Introduction to Solution Stoichiometry 
Whenever possible, reactions are carried out with all of the reactants in the same fluid phase. 
This means that it is preferable to have liquids
reacting with liquids, or gases reacting with gases. This is because no
matter what type of particle we talk about, in order for them to react
they must collide with each other. Solids have little or no movement
and therefore offer few opportunities for collision. Liquids and gases
are both fluids and offer much more opportunity for frequent collision.
Liquids are used in the lab because they are fairly easy to create and use.
Gases are used but specialized gas handing equipment is required. 
A solution is a uniform mixture of particles
of atomic, ionic, or molecular size. A minimum of two substances
are present. One is called the solvent and all the others
are called the solutes. The solvent is the fluid medium in
which all of the solutes are dissolved. A solvent can be a solid,
liquid or a gas but the most common solvent is water and therefore we deal
almost exclusively with aqueous solutions. Solutes are any
substance dissolved in the solvent. 
A dilute solution is one in which the ratio
of solute to solvent is very small, for example, a few crystals of sugar
in a glass of water. In a concentrated solution the ratio of solute
to solvent is large. Maple syrup is a concentrated solution of sugar
in water. A saturated solution is one in which no more solute
can be dissolved at a particular temperature. An unsaturated solution
is one in which the ratio of solute to solvent is lower than that of the
corresponding saturated solution. If more solute is added to an unsaturated
solution, at least some of it should dissolve. A supersaturated
solution is an unstable system in which the ratio of dissolved solute
to solvent is higher than that of a saturated solution. A supersaturated
solution can be made by gently cooling a hot saturated solution. At
a lower temperature the dissolved solute can be made to precipitate
out when a seed crystal is added. The process is called precipitation
and the substance that forms is the precipitate. 
The amount of solute needed to make a saturated
solution in a given quantity of solvent at a specific temperature is called
the solubility of the solution 
Units of Concentration (v/v, w/v, w/w and ppm) 
These units of concentration are most often seen
and used with commercial products. Expect for 'ppm' they are not used
often in the lab. 
Percent Concentration Volume/Volume (v/v): used with 2 liquids. 
% Concentration
= V_{solute} X 100% V_{solution} e.g.. 5 mL of vinegar are dissolved in 100 mL of vinegar solution. What is its v/v concentration. % concentration =
5 mL of vinegar X 100% = 5% 
Example #1 A photographic stop bath
contains 140 mL of pure acetic acid in a 500 mL bottle of solution.
What is the v/v concentration? 
% concentration
= 140 mL X 100% = 28% 500 mL 
Percent Concentration Weight/Volume (w/v): used with one solid and one liquid 
This means there is a certain mass, in grams,
in every 100 mL of solution. 
e.g.. a 3% H_{2}O_{2}
topical antibiotic solution means that there is 3 grams of H_{2}O_{2}
in every 100 mL of solution. 
Percent Concentration Weight/Weight (w/w): used with two solids 
Useful when dealing with alloys of precious metals.

e.g.. A ring with a mass of 12.0 grams contains
11.1 grams of pure silver. What is the w/w%?
% concentration = 11.1 grams X 100%
= 92.5% w/w of silver 
Parts per Million Concentration (ppm) 
Environmental solution are often very low in concentration.
We often use terms like: 1 part per million (ppm): 1 part out of 1 X 10^{6} parts 1 part per billion (ppb): 1 part out of 1 X 10^{9} parts 1 part per trillion (ppt): 1 part out of 1 X 10^{12} parts 1 ppm = 1 drop in a full bathtub 
We express ppm concentration in a variety of units
depending on what we need to use. But they are all interrelated.

ppm = 1 g
= 1 g
= 1 mg
= 1 mg =
1 microgram 10^{6} mL 1000 L 1 L 1 kg 1 g 
Example #2 Dissolved O_{2}
in water shows a concentration of 250 mL of water At SATP and 2.2 mg of
O_{2}. What is the concentration in ppm? 
ppm concentration = 1 mg =
2.2 mg = 8.8 mg/L = 8.8 ppm 1 L 0.25 L 
Go to the Concentration Unit Calculations Other than Molarity Worksheet 
Mole Fractions and Mole
Percents 

One way to describe
the relative numbers of molecules of all components of a solution is by
their mole fractions. The mole
fraction of any component in a mixture is the ratio of the number
of moles of it to the total number of moles of all the components present.
Expressed mathematically, 

X_{a} =
n_{a}
n_{a} + n_{b} + n_{c} + ... + n_{x} 

Where X_{a}
is the mole fraction of component 'a', and n_{a}, n_{b},
n_{c} ... n_{x}, are the numbers of moles of each component
in the mixture, 'a', 'b', 'c', ....'x', respectively. The sum of
all the mole fractions for a mixture must always be equal to 1. 

