1 2.0 STOICHIOMETRY, CALCULATIONS USING CHEMICAL

Stoichiometry
2.0
Jorge Ramos, Ph.D.
STOICHIOMETRY, CALCULATIONS USING CHEMICAL EQUATIONS
The word formula on its own has various meanings; here we are referring to the relative
amount of each element for a given amount of compound. There are various ways of
representing the relative amount of each element for a given amount of compound, these
lead to MOLECULAR formulas and EMPIRICAL formulas. There are some concepts we
should be familiar with before an experiment is carried out...
2.1
MASS PERCENTAGE
The percentage of γ =
[mass of γ in the whole]
-------------------------- X 100
[mass of the whole]
Example 2.1a
Compute the mass percentage of the elements in ammonium nitrate, NH4NO3
NH4NO3 = [14 X 2] + [1 X 4] + [16 X 3] = 80.0 g/mol
% N = [(14 X 2) / 80.0] X 100 = 35.0%
% H = [(1 X 4) / 80.0] = 5.00%
% O = [(16 X 3) / 80.0] = 60.0%
2.2
EMPIRICAL FORMULAS
The mass percentage was just obtained from an EMPIRICAL FORMULA, so the inverse
process would yield an empirical formula from mass percentage. An empirical formula is
the formula of a compound written with the smallest whole number subscripts.
Example 2.2a
A compound contains the following mass percentages: 69.6% O, and 30.4% N. What is
the empirical formula?
mol of O = [69.6 / 16] = 4.35
mol of N = [30.4 / 14] = 2.17
O 4.35 --> 4.35 / 2.17 = 2.00
N 2.17 --> 2.17 / 2.17 = 1.00
The ratio N:O is 1:2, so the empirical formula is NO2
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Stoichiometry
Jorge Ramos, Ph.D.
Example 2.2b
Determine the empirical formula of ibuprofen, a headache remedy that contains the
following mass percentages: 75.69% C, 8.80% H, and 15.51% O.
mol of C = [75.69 / 12] = 6.30
mol of H = [8.80 / 1] = 8.80
mol of O = [15.51 / 16] = 0.969
C 6.30 --> 6.30 / 0.969 = 6.50 --> 6.50 X 2 = 15
H 8.80 --> 8.80 / 0.969 = 9.08 --> 9.08 X 2 = 18
O 0.969 --> 0.969 / 0.969 = 1.00 --> 1.00 X 2 = 2
2.3
MOLECULAR FORMULAS
The empirical formula is sometimes the molecular formula of the compound, as is the
case in the above example. However, the molecular formula is unique for a particular
compound, while many compounds might have the same empirical formula.
Example 2.3a:
Certain compound is 39.9% C, 6.7% H, and 53.5% O. Determine the empirical formula.
The molar mass of this compound is 180.0 g/mol. What is its molecular formula?
mol of C = [39.9 / 12.0] = 3.3 mol of C
mol of H = [6.7 / 1.00] = 6.7 mol of H
mol of O = [53.5 / 16.0] = 3.3 mol of O
C 3.3 --> 3.3 / 3.3 = 1.0
H 6.7 --> 6.7 / 3.3 = 2.0
O 3.3 --> 3.3 / 3.3 = 1.0
The ratio C:H:O is 1:2:1, so the empirical formula is CH2O
Then we use:
Molar Mass = Number of Moles X Empirical Formula Mass
where the Number of Moles is the number of empirical units that fit into the molar mass
of the compound, and the empirical formula mass is the sum of molar masses of the
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Stoichiometry
Jorge Ramos, Ph.D.
component elements in one empirical unit:
C + [2 X H] + O = 12.0 + [2 X 1.00] + 16.0 = 30.0 g/mol
And the molecular formula can then be obtained. First we obtain the number of empirical
units that fit into the molar mass of the compound:
n = MM / EFM = 180.0 / 30.0 = 6.00
So the empirical unit fits six times in the molar mass of the compound.
CH2O X 6.00 = C6H12O6
Example 2.3b:
MSG has a molar mass of 169 g/mol and contains the following mass percentages:
35.51% C, 4.77% H, 37.85% O, 8.29% N and 13.60% Na. Determine the empirical and
molecular formula of MSG.
mol of C = [35.51 / 12.01] = 2.96 --> 2.96/0.592 = 5
mol of H = [4.77 / 1.008] = 4.73 --> 4.73/0.592 = 8
mol of O = [37.85 / 16.0] = 2.37 --> 2.73/0.592 = 4
mol of N = [8.29 / 14.01] = 0.592 --> 0.592/0.592 = 1
mol of Na = [13.60 / 22.99] = 0.592 --> 0.592/0.592 = 1
Molar mass of one empirical unit:
C(5) + H(8) + O(4) + N(1) + Na(1)
( 60.05 +
8.064 + 60.0 + 14.01 + 22.99 ) g/mol = 165 g/mol
165 and 169 are similar enough to conclude that the empirical and molecular formula of
MSG are the same: C5H8O4NNa
2.4
STOICHIOMETRY
This is a fundamental part of quantitative work in chemistry and is based on the notion
that an equal quantity of matter exists both before and after a reaction. To understand the
concept, let's consider the reaction:
WO3(s) + 3H2(g) --> W(s) + 3H2O(g)
Until more advanced concepts are covered, the physical state of the compounds (solid,
liquid, gas, plasma) are not to be considered important. Just notice that the equation is
balanced:
For the reactants,
There is a number 1 on front of WO <-- no number means 1
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Stoichiometry
Jorge Ramos, Ph.D.
