% Composition and Empirical formula
The calculation of % composition can be based on what we know today is the correct
formula for water, H2O. The molar mass of water is 18.0 g/mol. Of that 18.0 g, 2.0 g
are hydrogen and 16.0 g are oxygen. So the mass percents of each element could be
expressed as:
53
Empirical Formula and Molecular Formula
Consider the data obtained by analyzing a compound. A chemist determines that the
substance contains potassium, chlorine and oxygen.
Here is some sample data:
element
mass, g
potassium 0.479
chlorine
0.434
oxygen
0.588
What is the chemical formula of this compound? It might be tempting to say
K0.479Cl0.434O0.588 but of course that is not correct.
The numbers in chemical formulas are always integers since they represent actual
whole atoms which combine. So if we knew the moles of K, Cl and O we would be
that much closer to knowing the formula.
element
moles
potassium 0.0123
chlorine
0.0122
oxygen
0.0368
This is a little disappointing as we are no closer to integers than we were before. A
closer look at those decimals will show that there is a nearly integer ratio hidden in
54
the data. One way to get it out is to divide all of the values by the smallest value. That
makes the smallest number 1 and hopefully all the other values larger integers (or also
1).
So the formula for this compound is KClO3. Such a formula is often called the
empirical formula. It gives the smallest integer ratio which represents the proportions
in which the atoms combine in the compound. It may also represent the actual or
molecular formula.
Molecular formulas are either the same as the empirical formula or some integer
multiple. To know whether an empirical formula is also the molecular formula you
need to know the molar mass of the compound.
55
Problems:
56
57
Percentage concentration
The use of percentages is a common way of expressing the concentration of a
solution. The percentages can be calculated using volumes as well as weights, or even
both together.
Volume percent:
Is usually used when the solution is made by mixing two liquids.
For example, rubbing alcohol is generally 70% by volume isopropyl alcohol. That
means that 100 ml of solution contains 70 ml of isopropyl alcohol. That also means
that a liter (or 1000 ml) of this solution has 700 ml of isopropyl alcohol plus enough
water to bring it up a total volume of 1 liter, or 1000 ml.
Volume percent =
volume of solute
volume of solution
x 100
Weight Percent:
Another similar way of expressing the concentration of a solution is to express it in
weight percent (or mass percent, if you prefer).
Weight percent =
weight of solute
weight of solution
x 100
Example 1:
% composition of H in H2O =
% composition of O in H2O =
Atomic mass of H
molar mass of water
Atomic mass of O
molar mass of water
58
x 100 =
x 100 =
2x1
18
16
18
x 100 = 11%
x 100 = 88.9 %
Example 2:
What is the weight percent of glucose in a solution made by dissolving 4.6 g of
glucose in 145.2 g of water?
Solution:
To get weight percent we need the weight of the solute and the total weight of the
solution.
Determine total weight of solution:
4.6 g glucose
+ 145.2 g water
149.8 g solution
Calculate percent:
Weight % glucose =
4.6 g glucose x 100 = 3.1% glucose
149.8 g solution
Example 3:
How would you prepare 400. g of a 2.50% solution of sodium chloride?
Solution:
We need to find out how much salt is needed and how much water is needed
Determine weight of salt: 400. g x 2.50% salt = 10.0 g salt
400. g solution x
2.50 g salt
100 g solution
= 10.0 g salt
Determine weight of water:
400. g total
- 10. g salt
390. g water
Answer:
Dissolve 10.0 g salt in 390. g water.
59
Example 4
Take time now to answer the following questions. The third question, you will note,
has an extra twist to it. Take some time to do these now, get some help if you need it
and then check your answers below.
1.
What is the weight percent of ethanol in a solution made by dissolving 5.3 g of
ethanol in 85.0 g of water?
2.
How would you make 250. g of a 7.5% solution of glucose in water?
3.
A sample of a solution weighing 850.0 g is known to contain 0.223 moles of
potassium chloride. What is the weight percent of potassium chloride in the
solution?
