Empirical Formula
Definition
it gives the simplest or least or
smallest whole number ratio of different combining atoms of elements present in
a molecule or formula unit of a compound.
the word
'empirical' means derived from experimental data. It is the type
of formula which expresses only the relative number of each kind of atoms in a
molecule (or formula unit) of a compound.
it tells us which elements are present
and their atomic ratio in a molecule of a compound but it does not necessarily
give the actual number of atoms present in the molecule.
Example
1. The
empirical formula of benzene is CH i.e. the least ratio of C to H in its
molecule is 1:1. However the true or
molecular formula of benzene is C6H6. [Similarly the
empirical formula of ethyne (C2H2) is CH].
2. The empirical formula of glucose is CH2O i.e. the least ratio of C, H and O is 1:2:1. However the true or molecular formula of glucose is C6H12O6. [Similarly the empirical formula of Acetic acid (C2H4O2) is CH2O].
The
molecular formula of glucose is C6H12O6 showing
that number of atoms in it are 6, 12 and 6. To simplify the ratio, it is
divided by the largest number i.e. 6 to give its empirical formula. The
empirical formula of glucose becomes CH2O
C = 6/6 = 1
H = 12/6 = 2
O = 6/6 = 1
C6H12O6
6: 12: 6
1 : 2 : 1
C : H2 :O
Empirical
Formula of some important compounds
The term empirical
formula is used for ionic compounds and giant covalent or macromolecular
compounds like sand, diamond and graphite etc. It is also used for covalent
compounds as CH2O for glucose and acetic acid.
Explanation
Showing
Lowest whole number ratio
An
empirical formula shows only the lowest whole number ratio of combining
atoms of different elements in a molecule and it does not necessarily give the
actual or exact number of atoms in a molecule. Thus mostly it does not
express the true composition of a chemical compound.
For Example;
the simplest whole number ratio between C to
H in a molecule of ethyne or Acetylene is 1:1 giving its empirical formula CH
while true or molecular formula of ethyne is C2H2.
Reason for Calling Simplest Formula
It
is also called simplest formula as it gives least whole number ratio of
various atoms present in a molecule.
Used for Ionic or Inorganic
Compounds
Ionic
(or inorganic) compounds are usually represented by their empirical
formulae (showing the lowest whole number ratio of the ions in them).
For Example;
sodium chloride, which is an ionic compound,
is shown by its empirical formula NaCl showing that ratio of sodium ions to
chloride ions is 1:1.
Some
other examples of empirical formulae of few ionic compounds are MgO, CaO, KF,
KCl, KBr, KI, NaBr, NaI, NaF, LF, NaH, KH, NaOH, KOH, CuSO4, CaCl2,
K2SO4, Na2SO4, MgSO4, Na
NO3, KNO3
Different Compounds Showing Same
empirical formula
The empirical formula of two or more (covalent) compounds may be the
same.
For Example;
Benzene
and Acetylene both have same empirical formula of CH [but their molecular
formulas are C6H6 and C2H2
respectively].
Similarly
CH2O is the empirical formula of various compounds like glucose,
fructose, acetic acid, and formaldehyde.
Compounds having Same Empirical and Molecular Formulae
The
empirical formula of a compound may or may not the same as its molecular
formula. A chemical compound can have the same empirical
and molecular formula only when the simplest ratio and actual ratio of its
atoms are identical.
For
many simple covalent compounds have same empirical and molecular formula e.g. H2O,
CO2, NH3, CH4, HCHO or CH2O
(formaldehyde), C12H22O11 (sugar)
Found From Experiments
Converting Elements into known
Compounds
The first step in determining empirical formula is to find out mass of each element present in it. The mass of element is determined by converting it to the compound whose mass can be determined directly. Usually elements present in organic compounds are converted into following compounds:
(i). Carbon is converted into CO2.
(ii). Hydrogen is converted into H2O.
(iii). Sulphur is converted into BaSO4.