There is a number 3 on front of H2
For the products,
There is a number 1 on front of W
There is a number 3 on front of H2O
The chemical equation can be translated into other formats:
1 mol WO3(s) + 3 mol H2(g) --> 1 mol of W(s) + 3 mol of H2O(g)
231.84 g of WO3(s) + 6.00 g of H2(g) --> 183.84 g of W(s) + 54.00 g of H2O(g)
The equation is balanced, that is how we know that the given amounts will produce 1 mol
of W(s). If the amount of WO3(s) were increased to 4 mol while keeping the amount of
H2(g) constant we would still get 1 mol of W(s) because H2(g) is LIMITING. In order to
use up the 4 mol of WO3(s) we would need 12 mol of H2(g) so that none of the reagents
is limiting.
The predictive power of stoichiometry is derived from its ability to connect the
submicroscopic and the macroscopic world. Note that 3 mol of H2O weights 54.00 grams,
each mol amounts to 6.022 X 1023 molecules of H2O. There is nothing special about this
number, the concept was conceived before the actual value was determined. The
following two statements are equivalent and will better illustrate the concept:
THERE ARE 6 BEERS IN ONE SIX-PACK OF BEERS.
THERE ARE 6.022 X 1023 ATOMS IN ONE MOLE OF ATOMS.
None of these statements provides information about the mass of the six-pack of beers or
the mole of atoms. For example, if you go to the store and buy a six-pack of 11-ounce
beers, then the six-pack would weight 66 ounces. If you decide to get the 20-ounce beers
six-pack, then the six-pack would weight 120 ounces. And it is still a six-pack.
The same can be said about the mole of atoms: one mole of Neon atoms weights ~20
grams, one mole of Calcium atoms weights ~40 grams. And it is still one mole.
Example 2.4a:
The following chemical equation is balanced. The corresponding number of grams used
are as shown and the reaction goes to completion.
3 CCl4 +
2 ΨF3
4.00 g
4.61 g
Æ
3 CCl2F2
+
2 ΨCl3
3.14 g
Find the molar mass of the element Ψ.
Moles of CCl4 = 4.00 g ÷ [153.8 g/mol] = 0.02601 mol
By mol ratio: 3 CCl4: 2 ΨF3 then it is 3:2 ratio
Moles of ΨF3 = [0.02601 mol] X [2/3] = 0.01734 mol of ΨF3
Molar mass of ΨF3 = 4.61 g ÷ [0.01734 mol] = 265.9 g/mol
Molar mass of Ψ = 265.9 g/mol – [3 X 19.00 g/mol] = 208.9 g/mol
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Stoichiometry
Jorge Ramos, Ph.D.
Example 2.4b:
Excess HCl was added to an antiacid tablet weighting 0.50 grams. The active ingredient
in this type of tablet is Mg(OH)2. The chloride salt from this reaction was recovered and
its total mass was 90.0 mg. Compute the percent Mg(OH)2 by mass in the tablet. You can
refer to the following chemical equation:
Mg(OH)2 + HCl Æ Mg(Cl)2
+ 2H2O
mol of Mg(Cl)2 = 90.0 X 10-3 g / [95.2 g/mol] = 0.945 X 10-3 mol = mol of Mg(OH)2
because from the chemical equation, Mg(Cl)2 = Mg(OH)2
Mg + O2 + H2 = 24 + [16 X 2] + [1 X 2] = 58 g/mol
mass of Mg(OH)2 in the tablet = 0.945 X 10-3 mol X 58 g/mol = 54.81 10-3 g
Mg(OH)2 by mass in the tablet = [54.81 10-3 g / 0.50 g] X 100 = 11 %
Example 2.4c:
A sample of soil is known to contain Barium compounds. 5.00 g of this sample were
dissolved to make 1.00 L of solution. Excess sulfuric acid was added to 200.0 mL of the
solution. Answer question (a) assuming that the precipitate is entirely BaSO4:
Ba2+(aq) +
2-
SO4 (aq)
Æ
BaSO4(s)
(a) If 0.1600 g of precipitate were recovered, what is the percent Barium by mass in the
sample?
(b) Upon further analysis it was found that the soil is 3.0 % Calcium by mass so there
was a second reaction:
2+
Ca (aq) +
2-
SO4 (aq)
Æ
CaSO4(s)
Compute the percent Barium by mass in the soil taking into account the new data.
Fraction of barium in BaSO4 = [137.3] / [137.3 + 32.06 + (16 X 4)] = 0.5884 = 58.84 %
Mass of barium in precipitate = 0.5884 X 0.1600 g = 0.09414 g
Since 5.0 g of soil went into 1.00 L and only 200.0 mL were titrated, the barium amount is 5 times as large
Mass of barium in soil sample = [0.09414 g] X [1000.0 mL / 200.0 mL] = 0.4707 g
Percent barium by mass in soil sample = [0.4707 g / 5.00 g] X 100 = 9.41 %
But the soil contains 3.00 % calcium, so the precipitate contains CaSO4 as well
Mass of calcium in soil sample = 5.00 g X 0.0300 = 0.150 g
Mass of calcium in precipitate = 0.150 g X 0.2 = 0.0300 g (because only 200.0 mL of solution are titrated)
-4
# of mol of Ca = 0.0300 g / 40.08 g/mol = 7.49 X 10 mol
-4
Because there is 1 Ca in each CaSO4 unit, # of mol of CaSO4 = 7.49 X 10 mol
-4
mass of CaSO4 in precipitate = 7.49 X 10 mol X [40.08 + 32.06 + (16 X 4)] g/mol = 0.102 g
mass of BaSO4 in precipitate = 0.1600 g – 0.102 g = 0.0580 g
Mass of barium in precipitate = 0.0580 g X 0.5884 = 0.0341 g
Mass of barium in soil sample = 0.0341 g X [1000.0 mL / 200.0 mL] = 0.170 g
New percent barium by mass in soil sample = [0.170 g / 5.00 g] X 100 = 3.40 %
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