Answers Example 4
1-
The answer to "1" is 5.9 % ethanol. If you got 6.2% probably what you did
was to not add together the mass of the ethanol and the water to get the total
mass of the solution.
wt % =
2-
5.3
(85.0 + 5.3)
x 100 = 5.86 %
The answer to "2" brings up a couple points. One is that the question asks
"how would you prepare?" So the answer is not just 18.75g of glucose. Since
the question asks you how to do something, you need to answer with how you
would do it. You would dissolve 18.75 g of glucose in 231.25 g of water.
wt % =
wt of solute =
=
wt of solute
wt of solution
x 100
wt % x wt of solution
100
7.5 x 250g
= 18.75 g
100
Volume of water = 250g - 18.75g = 231.25g
60
3-
The answer to "3" is 1.956% potassium chloride. Depending on how you went
about doing your calculations and rounding off you may have gotten 1.95%
instead of 1.96%. Don't worry too much about that. One more point about the
last question here is that in order to do it you needed to change from moles of
KCl to grams of KCl.
wt % =
M=
wt of solute
wt of solution
x 100
wt
atomic weight of KCl
wt = 0.223 x (39.1 + 35.5) = 16.63g
wt % =
16.63g
850
x 100 = 1.956
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Stoichiometry
I-
The Mole & Avogadro's number
# Moles =
# Moles =
# atoms
=
# atoms =
For a solid acid or base:
mass of solute
molar mass of solute
Molarity (M) x
Volume (liter)
# Moles x Avogadro' s number
# Moles x 6.02 x 1023
# of equivalents =
eq.wt
=
mass
equivalent weight
MM
# of acidic H+ or OH -
For a solution of an
acid or base:
# of equivalents = N x VL
N stands for normality,
which is defined as:
N =
II-
# of equivalents
(i.e. equivalent per L)
VL of solution
Dilution Calculations
# of moles conc = # of moles dil
Molarity (Mconc) x Volume (Vconc) = Molarity (Mdil) x Volume (Vdil)
MV
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=
M' V'
III- Normality: Acids and Bases
# of equivalents conc = # of equivalents dil
Normality (Nconc) x Volume (Vconc) = Normality (Ndil) x Volume (dil)
NV
=
+
N' V'
-
N = M x (# of acidic H or basic OH groups)
Normality (N) = Molarity (M) x # of charge
IV- Molar concentration of a fluid
10 x % x density
Molarity (M) =
Mwt
10 x % x density
Normality (N) =
eq.wt
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The Mole
The term mole means a small mass. If the mass of a single 12C atom is 12.000 amu,
then one mole of these atoms would have a mass of 12.0 grams. By definition, a mole
of any substance contains the same number of elementary particles, as there are
atoms in exactly 12 grams of the 12C isotope of carbon.
Example: A single 12C atom has a mass of 12 amu, and a mole of these atoms would
have a mass of 12 grams.
A mole of any atoms has a mass in grams equal to the atomic weight of the element.
Each time we use the term, we refer to a number of particles equal to the number of
atoms in exactly 12 grams of the 12C isotope of carbon.
Avogadro's number (or Avogadro's constant) = 6.022 x 1023.
Is the number of elementary particles in a mole of any substance. For most
calculations, four significant figures for Avogadro's constant are enough: 6.022 x
1023.
A mole of any substance contains Avogadro's number of elementary particles. It
doesn't matter whether we talk about a mole of atoms, a mole of molecules, a mole of
electrons, or a mole of ions. By definition, a mole always contains 6.022 x 1023
elementary particles.
This concept enables chemists to essentially "count" atoms by massing them. For
example, if 12.0 g of carbon contains 6.02 x 1023 atoms then 6.00 g of carbon must
contain half that amount or 3.01 x 1023 atoms.
# of Moles =
mass of solute
molar mass of solute
# of atoms =
# of Moles x Avogadro' s number
# of atoms =
# of Moles x 6.02 x 1023
Converting between grams and moles can set up using unit analysis as the following
examples will show.
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Example:
What is the number of moles of carbon present in 6.0 g :
# of Moles =
# of Moles =
mass of solute
molar mass of solute
6.0 g x 1
12.0 g
= 0.5 Mole
It is also possible to determine the number of atoms (or molecules) in a sample in a
similar fashion:
What is the number of atoms in 6.0g of carbon:
number of atoms in 6.0 g C =
6.0 g x 1
12.0 g
x 6.02 x 1023 atoms
= 3.01 x 1023 atoms
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Practice:
66
67
Molarity
Another way of expressing concentration, the way that we will use most in this
course, is called molarity. Molarity is the number of moles of solute dissolved in one
liter of solution. The units, therefore are moles per liter, specifically it's moles of
solute per liter of solution.
