g
nC
Moles of C
calculated · mol
nH
Moles of H
calculated · mol
mO
Mass of O
calculated · g
nO
Moles of O
calculated · mol
rH
H per C
calculated
rO
O per C
calculated
Formulas
nC
=
mCO₂
44.01
0.008325
=
0.3664
44.01
nH
=
2
×
mH2O
18.02
0.01665
=
2
×
0.15
18.02
mO
=
m
−
12.01
×
nC
−
1.008
×
nH
0.1332 = 0.25 − 12.01 × 0.008325 − 1.008 × 0.01665
nO
=
mO
16.00
0.008327
=
0.1332
16.00
rH
=
nH
nC
2
=
0.01665
0.008325
rO
=
nO
nC
1
=
0.008327
0.008325
Assumptions
•
All the C ends in CO₂ and all the H in H₂O. O is what is left of the sample’s mass, since the O₂ it burns in can’t be told apart.
•
Molar masses: CO₂ 44.01, H₂O 18.02, C 12.01, H 1.008 and O 16.00 g/mol.
•
Ratios within 0.05 of a whole number round; otherwise multiply them all by 2, 3 … until they are whole (1.33 × 3 = 4).
Step-by-step
WE KNOW
m = 0.25 g, mCO₂ = 0.3664 g, mH2O = 0.15 g
FIND
moles of C (nC), moles of H (nH), mass of O (mO), moles of O (nO), H per C (rH), O per C (rO)
Step 1 · Find moles of C (nC)
Use nC = mCO₂ ÷ 44.01
Each mole of CO₂ holds one mole of C.
nC
=
0.3664
44.01
nC = 0.008325 mol
Step 2 · Find moles of H (nH)
Use nH = 2 × mH2O ÷ 18.02
Each mole of H₂O holds two moles of H.
nH
=
2
×
0.15
18.02
nH
=
0.3
18.02
nH = 0.01665 mol
Step 3 · Find mass of O (mO)
Use mO = m − 12.01 × nC − 1.008 × nH
Take the grams of C and of H from the sample: what is left is O.
mO = 0.25 − 12.01 × 0.008325 − 1.008 × 0.01665
mO = 0.25 − 0.09998 − 0.01678
mO = 0.13323 g
Step 4 · Find moles of O (nO)
Use nO = mO ÷ 16.00
Divide the grams of O by 16.00 g/mol.
nO
=
0.13323
16.00
nO = 0.008327 mol
Step 5 · Find H per C (rH)
Use rH = nH ÷ nC
Divide by the moles of C to count the H atoms for each C.
rH
=
0.01665
0.008325
rH = 2
Step 6 · Find O per C (rO)
Use rO = nO ÷ nC
Divide by the moles of C to count the O atoms for each C.
rO
=
0.008327
0.008325
rO = 1 → 1 C, 2 H and 1 O, so the empirical formula is CH₂O.
Check
Put the numbers back in. Both sides should match.
0.008325
=
0.3664
44.01
✓
0.01665
=
2
×
0.15
18.02
✓
0.13323 = 0.25 − 12.01 × 0.008325 − 1.008 × 0.01665 ✓
0.008327
=
0.13323
16.00
✓
2
=
0.01665
0.008325
✓
1
=
0.008327
0.008325
✓
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The molecular formula from the empirical formula
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