(iv). Chlorine is converted into AgCl.
(v). Bromine is converted into AgBr.
(vi). Iodine is converted into AgI.
(vii). Phosphorus is converted into Mg2P2O7
Steps for
Determination of Empirical Formula
1. Calculation of Mass of
Elements
2. Determination of the
percentage composition of each element
3. Finding
the number of gram atoms or mole dividing the % of elements by its relative
atomic mass
4. Determination of simplest
atomic ratio of each element by diving gram atom or mole by the
smallest number of mole.
If the atomic ratio is
in simple whole number, it gives the empirical formula otherwise multiply the
simplest ratios with a suitable number to get the whole number atomic ratio.
1. Calculation of Mass of each
Elements
If mass of elements is
not given, it is calculated by using following formulas:
For Example
2. Determination of %ages (percentage composition) of each Element
3. Determination of Mole Ratio of each Element
4. Determination of Simplest Ratio of each Element
The numbers or
quotients so obtained (i.e. mole fraction) are divided by their least value to
get simplest ratio. The values obtained in Step IV are then converted to
nearest whole number by rounding off.
If resulting figure is
not whole number is made a whole number by multiplying them all by a suitable
digit like 2 (in rare cases we might have to multiply by 3 or 4) to obtain
whole numbers. These whole numbers represents simplest ratios of element.
Percentage
Composition
If number of atom =1; then molecular mass = minimum molecular mass
Analysis
of a Compound
Qualitative Analysis
It involves identification of types of elements present in a compound.
Quantitative Analysis
It involves the determination of the masses of all the elements.
Methods to determine percentage composition
The percentage
composition obtained from a given formula of a compound is called theoretical
percentage composition.
Combustion analysis
The sequence of
combustion analysis is shown in the following diagram
By combustion
analysis, only those organic compounds can be analyzed which simply contain
carbon, hydrogen and oxygen.
Molecular Formula
Definition
“It is the type of formula which expresses the actual number of atoms of
each element present in a molecule of a substance”. Molecular formula gives actual atomic ratios in a molecule
of a compound. Molecular formula is same as empirical formula or it may be an
integral multiple of empirical formula.
For example the molecular formula of glucose is C6H12O6
and molecular formula of ethene is C2H4.
Explanation
1. A molecular formula not
only indicates the relative number of atoms but also the actual or total number
of atoms of each element in a molecule of the compound.
For example:
The molecular formula
of benzene is C6H6. This indicates that benzene molecule
is composed of 6 atoms of carbon and 6 atoms of hydrogen.
2. The molecular formula
and empirical formula may be the same (when actual number of atoms and simplest
ratio between them are same).
For example:
Molecular Formula and
Empirical Formula of water and methane are same i.e. H2O and CH4.
Work Out Molecular Formula
The molecular formula
of a substance is an integral multiple of the empirical formula i.e. It is
obtained by multiplying empirical formula by empirical units, n.
Molecular Formula =
(Empirical Formula)n
Where ‘n’ is any
simple whole number (i.e. 1, 2, 3 etc.). When value of ‘n’ is equal to 1, then
molecular formula and empirical formula are same.
Usually ionic
compounds (NaCl, MgCl2, MgO etc.) have same molecular and empirical
formula.
The value of integer
‘n’ is calculated by following formula:
Numericals on Empirical and Molecular Formula
Q1. 6.38 g of ethylene glycol gives 9.06 g of CO2 and
5.58 g of H2O. Its molecular mass is 62 a.m.u. Find out its
molecular formula.
Solution
Given
Mass of organic
compound = 6.38 g
Mass of CO2 = 9.06 g
Mass of H2O = 5.58 g
Molecular mass of
compound = 62 a.m.u
Required
Molecular formula of compound = ?