# Moles =
mass of solute
molar mass of solute
# Moles = Molarity (M) x Volume (liter)
So when you see M or M it stands for molarity, and it means moles per liter (not just
moles).
You must be very careful to distinguish between moles and molarity. "Moles"
measures the amount or quantity of material you have; "molarity" measures the
concentration of that material. So when you're given a problem or some information
that says the concentration of the solution is 0.1 M that means that it has 0.1 mole for
every liter of solution; it does not mean that it is 0.1 moles. Please be sure to make
that distinction.
Calculations Using Molarity
There are several types of calculations that you need to be able to do with molarity.
First, you should be able to calculate the molarity if you are given the components
of the solution.
Second, you should be able to calculate the amount of solute in (or needed to make)
a certain volume of solution.
Third, you might need to calculate the volume of a particular solution sample.
Fourth, you might need to calculate the concentration of a solution made by the
dilution of another solution. This and related calculations will be covered in a
separate page.
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In either of the first two cases, the amount of solute might be in moles or grams and
the amount of solution might be in liters or milliliters. Please note that with molarity
we are concerned with how much solute there is and with how much solution there is,
but not with how much solvent there is.
Examples (Ex. 1)
Calculating Molarity from Moles and Volume (Ex. 1a)
What is the molarity of a solution containing 0.32 moles of NaCl in 3.4 liters?
Molarity (M) =
Molarity (M) =
number of Moles
Volume (liter)
0.32 moles
3.4 L
= 0.094 M
Calculating Molarity from Mass and Volume (Ex. 1b)
What is the molarity of a solution made by dissolving 2.5 g of NaCl in enough water
to make 125 ml of solution?
To get molarity we still need to divide moles of solute by volume of solution. But this
time we're not given the moles of solute. We have to calculate it from the mass of
NaCl (M wt of NaCl = 23 + 35.5 = 58.5).
# of Moles =
mass of solute
molar mass of solute
# of Moles of solute =
Molarity (M) =
Molarity (M) =
2.5 g
58.5 g
= 0.0427 mol
# of Moles
Volume (liter)
0.0427
0.125 L
= 0.34 M
Calculating Mass of Solute from Molarity (Ex. 1c)
This question asks how you would prepare 400. ml of 1.20 M solution of sodium
chloride.
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In this case, what you need to find out is how much NaCl would have to be dissolved
in 400 ml to give the concentration that is specified. This amount is going to have to
be in grams because we don't have any balances that weigh in moles. So there is more
than one step to this problem.
Molarity (M) =
number of Moles
Volume (liter)
number of Moles = Molarity (M) x Volume (liter)
= 1.2 x 0.400 L = 0.480 mole
wt of solute
number of Moles =
M wt of solute
wt of solute = number of Moles x M wt of solute
=
0.480 x 58.5 = 28.08 g
Volume of water = 400 g water - 28.08 g NaCl = 371.92 g water
Calculating Moles of Solute from Molarity (Ex. 1d)
How many moles of salt are contained in 300. mL of a 0.40 M NaCl solution?
This question is a little easier. We do it the same way as the first step of the previous
problem and then we stop.
Molarity (M) =
number of Moles
Volume (liter)
number of Moles = Molarity (M) x Volume (liter)
= 0.40 M x 0.300 L
= 0.12 mole
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Molarity Calculations: Practice
1-
How would you prepare 100. mL of 0.25 M KNO3 solution?
2-
A chemist dissolves 98.4 g of FeSO4 in enough water to make 2.000 L of
solution. What is the molarity of the solution?
3-
How many moles of KBr are in 25.0 mL of a 1.23 M KBr solution?
4-
Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are
in 400. mL of this solution?
Answers
1-
Dissolve 2.53 g of KNO3 in enough water to make 100 ml of solution.
2-
0.324 M
3-
0.0308 mol
4-
120 g
4-
Select the letter of the best answer for each of the following questions.
i. What is the molarity of the solution which contains 5.0 g of NaOH (molecular
weight = 40 g/mole) dissolved in water to make 200 mL of solution?
a. 0.025 M
b. 0.13 M
c. 0.20 M
*d. 0.63 M
e. 1.6 M
ii. What is the weight percent of NaCl in a solution made by dissolving 20 g of NaCl
(FW = 58 g/mole) in 200 mL of water?
a. 0.16%
b. 0.17%
*c. 9.1%
d. 10%
e. 17%
iii. When 40.0 mL of 2.0 M NaCl (FW = 58) is diluted to 100.0 mL, what is the new
concentration?
a. 0.20 M
b. 0.50 M
*c. 0.80 M
d. 5.0 M
iv. How many moles of NaOH (FW = 40 g/mole) are there in 50 mL of 0.30 M
NaOH solution?
