How to solve for molality, one step at a time:
If you are asked to solve for molality, there are 4 small steps you should do to get there.
The small steps listed below are going to be used to solve the following 2 example
problems:
 28 grams of CaBr2 is dissolved in 500 g of H2O. What is the molality of the solution?
 115 grams of Al(NO3)3 is dissolved in 1.35 kg of ethanol. What is the molality of the
solution?
1. Formula
Mass
Given:
 A chemical formula
 A periodic table
(also known as molar mass,
molecular mass, etc)
1st Example
2nd Example
2. Moles of
solute
Formula = CaBr2
(given in problem)
Ca = 40.08 g/mol
Br = 79.90 g/mol
(given on periodic
table)
Formula: Al(NO3)3
(given in problem)
Al = 26.98 g/mol
(given on periodic
N = 14.01 g/mol
table)
O = 16.00 g/mol
Given:
 Grams of solute
 Formula mass of the solute
1st Example 28 g CaBr2
FM CaBr2 =
199.88 g/mol
(given in problem)
(found in step 1)
A. Add together the average masses
(given on the periodic table) for the
atoms in the formula.
B. Unless there are parentheses in the
formula, subscripts only refer to the
symbol next to them.
40.08 + 2(79.90) = 199.88 g/mol
26.98 + 3(14.01) + 9(16.00) = 213.01 g/mol
Or:
26.98 + 3(14.01 + 16.00) = 213.01 g/mol
(Both methods are equally true.)
A. Grams ÷ formula mass = moles
B. Make sure you don’t mix solute &
solvent together. Solute is the stuff
being dissolved in a liquid.
This can be solved on the line:
28 g CaBr2 | 1 mol CaBr2 = 0.14 mol CaBr2
| 199.88 g CaBr2
Or with this equation: g/FM = moles
28/199.88 = 0.14 moles
2nd Example 115g Al(NO3)3
FM Al(NO3)3 =
213.01 g/mol
LPChem:Wz:1415
(given in problem)
(found in step 1)
(Mathematically the two methods are
identical, although “the line” does a better
job keeping track of the units.)
On the line:
115g Al(NO3)3 | 1 mol
= 0.540 mol
|
|213.01 g
Unit 16: Solutions
3. Kilograms
Solvent
Given:
 Grams of solvent
OR
 Kilograms of Solvent
1st Example 500 g H2O
(given in problem)
Please note: the identity and
formula of the solvent (H2O)
didn’t actually have anything to
do with the math!
2nd Example 1.35 kg ethanol
(given in problem)
Again, the identity of the solvent (ethanol)
doesn’t actually affect the math.
Given:
 Moles solute
 Kilograms solvent
1st Example 0.14 mol CaBr2
(found in step 2)
4. Molality
kg solvent = 0.5 kg
2nd Example 0.540 mol Al(NO3)3
kg solvent = 1.35 kg
(found in step 3)
(found in step 2)
(found in step 3)
A. If the problem gives kilograms, this
step is already done. Move on.
B. If the problem gives grams of
solvent, convert grams to kilograms.
(1000 g = 1 kg)
This can be solved on the line:
500 g H2O | 1 kg
| 1000 g
= 0.5 kg H2O
Or with this equation: g/1000 = kg
500/1000 = 0.5 kg
(Mathematically the two methods are
identical, although “the line” does a better
job keeping track of the units.)
The number given is in kg, so I don’t have to
do anything.
kg solvent = 1.35 kg ethanol
A. Moles solute ÷ kg solvent = molality
B. Molality is its own unit (mol/kg = m)
C. It must be a lowercase m!
Moles solute ÷ kg solvent = molality
0.14 mol CaBr2 ÷ 0.5 kg H2O = molality
0.14 / 0.5 = 0.28 m
It’s a 0.28 molal solution.
Moles solute ÷ kg solvent = molality
0.540 mol Al(NO3)3 ÷ 1.35 kg ethanol =
molality
0.540 / 1.35 = 0.400 m
It’s a 0.400 molal solution.
All together now! 120. g K2SO4 is dissolved in 45 g H2O. What is the molality?
Given:
1) Find FM of solute using periodic table:
 g of solute = 120. g K2SO4
2(39.10) + 32.06 + 4(16.00) = 174.26 g/mol K2SO4
 Formula of solute = K2SO4
2) Find moles of solute (g ÷ FM = mol)
 g of solvent = 45 g H2O
120.g ÷ 174.26 g/mol = 0.689 mol K2SO4
3) Find kg solvent (g ÷1000 = kg)
45 g ÷ 1000 = 0.045 kg H2O
4) Solve for molality (molsolute ÷ kgsolvent = m)
0.689 molK2SO4 ÷ 0.045 kgH2O = 15.3 m K2SO4 solution
(This would be read “15.3 molal potassium sulfate solution”)
LPChem:Wz:1415
Unit 16: Solutions