SME
Grade 10 Science · Problem type
Liters of gas at STP
Use this for “How many moles, grams and molecules are in 11.2 L of O₂ at STP?”
Diagram
Mass = 0.5 mol × 32 g/mol = 16 g.
Particles = 0.5 mol × 6.022 × 10²³ = 3.011 × 10²³.
Volume at STP = 0.5 mol × 22.4 L/mol = 11.2 L.
Every path goes through moles: the filled box is the value given.
This is an example. Type your own values to start a new problem.
V
L
÷
22.4
L/mol
=
n
mol
m
Mass
calculated · g
N
Particles
calculated
Formulas
V
=
22.4
×
n
11.2 = 22.4 × 0.5
m
=
32.00
×
n
16 = 32.00 × 0.5
N
=
6.022 ×
10
23
×
n
3.01 ×
10
23
=
6.022 ×
10
23
×
0.5
Assumptions
•
22.4 L per mole holds only for a gas at 0 °C and 1 atm (STP).
•
Oxygen gas, O₂, has a molar mass of 32.00 g/mol.
Step-by-step
WE KNOW
V = 11.2 L
FIND
amount (n), mass (m), particles (N)
Step 1 · Find amount (n)
Use V = 22.4 × n
Divide the liters by 22.4 L per mole.
n
=
V
/22.4
n
=
11.2/22.4
n
=
0.5
mol
Step 2 · Find mass (m)
Use m = 32.00 × n
One mole of O₂ has a mass of 32.00 g.
m
=
32.00
×
0.5
m
=
16
g
Step 3 · Find particles (N)
Use N = 6.022 × 10²³ × n
One
mole
is
6.022 ×
10
23
particles.
N
=
6.022 ×
10
23
×
0.5
N
=
3.01 ×
10
23
Check
Put the numbers back in. Both sides should match.
11.2 = 22.4 × 0.5 ✓
16 = 32.00 × 0.5 ✓
3.01 ×
10
23
≈
6.022 ×
10
23
×
0.5
✓
Topic check
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