71
a. 0.006
*b. 0.015
c. 0.60
d. 6
e. 15
v. What is the species concentration of Na+ in a 0.50 M solution of Na2SO4?
a. 0.20 M
b. 0.25 M
c. 0.50 M
*d. 1.0 M
e. 2.0 M
vi. How many moles of chloride ion are there in 100 mL of 0.50 M AlCl3?
a. 0.050
b. 0.10
*c. 0.15
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d. 0.5
e. 1.5
Dilution Calculations
Quite often, however, solutions are prepared by diluting a more concentrated solution.
For example, if you needed a one molar solution you could start with a six molar
solution and dilute it.
The number of moles of solute in the concentrated solution is equal to the number of
moles in the dilute solution. You have simply increased the amount of solvent in the
solution.
# of moles conc = # of moles dil
Molarity (Mconc) x Volume (Vconc) = Molarity (Mdil) x Volume (Vdil)
MV
M' V'
=
Examples (Ex. 2)
Calculating New Concentration (Ex. 2a)
A chemist starts with 50.0 mL of a 0.40 M NaCl solution and dilutes it to 1000. mL.
What is the concentration of NaCl in the new solution?
Mconc x Vconc = Mdil x Vdil
0.40 x 50 ml = Mdil x 1000 ml
Mdil =
0.40 x 50
1000
= 0.02 M
Calculating Initial Volume (Ex. 2b)
A chemist wants to make 500. mL of 0.050 M HCl by diluting a 6.0 M HCl solution.
How much of that solution should be used?
Mconc x Vconc = Mdil x Vdil
Vconc =
Mdil x Vdil
Mconc
=
0.05 x 500 ml
= 4.2 ml
6.0
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Dilution Calculations: Practice
Now you should practice working on dilution problems by answering the following
questions. Do those now and check your answers before you continue.
1-
How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M
NaCl solution?
2-
What would be the concentration of a solution made by diluting 45.0 mL of
4.2 M KOH to 250 mL?
3-
What would be the concentration of a solution made by adding 250 mL of
water to 45.0 mL of 4.2 M KOH?
4-
How much 0.20 M glucose solution can be made from 50. mL of 0.50 M
glucose solution?
Answers
1-
19 mL (2 s.d. from 18.75 mL)
2-
0.76 M (2 s.d.)
3-
0.64 M
4-
130 mL (2 s.d. from 125 mL)
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Acid –Base Titration
Normality: Acids and Bases
According to the Arrhenius definition, an acid liberates H+ in solution and a base
liberates OH-.
The equivalent weight of an acid is the mass of the acid that can supply 1 mole of
H+ .
For example:
* Nitric acid, HNO3 has molar mass 63.02 g / mol. Since an HNO3 unit contains one
acidic H, one mole of HNO3 could supply one mole of H+, so the equivalent weight of
HNO3 is 63.02 g / equiv ("grams per equivalent" -- note the units).
* Sulfuric acid, H2SO4, has molar mass 98.08 g/mol. Since an H2SO4 unit contains
two acidic H's, one mole of H2SO4 could supply two moles of H+, so the equivalent
weight of H2SO4 (i.e. the mass of H2SO4 that could supply one mole of H+) is not
98.08 g but 98.08 g /2 = 49.04 g / equiv.
For an acid, then, the general formula for equivalent weight is
Equivalent weight =
molar mass
(unit g / equiv.)
+
# of acidic H s
The equivalent weight of an Arrhenius base is completely analogous to that of an
Arrhenius acid, except that the base has OH- ions instead of H+ ions.
* The equivalent weight of a base is the mass of the base that can supply 1 mole
of OH-.
Equivalent weight =
molar mass
(unit g / equiv.)
# of acidic OH+s
In an acid-base titration, the equivalence point is where the moles of H+ equals the
moles of OH-.
"Equivalents of H+ or OH-" means the same thing as "moles of H+ or OH-", so the
equivalence point is where the equivalents of H+ = equivalents of OH- (that's why
they call it the "equivalence point").
