Chemistry for Engineers 1A (CHEM 181) – UKZN – School of Chemistry , Howard College Campus,
compiled by A.Bissessur
2011
MASS RELATIONSHIPS IN CHEMICAL
REACTIONS
1. The mole, Avogadro’s number and molar
mass of an element
2. Molecular mass (molecular weight)
3. Percent composition of compounds
4. Empirical and Molecular formulas
5. Stoichiometry
6. Limiting Reagents
7. Reaction Yields
8. Combustion reactions
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Chemistry for Engineers 1A (CHEM 181) – UKZN – School of Chemistry , Howard College Campus,
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THE MOLE (unit mol, symbol n)
It is the amount of a substance that contains as many elementary
entities (atoms, molecules or other particles) as there are atoms in
exactly 12 g of the carbon-12 isotope.
Consider: 1 mol = 6.022 x 1023 and 1 mol of C = 12 g of C
therefore the actual number of atoms in 12 g of the carbon-12
isotope = 6.022 x 1023 atoms called the Avogadro’s Number
1 mol of C = 12 g = 6.022 x 1023 atoms
Converting between moles and number of atoms
Example 1.
Given:
Solution:
Example 2.
Given:
Solution:
1 mol atoms
6.022 x 1023 atoms
3.5 mol He
3.5 mol He x
6.022 x 1023 atoms
1 mol atoms
or
Find: He atoms
6.022 x 1023 He atoms
1 mol He atoms
1.1 x 1022 Ag atoms
1.1 x 1022 Ag atoms x
= 2.1 x 1024 He atoms
Find: mols Ag
1 mol Ag atoms
6.022 x 1023 Ag atoms
= 1.8 x 10-2 mols Ag
2
Chemistry for Engineers 1A (CHEM 181) – UKZN – School of Chemistry , Howard College Campus,
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CONVERTING BETWEEN GRAMS AND MOLES OF AN ELEMENT
The mass of 1 mol of atoms of an element is called the molar mass
The value of an element’s molar mass in grams per mole is numerically equal to the
element’s atomic mass in atomic mass units (amu)
For example Copper has a atomic mass of 63.55 amu, therefore 1 mol of
copper atoms has a mass of 63.55 g and the molar mass of copper = 63.55
g/mol or g mol-1.
32.07 g of sulfur = 1 mol sulfur
12.01 g of carbon = 1 mol carbon
= 6.022 x 1023 S atoms
= 6.022 x 1023 C atoms
6.94 g of lithium = 1 mol lithium = 6.022 x 1023 Li atoms
Therefore the molar mass of any element becomes a conversion factor between grams of
that element and moles of that element:
Example for carbon:
12.01 g C
12.01 g = 1 mol C =
1 mol C
Example 1.
Given:
Solution:
Find:
0.58 g of C
0.58 g x
Mass of element (g)
Mass of element (g)
1 mol C
or
1 mol C
12.01 g C
mols of C
=
4.8 x 10-2 mols
12.01 g C
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
→
mols of element (n)
n x molar mass
←
mols of element (n)
N = no of atoms, N A = Avogadro’s number, n = no. of moles
n x 𝑁𝑁𝐴𝐴
→
N/𝑁𝑁𝐴𝐴
←
Number of atoms (N)
Number of atoms (N)
3
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COUNTING MOLECULES BY THE GRAM
For elements, molar mass is the mass of 1 mol of atoms of that element.
For compounds, the molar mass is the mass of 1 mol of molecules or formula
units of that compound (also known as molecular mass or molecular weight)
If the atomic masses of component atoms, the mass of the molecule can be
calculated.
For example the molar mass /molecular mass/molecular weight or formula
mass of CO 2 is:
Molecular mass
=
1(atomic mass of C) + 2(atomic mass of O)
=
44.01 amu
=
1(12.01 amu) + 2(16.00 amu)
The molar mass of CO 2 = 44.01 g/mol or g mol-1
CONVERTING BETWEEN GRAMS AND MOLES OF A COMPOUND
Given:
Solution:
Find:
22.5 g of CO 2
22.5 g x
1 mol CO2
44.01 g
= 0.511 mol CO 2
mols CO 2
CONVERTING BETWEEN GRAMS OF A COMPOUND AND NUMBER OF
MOLECULES
Example 1.