Calculation
1.Calculation of
mass of C and H
2. Calculation of
%age of each element
3. Calculation of Mole Ratio (M.R.) of each element
4. Calculation of simplest atomic ratio (SAR) of each element
5. Calculation of integral n
Molecular mass of compound = 62 g
Emp. formula mass of CH3O = 12 + 3 + 16 = 31 g
Integer (n) = molar mass/empirical formula mass
= 62/31 =2
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Q2. 1.0 g of a gaseous hydrocarbon occupying a volume of 0.386 dm3
at STP, was combusted in air to
yield 3.03 g CO2 and 1.55 g H2O. Find the molecular
formula of the hydrocarbon.
Solution
Given
Required
Molecular formula of hydrocarbon = ?
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Calculation of Molecular mass
Calculation of Molecular Formula
Q3. The empirical formula of compound is CO2H. 1.8 g of this
compound in gaseous state occupies 448
cm3 at STP. Find its molecular formula. [K.B. - 2015]
Solution
Given
Required
Molecular formula of compound = ?
Calculation of Molecular mass
Calculation of Molecular Formula
Molecular
mass of compound = 90 amu
Empirical
formula mass of CO2H =12 + 16 x 2 + 1 = 45 g
M.F. = (E.F.)
x n = (CO2H)2 =
C2O4H2
Q4. An organic compound producing air pollution contains 8.73%
carbon, 77.45% chlorine and 13.82%
fluorine. Find the molecular formula of the compound if its molecular mass is
137.5 amu. (KB-2021)
Solution
Assignment
Problems on E.F., M.F. and Percentage Composition
Four Steps Numericals on Calculating Empirical Formula
Q1. 1.367
g of an organic compound containing C, H and O was combusted in air to give
3.002 g CO2 and 1.64 g H2O. Determine its empirical
formula. (Ans: C3H8O)
Q2. 0.2475
g of an organic compound containing C, H and O gave on combustion 0.4950 g CO2
and 0.2025 g H2O. Calculate
its empirical formula. (Ans:
C2H4O)
Q3. 0.638 g of an organic compound on
combustion gave 0.594 g H2O and 1.452 g CO2. Find its
E.F. (Ans:
C3H6O)
Q4. 6.38
g of ethylene glycol gives 9.06 g of CO2 and 5.58 g of H2O.
Its molecular mass is 62 a.m.u. Find out its molecular formula. (Ans:
EF = CH3O, M.F = C2H6O2)
Q5. A combustion device was used to determine the empirical formula of a compound containing only C, H and O. A 0.6349 g sample of the unknown produced 1.603 g of
CO2 and 0.2810 g H2O. Determine the empirical formula of
the compound.
Q6. 5.325 g sample of
methyl benzoate, a compound used in the manufacture of perfumes, is found to
contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is
the empirical formula of this substance? (Answer; C4H4O)
Q7. isopropyl alcohol
is known to contain only C, H and O. Combustion of 0.255 g of isopropyl alcohol
produces 0.561 g of CO2 and 0.306 g of H2O. Find out its
EF. (Answer; C3H8O )
Q8. Fructose is a
type of sugar found in fruits and honey. Analysis of a 2.00 g sample shows that
it contains 0.80 g of C, 1.06 g of O, and 0.14 g of H. An independent analysis
shows that fructose has a molar mass of 180 g/mol. What is the molecular
formula for fructose? (Answer;
Q9. Menthol, which is
present in mentholated cough drops, is an organic compound containing only C,
H, and O. When a 0.2010 g sample is analyzed by combustion, 0.5658 g of CO2
and 0.2318 g of H2O are obtained. What is the empirical formula of
menthol? (Answer; C10H20O)
Q10. Eugenol is the
compound responsible for the odor of cloves. Analysis of a 0.0188-g sample of
eugenol by combustion gives 0.0506 g of CO2 and 0.0124 g of H2O.
