C l l i
Calculations
based
b d on Molarity
M l i
Molarity is the most common unit used in chemistry to express
the concentration of chemicals in solution.
The definition of molarity is:
Moles of solute
Molarity = M =
Liters of solution
This tutorial takes you through several basic calculations
using molarity.
E
Example
l 1:
1
If I have 0.075 moles of methanol dissolved in enough water to
make 150 mls of solution, what is the molarity of methanol in
th solution?
the
l ti ?
Since the molarity calculation is moles of solute divided my liters
of solution, you need to know how many moles of solute there are,
and how many liters of solution.
The moles of solute here is easy. It is given in the problem
as 0.075 moles of methanol.
If I have 0.075 moles of methanol dissolved in enough water
to make 150 mls of solution,
solution what is the molarity of the
solution?
The liters of solution is almost as easy.
y The pproblems states that
you have 150 mls of solution. Your only problem is to do the
unit conversion between mls and liters.
1 mls = 1x10-3 liters
150 mls x (1 x 10-3 liters/1 ml) = 0.150 liters
If I have 0.075 moles of methanol dissolved in enough water
to make 150 mls of solution,
solution what is the molarity of the
solution?
Moles of solute
Molarity = M =
Liters of solution
Molarity of methanol = .075 moles / .150 liters = .50 M
or .50 moles/liter
Note: We will often use the following shortcut. Instead of
writing the phrase ‘Molarity of X’ we simply write [X]
instead The square brackets are used to refer to the molar
instead.
concentration of a substance. So [Methanol]=.50 M.
C l l i 2
Calculation
Whatt iis th
Wh
the molarity
l it off a solution
l ti preparedd by
b dissolving
di l i
150 grams of NaCl in enough water to make 900 mls
of solution?
Molarity = moles/liters, so I need to find out how many moles
are in 150 g of NaCl, and I need to find out how many
liters are in 900 mls .
What is the molarity of a solution prepared by dissolving
150 grams of NaCl in enough water to make 900 mls
of solution?
Moles of NaCl:
NaCl has a molar mass of 22.99 (Na) + 35.45 (Cl) for a
total of 58.44 grams/mole.
150 grams x (1 mole/58.44grams) = 2.567 moles
Note: Because this in an intermediate part of my overall
molarity calculation I will keep several extra significant
figures at this point, on only round down to the right number
of significant figures after the final calculation.
What is the molarity of a solution prepared by dissolving
150 grams of NaCl in enough water to make 900 mls
of solution?
Liters of solution:
1 ml = 1x10-33 liters
900 mls x (1 x10-33 liter/ml) = .900
900 liters
Note: Again I keep a few extra significant figures because
this is an intermediate part of a complete calculation.
What is the molarity of a solution prepared by dissolving
150 grams of NaCl in enough water to make 900 mls
of solution?
Molarity = moles solute/liters of solution:
[Na+]=2.567 moles/.900 liters
=2.8519
=2
8519 M
M, round
[Na+]= 2.9M (rounding to 2 sig fig.)
Only round to the right number of sig figs after the last
calculation. If you want to be a real stickler for sig figs,
900 only has 1 sig fig, so maybe 3M is a better answer.
What is the molarity of a solution prepared by dissolving
150 grams of NaCl in enough water to make 900 mls
of solution?
In this example you calculated moles and liters in two separate
calculations and then combined them in a third calculation.
The preferred method is to do a single large calculation because
it is faster, and less prone to roundoff error.
Th it is
Thus
i better
b tt form
f
to
t do:
d
150 g × (1mole / 58.44 g )
Molarity =
900mls × (1x10 − 3 liters / mls)
To get your final answer.
E
Example
l 3:
3
I wantt to
t make
k 250 mls
l off 1.75M
1 75M Mg(NO
M (NO3)2. How
H many
grams of Mg(NO3)2 do I need to place in a 250 ml volumetric
flask to make this solution?
I have turned the calculation around in this problem. Instead
of giving you grams or moles, and making you calculate
molarity, I give you molarity and make you calculate moles
and grams.