Now, here some very important relationships:
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For a solid acid or base: # of equivalents =
For a solution of an
acid or base:
mass
equivalent weight
# of equivalents = N x VL
# of equivalents
(i.e. equivalent per L)
N stands for normality, N =
VL of solution
which is defined as:
Note that normality is defined analogously to molarity, which is moles/L.
It is easy to convert between normality and molarity. For an acid or base:
N = M x (# of acidic hydrogens or basic OH groups)
For example, a 1.0 M solution of H2SO4 is actually 2.0 N in H+. A 0.0050 M solution
of Ca(OH)2 is 0.010 N in OH-.
Example I
When 0.155 g of oxalic acid dihydrate (H2C2O4•2H2O) was dissolved in water, 26.22
mL of a NaOH solution were required to reach the equivalence point. What is the
molarity of the NaOH?
Solution
At the equivalence point,
# equivalents from acid = # equivalents from base
mass
0.155 g
= 0.00246 eqvs
=
eq. wt
63.03 g / eqv
For the solid acid:
# of eqvs =
For NaOH solution:
# of eqvs = Nb x (0.002622 L)
At the equivalent point:
Nb x (0.002622 L) = 0.00246 eqvs
so Nb
=
0.0938 eqvs / L
Since NaOH has only 1 OH group, N = M and therefore [NaOH] = 0.0938 M.
Example II
Suppose that 25.00 mL of an H3PO4 solution required 17.22 mL of 0.675 M KOH to
titrate completely. What is [H3PO4]?
Solution
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To solve this problem, you could write the balanced equation, find moles of KOH,
convert to moles H3PO4, then divide by 0.02500 L to get [H3PO4]. It’s much easier to
do using normality:
At the equivalence point, # equivalents from acid = # equivalents from base i.e.
Na
Va = Nb Vb
Since KOH has only one OH- group
Therefore we have
Nb = Mb = 0.675 N
Na (25.00 mL) = (0.675 N)(17.22 mL)
Finally, then, [H3PO4]
Na = 0.465 N
0.465 eqv/L
= Na/3 =
3 eqv/mol
= 0.155M
Problem 1:
Calculate the molarity of an acetic acid solution if 34.57 mL of this solution are
needed to neutralize 25.19 mL of 0.1025 M sodium hydroxide.
CH3COOH (aq) + NaOH (aq)
Na+(aq) + CH3COOH-(aq) + H2O (l)
Macid x Vacid = Mbase x Vbase
Mbase x Vbase
Macid =
Vacid
0.1025 x 25.19
= 0.075 mole
34.57
=
SOLUTION
A solution is a mixture of a fluid (often water, but not always) and another material
mixed in with it. The material mixed in with it is called the solute. The concentration
can be expressed in a number of ways, the most common in chemistry is the M,
molar. One molar is one mol of solute in a liter of fluid.
Concentration of pure material (massg) = densityg/ml x volumeml
Molarity = # of moles = mass
= % x d x Vml x 1000
V
molar mass x VL 100 x molar mass x Vml
Molarity =
10 x % x d
Molar mass
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Examples:
1- Commercial HCl has a density of 1.19 g/ml and a mass percent of 38% HCl.
Determine the molarity of the solution:
Molarity =
10 x % x d = 10 x 38 x 1.19 = 12.4 M
Mwt
36.5
2- How could you prepare 500 ml of 10% HCl utilizing the above data of
commercial HCl?
N = M x charge of +ve hydrogen ion
NHCl = 12.4 x 1 = 12.4 N
N V = mass / molar mass
VHCl needed to prepare 100 ml solution = 10g / 36.5 x 12.4 = 0.02209 L = 22.9 ml
VHCl needed to prepare 500ml = 22.9 x 500 /100 = 114.5 ml and continue to 385.5 ml
water to fill a 500ml measuring flask.
3-
You want to make only 250 mL of a 1 N H2SO4 solution that will be used to
adjust the pH of BOD samples prior to analysis. How many milliliters of
concentrated sulfuric acid do you need to make 250 mL of a 1 N solution?
(density = 1.84 g/ml, concentration = 97%)
Normality of H2SO4 =
N V = N’ V ’
10 x % x d = 10 x 97 x 1.84 = 36.42 N
Eq.wt
49
Vml = 1 x 250 ml / 36.42 = 6.86 ml
If you took 6.86 mL of concentrated sulfuric acid and diluted it to 250 mL, you would
have a 1 N H2SO4 solution.
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