Given that eugenol is known to contain only carbon, hydrogen, and oxygen, what
is its empirical formula? (Answer; C5H6O)
Two Steps Numericals on Empirical Formula
Q1. Diethyl
zinc is a chemical used in the library to protect the books from the worms. Its
composition is 53% zinc, 38.9 C% and 8.1% hydrogen. Find the empirical formula
of the compound. [K.B. – 2007] (Ans;
C4H10Zn or (C2H5)2Zn)
Q2. An
organic compound contains 33.35% C, 3.705 % of H and 12.09% of N. The molecular
mass of the compound is 108. Determine the empirical and molecular formulae of
the compound. [The percentages given are wrong and they should be C = 66.7%, H
= 7.41% and N = 24.18%]. [K.B.
– 2006] (Ans; C3H4N and C6H8N2)
Q3.
Find the empirical formula of an
organic compound whose composition is given below:
C = 51.80%, H = 13.12% and O = 35.08%. [K.B.
– 2001] (Ans;
C2H6O)
Q4. Find the simplest formula of compound
containing: (Ans:
C2H5NO)
C = 40.48%; H = 8.43%; N =
23.73% and O = 27.36%.
Q5. What
is the empirical formulate for isopropyl alcohol (which contains only C, H and
O) if the combustion of a 0.255 grams isopropyl alcohol sample produces 0.561
grams of CO2 and 0.306 grams of H2O? (Ans:
C3H8O)
Q6.
An organic compound on analysis
gave 10.06% C, 0.84% H and 89.10% Cl. Find out its EF. (Ans: CHCl3)
Q7.
An O.C. on analysis gave 23.76%
C, 5.94% H and 70.3% Cl. Find out its empirical formula. (Ans: CH3Cl)
Q8.
An O.C. contains 4.74% C, 0.4% H
and 94.86% Br. Find its empirical
formula. (Ans:
CHBr3)
Q9.
An O.C. contains 77.42% C, 7.53%
H and 15.5% N. Find its empirical
formula. (Ans:
C6H7N)
Q10.
On analysis an O.C. gave 68.9% C,
4.8% H and 26.3% O. Find its E.F.
[Multiply by 2 in last step]. (Ans:
C7H6O2)
Q11.
An O.C. contains 60.8% C, 3.02% H
and oxygen. Find its E.F. (Ans: C5H3O2)
Q12.
An O.C. contains 40% C, 6.67% H and
53.33% O. Find its E.F. (Ans:
CH2O)
Q13.
A given compound contains C = 60%, H
= 13% and O = 27%. Calculate its
E.F. (Ans:
C3H8O)
Q14.
Calculate E.F. from the given
percentage composition; C = 68.8%, H = 5%, O = 26.2%. (Ans:
C7H6O2)
Q15.
Calculate the empirical formula of
compound which contains 52.07% C, 13.04% H and 34.88% O. (Ans: EF = C2H6)
Q16. Determine
the empirical formula of the following compounds that underwent combustion
analysis.
(a).
Toluene is composed of C and H
and yields 5.86 mg of CO2 and 1.37 mg of H2O after
combustion. (Ans; C7H8)
(b).
0.1005 g of menthol, which is
composed of C, H, and O, yields 0.2829 g CO2 and 0.1159 g H2O
after combustion. (Answer;
C10H20O)
Q17. Ibuprofen, a pain
relief medication, is 75.69% C, 8.80% H, and 15.51% O, by mass. What is the
empirical formula for ibuprofen. (Answer;?(Answer; C13H18O2)
Q18. Epinephrine, also
known as adrenaline, is 59.0% C, 7.1% H, 26.2% O, and 7.7% N, by mass. What is
the empirical formula for epinephrine? (Answer; C9H13O3N)
Q19. Caffeine contains
only C, H, N, and O. When a 0.376 g sample is analyzed by combustion, 0.682 g
of CO2, 0.174 g of H2O, and 0.110 g of N2 are
obtained. What is the empirical formula for caffeine? (Answer;
C4H5N2O)
Q20. Lysine, which is an
essential amino acid, contains only C, H, N, and O. In one experiment, the
complete combustion of 2.175 g of lysine produces 3.93 g of CO2 and
1.87 g of H2O. In a separate experiment, 1.873 g of lysine produces
0.436 g of NH3. What is the empirical formula of lysine? (Answer;
C3H7NO)
Q22. What is
the empirical formula of each of the following compounds?
a.
