Using Stoichiometry to Determine Chemical

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Q3U2 :Theoretical Yield, Mass Percents, and Empirical Formulas
Theoretical Yield: The amount of product
formed through a chemical reaction, as
predicted by Stoichiometry
 Actual Yield: The amount of product formed
when a reaction is performed in a lab, usually
less than theoretical yield.
Why are they different?
 The collection techniques, apparatus used,
time, and skill of the chemist can effect the
actual yield.

When actual and theoretical yields are
different, you express the efficiency of the
reaction using percent yield.
 Percent yield is the ratio of actual yield to
theoretical yield, expressed as a percent.
Percent yield = (actual yield) X 100
(theoretical yield)




Chemical formulas tell you the elements
involved in a reaction, the number of atoms
of each and the number of moles of reactants
and products
What is the major element in a reaction, by
mass?
To answer this, determine the mass percents
of each element
Steps
1. Determine the molar mass of the element in
the compound
2. Multiply the mass of 1 mole times the
number of moles in the compound
3. Now use that mass to find the mass percent
in the sample

In 154 grams of C10H18O (Geraniol), How many grams does
the Carbon atom contribute?
 1 molecule of C10H18O contains 10 atoms of Carbon
 Therefore 1 mole of C10H18O contains 10 moles of carbon
atoms
 Multiply the mass of 1 mole of carbon by 10 to get the
mass of carbon in one mole C10H18O
 (10 mol) (12.0gC) = 120g C
(1 mol)
 Use mass to determine mass percent
mass % of C = ( 120 gC )
X 100 = 77.9%
(154 g C10H18O)
Given 154 g of C10H18O
What is mass% of H?
18 mol H X 1.01 g/mole = 18 g H

Mass% H = mass H X 100
mass C10H18O
= 18.0 g H
X 100 = 11.7% H
154 g C10H18O

What is the Mass% of O?
What is the Mass% of O?
 (1 mol O) (16 g O) = 16 g O
( 1 mol)
16 g O
X 100 = 10.4% O
154g C10H18O

The previous steps will allow you to determine
the composition of any compound, as long as
you know the formula.
Try the Following:
1. Hydrogen fuels are rated with respect to their
hydrogen content. Determine the percent
hydrogen for the following fuels.
a) Ethane, C2H6
b) Methane, CH4
c) Whale Oil, C32H64O2
a)
b)
c)
Ethane, C2H6 2 mol C, 6 mole H
X 12.0g X 1.0 g
24.0g
6.0 g
(6.0 g/30.0 g ) X 100 = 20%
= 30.0g
Methane, CH4 1 mol C , 4 mol H
X 12.0g X 1.0 g
12.0 g
4.0 g
(4.0 g/16.0 g ) X 100 = 25%
Whale Oil, C32H64O2 32 mol C, 64 mol H, 2 mol O
X 12.0g X 1.0 g X 16.0 g
384. 0g
(64.0 g/480.0 g ) X 100 = 13%
64.0 g 32.0 g
Calculate the Percent Composition of oxygen
in the following compounds
a. SO3
b. CH3COOH
c. Ca(NO3)2
d. Ammonium Sulfate, (NH3) SO2

a.
b.
c.
d.
60.00%
53.29%
58.50%
48.43%

Hydrogen peroxide (H2O2), what is the mass
% of Oxygen?

Sodium nitrate (NaNO3), what is the mass %
of Oxygen?
a) H2O2
2 mol O, 2 mole H
X 16.0g X 1.0 g
32.0g
2.0 g
(32.0 g/34.0 g ) X 100 = 94.1%
b)
NaNO3
= 34.0g
1 mol Na, 1 mol N, 3 mole O
X 23.0g X 14.0 g X 16.0 g
23.0 g 14.0g
48.0 g
(48.0 g/85.0 g ) X 100 = 56.4%

Fish in some lakes have been found to contain a mercury
compound, possibly a contaminant from the making of
paper.

Analysis of this compound gives the following mass
percentages: Carbon, 5.57% ; hydrogen, 1.40% ; and
mercury, 93.03%.

Using this information, determine the Empirical
Formula of the compound.