Even though mole fractions
have no formal units, always remember that they stand for the ratio of
moles of a component to total moles in the mixture. Sometimes the mole fraction is multiplied by 100% and the result is referred to as the mole percent. 

Example of how to calculate mole fractions and
mole percent. 

What are the mole fractions of each component
in a solution that consists of 1.00 mole of C_{2}H_{5}OH,
0.500 mole of CH_{3}OH,
and 6.00 mole of H_{2}O?


Molar Concentrations and Molarity (M) 

Molarity is a way of specifying the amount of
solute in one litter of solvent. Molarity is also known as
the concentration of a solution. The concentration of a
solution is the ratio of solute to a given quantity of solvent. We can
use whatever units we wish but the most commonly used units are moles
of solute per litter of solution. The special name for this ratio
is the molar concentration or molarity which is abbreviated
'M'.
M =
mol solute =
mol solute 

Suppose we had a bottle with the label "0.5 M
KBr". This means that it contains a solution of potassium bromide
with a concentration of 0.10 mol of potassium bromide per litter of solution
(or per 1000 mL of solution). It does not tell
us the amount of solution in the bottle. 

Solutions are considered to be homogenous
substances once the solute has completely dissolved. 

Sample Problem  
A student requires 0.250 moles of NaCl for an
experiment. The only thing available to them is a bottle with a
solution labeled "0.400 M NaCl." What volume of the solution should
be used? Give the answer in milliliters. 

Solution:  
Use the information on the bottle. There
is 0.400 moles of solute per litter. We need 0.250 moles.
0.400 moles = 0.400 moles = 0.250 moles
x = 625 mL. Thus 625 mL of 0.400 M NaCl contains 0.250 mol of NaCl.


Practice Problem:  
A glucose solution with a molar concentration
of 0.200 M is available. What volume of this solution must be measured
to obtain 0.001 mol of glucose? 

Preparation of Solutions  
Another very common problem involving molar concentration
is the calculation of the number of grams of solute needed to make a given
volume of solution having a specific molarity. The best way to
see this is by example: 

Problem: How can 500 mL of 0.150 M Na_{2}CO_{3}
solution be prepared? 

Solution: This is stated in a way that normally
arises in the lab? What we really want to know is how many
grams of Na_{2}CO_{3} that are going to be
in 500 mL of 0.150 M Na_{2}CO_{3} solution. Although the label reads in M, the balance reads in grams. Before we can calculate the number of grams we need to know the number of moles.
0.150 M = 0.150 mol = 0.150 mol
= x x = 0.075 mol of solute. Therefore we need to weigh out 0.075 mol of Na_{2}CO_{3}. The number of grams of Na_{2}CO_{3} are therefore:
g = n * mm 

To answer the question: Weigh out
7.95 grams of sodium carbonate. Add it to 100 mL of water in a 500
mL volumetric flask. Swirl to dissolve. Top the
flask up with 400 mL more water up to the 500 mL meniscus mark. 

Practice problem: How can we prepare 250
mL of 0.200 M NaHCO_{3}? 

Go to the Molarity and Solution Creation Worksheet  
Go to the Stoichiometry of Solution Molarity Worksheet 
Molality or Molal Concentration (m) 

The number of moles
of solute per kilogram of solvent
is called the molal concentration
or the molality of a solution.
The usual symbol for molality is m. molality = m = moles of solute kg of solvent If for example, we dissolve 0.5000 moles of sugar in 1000 g of water (1.000 kg), we have made a 0.500 molal solution of sugar. The need to use a volumetric flask is gone since the water was weighted. 

The molality of a solution
is an indirect expression of the ratio of moles of solute to moles of
solvent, because the mass of the solvent is known. If needed, we
could convert the 1 kg of solvent into moles of solvent. 

To contrast molality
with molarity, suppose we put the same quantity of sugar 0.500 moles, into
a one liter volumetric flask and add water to the 1L mark, making sure that
everything is dissolved first. The calibration mark on the volumetric
flask correlates with the flask's specified volume only at a given temperature, which is
etched on the flask. If our solution were at this temperature,
then we would have a 0.5000 molar solution. But we would not now the mass of water used,
and it would very unlikely that this mass of water would be 1.00 kg.
Unless we actually measured the mass of water used to make the 0.500 molar
solution, or knew the density of the solution, we would have no way to calculate
its molality. 