Talc by mass composition
contains 19.2% Mg, 29.6% Si, 42.2% O and 9.0% H. (Answer;
Mg3Si4O10H34)
b.
Saccharin has by mass
composition 45.89% C, 2.75% H, 7.65% N, 26.20% O and 17.50% S. (Ans;
C7H5NO3S)
c.
Salicylic Acid, used in
aspirin, contains 60.87% C, 4.38% H, and 34.75% O by mass
composition.
(Answer; C7H6O3)
d.
L-Dopa, a drug used for the
treatment of Parkinson's disease, is 54.82% C, 5.62% H, 7.10% N, and 32.46% O
by mass
composition. (Answer; C9H11NO4)
Q23Mercury
forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by
mass. What is the empirical formula? (Ans:
HgCl2)
Q24.
Calculate the empirical formula of
an oxide of iron that contains 70% Fe by mass. (Ans:
Fe2O3)
Four Steps Numericals on Calculating Empirical Formula
Two Steps Numericals on Empirical Formula
Four Steps Numericals on Calculating Molecular Formula
Six Steps Numericals on Calculating Molecular Formula
1. Acetic acid contains C, H and O. 2.14 g
of a sample of acetic acid on complete combustion gave 3.105 g of CO2
and 1.27 g of water. The molecular mass of acetic acid is 60; find the
empirical and molecular formulae.
[K.B. – 2003,PM] (Ans: CH2O
and C2H4O2)
2. 4.2 g of sample of an organic compound
on complete combustion gave 6.21 g of CO2 and 2.54 g of H2O. Its molecular mass is 60. Find its molecular formula.
(Ans: C2H4O2)
3. 1 g of sample of hydrocarbon on
combustion gave 3.03 g CO2 and 1.55 g H2O. If molecular weight of it is 58, find out its
molecular formula.
(Ans: C4H10)
4. 0.5
g of a hydrocarbon gave on combustion 1.515 g CO2 and 0.77 g H2O.
If molecular weight is 58, find its molecular formula.
(Ans:
C4H10)
5. On
combustion 0.2000 g of sample of a carbohydrate produces 0.2933 g of CO2
and 0.1200 g of H2O.
Molecular weight of the compound is 180.
Calculate its molecular formula.
(Ans:
C6H12O6)
6. 0.94 g of an O.C. gave on combustion
1.32 g CO2 and 0.568 g H2O. Its molecular mass was found
to be 180; determine its molecular formula.
(Ans: C6H12O6)
7. 0.1802 g of an O.C. gave on combustion
0.2641 g of CO2 and 0.1081 g H2O. Its molecular mass was found to be thrice of
acetic acid. Find its molecular formula.
(Ans: C6H12O6)
8. 0.30 g of an organic compound
containing C,H and O on combustion
yields 0.44g CO2 and 0.18g H2O.
If one mole of compound weighs 60g, then find out its molecular formula of
the compound.
(Ans: C2H4O2
9. The combustion of 40.10 g of a compound
which contains only C, H, Cl and O yields 58.57 g of CO2 and
14.98 g of H2O. Another sample of the compound with a mass of 75.00
g is found to contain 22.06 g of Cl. What is the empirical formula of the
compound?
(Ans: C4H5ClO2)
10. The combustion of 1.38 grams of a
compound which contains C, H, O and N yields 1.72 grams of CO2 and
1.18 grams of H2O. Another sample of the compound with a mass of
22.34 grams is found to contain 6.75 grams of O. What is the empirical formula
of the compound?
[Hint; % of N = 100 −
(34.01 + 9.57 + 30.215) = 26.205%) (Ans: C3H10N2O2)
Two Steps Numericals on Calculating Molecular Formula
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