Empirical Formula: the formula of a compound that
has the smallest whole-number ratio of atoms in the
compound
 Steps
1. Fin d the relative numbers of atoms in the
2.
3.
formula unit of the compound
Use the molar mass to find the number of
moles of each element
Divide the mole numbers by the smallest one,
to find the whole number ratios
Empirical Formula: the formula of a compound
that has the smallest whole-number ratio of
atoms in the compound

Analysis of this compound gives the following mass
percentages: Carbon, 5.57% ; hydrogen, 1.40% ; and mercury,
93.03%.
 Using this information, determine the Empirical Formula of
the compound.
Carbon: 5.57g
12.01 g/mol = 0.464 mol
ratio = 1
Hydrogen: 1.40 g
1.008 g/mol = 1.39 mol
ratio: 1.39/ 0.464 = 3
Mercury: 93.03g
200.59 g/mol = 0.464 mol
ratio = 1
Empirical formula is CH3Mg, methyl mercury

An unknown compound is 18.8% Na, 29.0% Cl
and 52.2% O, what is the empirical formula of
the unknown?

Use molar mass to find the number of moles of each element.
 (18.8 g Na) (1mol Na) = 0.817 mol Na
(23.0 g Na)
 (29.0 g Cl) (1mol Cl) = 0.817 mol Cl
(35.5 g Cl)
 (52.2 g O) (1mol O) = 3.26 mol O
(16.0 g O)

Divide the mole numbers by the smallest one.
 0.817/ 0.817 = 1 mol Na ratio = 1
 0.817/ 0.817 = 1 mol Cl
ratio = 1
 3.26/ 0.817 = 3.99 mol O ratio = 4
The Empirical formula is NaClO4

Hydrogen = 5.17%, nitrogen is 35.9% and
sodium is 58.9%, what is the empirical
formula?



5.17 g H X 1 mol H/ 1.01 g H = 5.12 mol H
35.9 g N X 1 mol N/ 14.0 g N = 2.56 mol N
58.9 g Na X 1 mol Na/ 23.0 g Na = 2.56 mol Na
Ratio of H:N:Na is 2:1:1
Empirical formula is NaNH2
a.
15.8% carbon and 84.2% sulfur
b.
43.6% phosphorus and 56.4% oxygen
c.
28.7% K, 1.5% H, 22.8% P and 47.0% O
a.
b.
c.
CS2
P2O5
KH2PO4
Calculate the empirical formula for the following
compounds from percent composition
a. 0.0130 mol C, 0.0390 mol H, 0.0065 mol O
b. 11.66 g iron, 5.01 g oxygen
c. 40.0 percent C, 6.7 percent H, and 53.3
percent O by mass
a.
b.
c.
C2H6O
Fe2O3
CH2O


For most ionic compounds the two are the same
Covalent compounds can be a very different
case.
 The ratio between moles is the same, but the atoms
can share electrons differently, so the formula may be
different.
▪ Ex. C6H12O6 and C3H6O3 and C2H4O2
glucose
lactic acid
acetic acid
Different compounds with different properties, same molar
ratios, different chemical formulas!



You need the molar mass of the compound,
change to molecular mass (amu)
Divide the molecular mass of the compound
by the molecular mass of the empirical
formula unit
This will tell you how many of the empirical
formula units are in the chemical formula
40.0% C, 6.7% H, 53.3% O
Find the empirical formula

Is this the same as the chemical formula, if the
molar mass is 90.0 g/mol
3.33 mol C = 1 mol C
6.7o mol H = 2.01 mol H
3.33 mol O = 1.o mol O
Empirical formula is CH2O



90.0 g/mol = 90.0 amu molecular mass
CH2O molecular mass= 12+2 + 16= 30 amu
90.0/30.0 =3 3 empirical formula units
C3H6O3, lactic acid
a. empirical formula CH , molar mass = 78 g/mol
b. empirical formula NO2 , molar mass = 92.02 g/mol
c
caffeine, 49.5% C , 5.15% H , 28.9% N , 16.5% O
by mass, molar mass = 195 g.

a. C6H6

b. N2O4

c. Use mlar mass to find ratios, and empirical
formula first, divide molecular and empirical
formula masses to determine number of
empirical formula units in sample grams
C8H10N4O2

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