I want to make 250 mls of 1.75M Mg(NO3)2. How many
grams of Mg(NO3)2 do I need to place in a 250 ml volumetric
flask to make this solution?
When you gett a problem
Wh
bl lik
like this
thi that
th t is
i turned
t
d around,
d you
have to do some algebra to turn the equation around to match
the problem.
problem
In the first 2 examples we had:
Moles (Given in problem)
X (unknown molarity ) =
Liters (Given in problem)
I want to make 250 mls of 1.75M Mg(NO3)2. How many
grams of Mg(NO3)2 do I need to place in a 250 ml volumetric
flask to make this solution?
Here we have:
X ( Moles unknown)
Molarity ( given in problem) =
Liters (Given in problem)
As in most algebra problems, the first thing you have to do
i to
is
t gett X,
X the
th unknown,
k
isolated
i l t d on one side
id off the
th equation.
ti
I want to make 250 mls of 1.75M Mg(NO3)2. How many
grams of Mg(NO3)2 do I need to place in a 250 ml volumetric
flask to make this solution?
X ( Moles unknown)
Molarity ( given in problem) =
Liters (Given in problem)
To gget X byy itself on the right,
g , we need to multiply
p y both sides
of the equation by liters so we end up with the equation:
Molarity ( given in problem) × Liters (Given in problem) = X ( Moles unknown)
I want to make 250 mls of 1.75M Mg(NO3)2. How many
grams of Mg(NO3)2 do I need to place in a 250 ml volumetric
flask to make this solution?
Molarity ( given in problem) × Liters (Given in problem) = X ( Moles unknown)
Now that we have the equation we want, we can start to fill in
the blanks on the equation to solve it.
Molarity = 1.75M
Liters = 250ml x (1x10-3 l/ml) = .250 liters
X = 1.75 mole/liter x .250 liters = .4375 moles Mg(NO3)2
(I include extra significant figures in this intermediate calculation)
I want to make 250 mls of 1.75M Mg(NO3)2. How many
grams of Mg(NO3)2 do I need to place in a 250 ml volumetric
flask to make this solution?
To change .4375 moles Mg(NO3)2 to grams I need the molar mass
of Mg(NO3)2
= 24.31(Mg)
24 31(Mg) + 2x14
2x14.01(N)
01(N) + 6x16
6x16.00
00 (O)
= 148.33 g
.4375 moles x (148.55g/1 mole) = 65 g Mg(NO3)
P bl
Problems
to try
1. If I take 5 grams of glucose (C6H12O6) and dissolve it in
enough
g water to make 100. mls of solution, what is the
[glucose] in this solution?
22. I want to make 22.5
5 liters of 22.25M
25M Iron (II) chloride
chloride.
How many grams of Iron(II) chloride will I need to make up
this solution?
As usual try to figure the answers yourself before you see my
answers on the next page or two.
If I take 5 grams of glucose (C6H12O6) and dissolve it in
enough water to make 100
100. mls of solution,
solution what is the
[glucose] in this solution?
Molar mass of glucose = 6(12.01) + 12(1.008) + 6(16.00) = 180.156
Moles Glucose = 5 g x (1 mole/180.156 g) = .02775 moles
Lit off solution
Liters
l ti = 100.
100 mls
l x (1x10-3
(1 10 3 liters/ml)
lit / l) = .100
100 l
[Glucose] = .02775
02775 moles/.100
moles/ 100 l = .278M
278M
I want to make 2.5 liters of 2.25M Iron (II) chloride.
How many grams of Iron(II) chloride will I need to make up
this solution?
Molarity = Moles/liters; 2.25M
2 25M = X moles/2.5
moles/2 5 liters
X moles
o es = 2.25M
. 5 x 2.5
.5 liters
es
Moles Iron (II) chloride = 5.625 moles
Iron (II) chloride = FeCl2
so Molar mass =55.85 + 2(35.45) = 126.75 g
Grams Iron (II) chloride = 5.625 moles x (126.75 g/1 mole)
=710 grams