It is very important
that you learn the distinction between molarity and molality. Molarity is useful in
determining the moles of solute obtained when a specified volume of solution
is poured. Molality is useful is describing the ratio of moles
of solute to moles of solvent. The value of a concentration given
in units of molality does not change with temperature, but that of molarity
does. 

Ex. 1
An experiment calls for an aqueous 0.150 m solution of sodium chloride, To prepare
a solution with this concentration, how many grams of NaCl would have
to be dissolved in 500 g of water? Step 1: Prepare two conversions factors: 0.150 m is the same as 0.150 g NaCl or 1 000 g H_{2}O 1000 g H_{2}O 0.150 g NaCl 1 kg of water was changed to 1000 g! Step 2: Multiple by the mass needed using the first of the conversion formulas above so that the units cancel out. 500 g H_{2}O X 0.150 g NaCl = 0.0750 mol of NaCl 1000 g H_{2}O Step 3: Convert moles into grams. g = n X mm = 0.0750 moles X 58.5 g/mol = 4.39 g of NaCl By placing 4.39 g of NaCl into 500 g of water we obtain a 0.150 m solution. 

Converting from Weight
Percent to Molality 

What is the molality
of a 10.0% w/w NaCl solution? Solution: 10% w/w means we have a ratio of 10.0 g of NaCl to 100 grams of solution. We need to convert w/w concentration units into mol/kg of solvent units. Step 1: Calculate the moles of NaCl in 10.0 g of NaCl. n = g/mm = 10.0g / 58.5 g/mol = 0.171 mol of NaCl. Step 2: Calculate the kilograms of H2O in 100 grams of this solution. (Remember it is grams of solution with percent concentration, but kilograms (or 1000 g) of solvent with molality. There is 10.0 grams of NaCl in 100.0
grams of solution. So there must be 100.0 g  10.0 g = 90.0
grams of water present. 90.0 grams = 0.090 kg of water.
Step 3: Calculate molality, m, the ratio of moles to kg of solvent. m
= moles of solute / kg of solvent
= 0.171 mol of NaCl / 0.090 kg of H_{2}O = 1.90 m Therefore a 10.0% w/w NaCl solution is also 1.90 m NaCl 

Converting From Weight Percent
to Mole Fractions 

What are the mole fractions
and mole percents of the components in 10.0% w/w NaCl? The amount of solute from the example above is 0.171 mole of NaCl. We also found from the above example that there is 90.0 g of water in 100 grams of the solution. Moles of water is therefore n= g/mm = 90.0 g/18.02 g/mol = 5.00 mol of water. So the total number of moles is 0.171 moles + 5.00 moles = 5.171 moles total Calculate the mole fractions and mole percent.


Calculating Percents by Weight
from Mole Fractions 

So called "100proof"
alcohol has the following mole fractions of components: for water, 0.765,
and fr ethyl alcohol, 0.235. Calculate the percents (w/w) of water
and ethyl alcohol in 100 proof ethyl alcohol. The formula weights
are: H_{2}O: 18.0 and ethyl alcohol, C_{2}H_{5}OH,
46.1 AS in all these kinds of calculations, we work from the meanings of the given concentration units toward the final units. To work toward percents by weight, we have to work toward a ratio calculated from grams of components. Let's start with water. When the mole fraction of water is 0.765, we know that there are 0.765 mole of H_{2}O in a total of 1 mole of all components. So, in terms of water we have: g = n X mm
= 0.765 mol X 180.8 g/mol = 13.8 g of H_{2}O Next, we look at the ethyl alcohol. g = n X mm
= 0.235 mol X 46.1 g/mol = 10.8 g of C_{2}H_{5}OH Thus, in "1.0 mole" of this ethyl alcohol solution there are 13.8 g of H_{2}O and 10.8 g of C_{2}H_{5}OH. To convert to percents by weight, we need the total mass of the solution, which is 13.8 g + 10.8 g = 24.6 g. For the percents:


Converting Molality into Mole Fractions 

Ammonium nitrate is
sometimes a component in nitrogen fertilizers. Suppose that an aqueous
solution of NH_{4}NO_{3} has a concentration of 2.48 m. What are the mole fractions of
NH_{4}NO_{3} and water in this solution? Solution: The given molal concentration means that we can create two conversion units. 2.48 m means 2.48 mol NH_{4}NO_{3} and 1 kg water 1 kg water 2.48 mol NH_{4}NO_{3} We need to work our way toward a ratio of moles of solute to total number of moles, so we need to find the moles of water in 1 kg. Remember to treat the 1 kg as exact, therefore 1 kg of water = 1000 g of water and water has a molecular mass of 18.02. To find moles of water : n = g / mm = 1000 g / 18.0 g/mole = 55.6 mol of H_{2}O So the total moles, water + NH_{4}NO_{3}, is 2.48 moles + 55.6 moles
= 58.1 mol (rounded correctly)
Now we can calculate the mole fraction of NH_{4}NO_{3} and H_{2}O X_{NH4NO3} = 2.48 moles = 0.0427 58.1 moles The mole fraction of water is X_{H2O} = 1.000  0.0427 = 0.9573 

Converting a Percent by Weight into a Molar Concentration 

A certain supply of
concentration hydrochloric acid is 36.0% (w/w) in HCl. The density
of the solution is 1.19 g/mL. Calculate the molar concentration
of HCl in this solution. Solution: We are converting from w/w to mol/L. We need the density to get from g solution into mL solution. The density is given as 1.19 g/ mL and from this we get two possible conversion units. either 1.19 g solution or 1 mL solution
1 mL solution 1.19 g solution Since we have to deal with 1000 mL of solution as we work from grams of solution toward the molar concentration, we now calculate how many grams of this solution are in 1000 mL. 1000 mL solution X 1.19 g solution = 1190 g solution 1 mL solution But this is the mass of a 36.0% w/w in HCl; 36.0% of this mass is pure HCl. So the quantity of pure HCl in 1190 g of solution is: 1190 g X 36.0 = 428 g of HCl 100 Thus, there is 428 grams of HCl in 1190 grams of solution. But, remember that this mass of solution has a volume of 1000 mL or 1 L. To get the molarity we need to convert 428 grams of HCl into moles of HCl. The formulas weight os HCl is 36.5 so: n = g / mm
= 428 g / 36.5 g/mol = 11.7 mol of HCl Therefore the molarity of the HCl solution is 11.7 mol/L or 11.7 M 

Converting From Molarity to Percent by Weight 

Perchloric acid, HClO_{4},
(formula weight, 100.5), can be purchased at a concentration of 9.20
M. The density of this solution is 1.54 g/mL. What is the percent
by weight of perchloric acid in this solution? 

When the molarity is
9.20 M, we have the following ratio. 9.20 mol HClO_{4 } 1000 mL solution 

What we want is the
ratio of grams of HClO_{4} to grams of solution, taken as a percent.
this is what a percent by weight means. So, in 9.2 mol of HClO_{4},
we have: 9.20 mol HClO_{4} X 100.5 g HClO_{4} = 925 g of
HClO_{4}
1 mol HClO_{4} And in 1000 mL solution we have the following mass of solution. 1000 mL solution X
1.54 g solution
= 1540 g of solution
1 mL solution So the ratio of grams of solute to grams of solution, taken as a percent, is 925 g of HClO_{4} X 100 % = 60.0% (w/w) 1540 g (solution) Thus, 9.20 M HClO_{4} is 60.0% (w/w) HClO_{4}. 

Go to the Molality, Mole Fraction and Mole Percent Worksheet 


Stoichiometry of Reactions in Solution 
Up until now we have used pure substances and
added them to water in order to make solutions. The point of using
solutions is that they are often easier to use in the lab than pure substances.
We will now find out how to do stoichiometry with these solutions.

Problem: What volume
of 0.556 M HCl has enough hydrochloric acid to combine exactly with 24.5
mL of aqueous sodium hydroxide with a concentration of 0.458 M?
The equation for the reaction is: 
HCl(aq) + NaOH(aq) > NaCl(aq) +
H_{2}O 
Solution: This is a simple mL > mol
> mol > mL solution. 
Step #1 Start with what you know: You have 24.5 mL of 0.458 M NaOH. How many moles is this?
0.458 mol =
x

