SME
Analytical Chemistry · Problem type
Percent from the mass lost on heating
Use this for “A 0.9000 g magnesite sample weighs 0.5172 g after heating to constant mass, the loss being CO₂ (44.01 g/mol). What percent is MgCO₃ (84.31 g/mol)?”
Chart
Residue: 0.735 g
Gas lost: 0.515 g
This is an example. Type your own values to start a new problem.
w (%)
=
mₛ (g)
−
mheated (g)
mₛ (g)
×
Mcompound (g/mol)
Mgas (g/mol)
×
100
mₗₒₛₜ
Mass lost
calculated · g
mcompound
Mass of the compound
calculated · g
Formulas
mₗₒₛₜ = mₛ − mheated
0.515 = 1.25 − 0.735
mcompound
=
mₗₒₛₜ
×
Mcompound
Mgas
1.171
=
0.515
×
100.09
44.01
w
=
100
×
mcompound
mₛ
93.7
=
100
×
1.171
1.25
Assumptions
•
Heating drives off one gas completely, and only that gas: one mole of it for each mole of the compound.
•
Nothing else in the sample loses mass at that temperature, and the residue is weighed dry and cool.
Step-by-step
WE KNOW
mₛ = 1.25 g, mheated = 0.735 g, Mcompound = 100.09 g/mol, Mgas = 44.01 g/mol
FIND
mass lost (mₗₒₛₜ), mass of the compound (mcompound), compound in the sample (w)
Step 1 · Find mass lost (mₗₒₛₜ)
Use mₗₒₛₜ = mₛ − mheated
The gas driven off is the mass the sample lost on heating.
mₗₒₛₜ = 1.25 − 0.735
mₗₒₛₜ = 0.515 g
Step 2 · Find mass of the compound (mcompound)
Use mcompound = mₗₒₛₜ × Mcompound ÷ Mgas
Grams of gas over its molar mass give its moles, the same as the moles of compound; times the compound’s molar mass gives its grams.
mcompound
=
0.515
×
100.09
44.01
mcompound
=
51.546
44.01
mcompound = 1.171 g
Step 3 · Find compound in the sample (w)
Use w = 100 × mcompound ÷ mₛ
The compound’s grams over the sample’s grams, as a percent.
w
=
100
×
1.171
1.25
w
=
117.1
1.25
w
=
93.7%
Check
Put the numbers back in. Both sides should match.
0.515 = 1.25 − 0.735 ✓
1.171
=
0.515
×
100.09
44.01
✓
93.7
=
100
×
1.171
1.25
✓
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