Step #2 The HCl and NaOH react
on a 1:1 basis. Therefore
1 HCl = 1 NaOH
x = 0.011 mol of HCl are needed. 
Step #3 We need 0.011 moles of HCl
and the solution we have available is 0.556 M. Therefore
0.556 mol = 0.011 mol
x = 19.78 mL are needed. If we measure out 19.78 mL of the HCl solution it will
exactly neutralize the NaOH. 
Example Problem #2:
How many milliliters of 0.114 M H_{2}SO_{4} solution
provide the sulfuric acid required to react with the sodium hydroxide
in 32.2 mL of 0.122 M NaOH according to the following equation?
H_{2}SO_{4}(aq) + 2 NaOH(aq) >
Na_{2}SO_{4}(aq) + 2 H_{2}O 
Solution: Again we need the mL > mol
> mol > mL solution. 
Step #1 Start with what you know. You have 32.3 mL of 0.122 NaOH.
0.122 mol =
x
x = 0.004 mol of NaOH are present. 
Step #2
1 H_{2}SO_{4} = 2 NaOH x 0.004 mol
x = 0.002 mol of H_{2}SO_{4} are required. 
Step #3 Therefore
we need:
0.114 mol = 0.002 mol
Therefore we need 17.54 mL of 0.114 M H_{2}SO_{4}
to exactly neutralize 32.3 mL of a 0.122 M solution of NaOH. 
Practice Problems 
1. What volume of 0.337 M KOH provides enough
solute to combine with the sulfuric acid in 18.6 mL of 0.156 M H_{2}SO_{4}?
The reaction is:
2 KOH(aq) + H_{2}SO_{4}(aq) >
K_{2}SO_{4}(aq) + 2 H_{2}O 
2. Hydrochloric acid reacts with sodium
carbonate as follows:
2 HCl(aq) + Na_{2}CO_{3}(aq) > 2 NaCl(aq) + CO_{2}(g) + H_{2}O What volume of 0.224 M HCl is neutralized
by the Na_{2}CO_{3} in 24.2 mL of 0.284 M Na_{2}CO_{3}?

Dilutions from Concentrated Stock Solutions 
When a dissolved solvent is added to a solution,
the solute is spread out through the large volume and the number of moles
per unit volume decreases. The solution is said to be diluted.
Such dilutions are a natural part of chemistry in the lab. If two
solutions of different solutes are mixed, the total volume increases and
becomes occupied by both solutes. As a result the concentration
of both solutes is less. At other times dilution is deliberate and
essential. Stock solutions of common chemicals are often prepared
in large volumes in preset concentrations. The chief supply of HCl
in the lab may be 1.00 M HCl. In your experiment, you may require
for a much less concentrated solution, so the concentrated solution must
be diluted first. 
When carrying out a dilution with an acid.
Always add acid to water. Never add water to
the acid. When acid and water mix a great deal of heat
is released. Acid added to water gives a large heat sink for the heat
to dissipate in. If you add water to acid, the heat generated will
cause the water to boil, and may splatter you with the hot acid solution.

Dilution Calculation 
All the calculations that you need to do are based
on a simple fact. As a solution is diluted, the number of moles of
solute doesn't change; the solute simply spreads out through a larger volume.

We will use the following symbols: V for volume
and M for molarity. Therefore V_{c} and M_{c} are the
concentrated solutions volume and molarity. V_{d} and M_{d}
are then the diluted solutions volume and molarity.
V_{c }M_{c} = V_{d }M_{d}

Problem: How can we prepare 100 of 0.040
K_{2}Cr_{2}O_{7} from 0.200 M K_{2}Cr_{2}O_{7}?
Solution: 
First assemble the data into a table: V_{c} = ? V_{d} = 100 mL M_{c} = 0.200 M M_{d} = 0.040 M Next, use the formula from above. V_{c} X 0.20
M = 100 mL X 0.040 M Place 20.0 mL of 0.200 M K_{2}Cr_{2}O_{7}
into a 100 mL volumetric flask and then add water until the final volume
is 100 mL. 
Practice Problem: Describe how to make 500
mL of 0.20 M NaOH from 0.50 M NaOH. 
Problem: 
How many milliliters of water would have to be
added to 100 mL of 0.40 M HCl to give a solution with a concentration of
0.10 M? 
First assemble the data.
V_{c} = 100 mL
V_{d} = ? M_{c} = 0.40 M M_{d} = 0.10 M Use the equation: 100 mL X 0.40 M = V_{d}
X 0.10 M The final volume must be 400 mL. But since we started
with 100 mL we will add 300 mL of water until the volume reaches 400 mL.

Practice Problem: How many milliliters of water
would have to be added to 300 mL of 0.5 M NaOH to give a solution with
a concentration of 0.2 M? Go to the Solution Dilution Worksheet 