Given:
22.5 g of CO 2
Solution: 22.5 g x
1 mol CO2
44.01 g
x
Find: number of CO 2 molecules
6.022 x 1023 CO2 molecules
= 3.08 x 1023 CO 2 molecules
mol CO2
4
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2011
Example 2
Given: 4.78 x 1023 NO 2 molecules
Solution:
Find: grams NO 2
Molar mass of NO 2 = 14.01 + 2(16.00) = 46.01 g mol-1
4.78 x 1023 NO 2 molecules X
46.01 g NO2
1 mol NO2
X
=
23
6.022 x 10
NO2 molecules
1 mol NO2
36.5 g NO 2
Practice Exercise
1. Calculate the molar mass of the following compounds:
(a) N 2
(b) CO
(c) MgCl 2
(f) (NH 4 )SO 4. FeSO 4 .6H 2 O
(d) CaCl 2 . 6H 2 O (e) C 6 H 5 OH
(g) CH 3 CH 2 CH 2 CH 2 CH 2 COOH
Percent Composition of Compounds
It is the percentage mass of each element in a compound.
General formula:
% composition of element X =
n x molar mass of element X
molar mass of compound
Where n = number of moles of the element in 1 mole of the compound
x 100%
5
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Example 1.
Calculate the mass % of H and O in H 2 O 2 .
Solution:
1 mol of H 2 O 2 = 34.02 g contains 2 mols of H and 2 mols of O
Molar masses (in g mol-1): H 2 O 2 = 34.02, H = 1.008, O = 16.00
%H=
%O=
2 x 1.008 H g/mol
34.02 H 2 O 2 g/mol
2 x 16.00 O g/mol
34.02 H 2 O 2 g/mol
x 100 = 5.926 %
x 100 = 94.06 %
Calculation of the percent composition of H and O in the Empirical formula of
H 2 O 2 which is HO
1 x 1.008 H g/mol
%H=
%O=
17.01 HO g/mol
1 x 16.00 O g/mol
17.01 HO g/mol
x 100 = 5.926 %
x 100 = 94.06 %
Both molecular formula and empirical formula gives the same percent
composition.
Practice Exercise
1.
Calculate the percent composition of Cl in CCl 2 F 2 .
3.
Calculate the percent composition of K and Mn in KMnO 4 .
2.
4.
Calculate the percent composition of each element in C 12 H 23 O 6 Br
Calculate the percent composition of H 2 O in CaCl 2 ⋅ 2H 2 O
6
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CALCULATING EMPIRICAL AND MOLECULAR FORMULAS FROM
REACTION DATA
(a)
Calculating Empirical Formulas
Example 1.
Calculate the empirical formula of a compound that is made up of 69.58% Ba,
6.090 % C and 24.32 % O only.
Solution
Mass % →mass of element →moles of each element → mole ratio → empirical formula
Assume 100g of the compound
Therefore: Ba = 69.58 g,
Mass (g)
Molar mass (g mol-1)
Moles (mols)
C = 6.090 g ,
O = 24.32 g
Ba
C
O
69.58
6.090
24.32
6.090
24.32
137.3
12.01
0.5068
0.5071
0.5068
0.5071
1
1
69.58
137.3
Divide by the smallest mols
0.5068
Round off to the nearest whole number
1
12.01
0.5068
1
Empirical formula = BaCO 3
16.00
16.00
1.520
1.520
0.5068
2.999
3
7
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2011
Practice Exercise 1 .
A compound containing nitrogen and oxygen is decomposed in the laboratory
and produces 24.5 g of nitrogen and 70.0 g of oxygen. Determine the empirical
formula of the compound.
Solution:
Mass (g)
Molar mass (g mol-1)
Moles (mols)
N
24.5
O
70.0
14.01
16.00
1.75
4.38
24.5
14.01
Divide by the smallest mols
1.75
1.75
1
70.0
16.00
4.38
1.75
2.5
Multiply by a factor# to convert to a whole number
# See notes below
N 1 O 2.5 x 2 = N 2 O 5
Practice Exercise 2.
A laboratory analysis of aspirin determined the following mass percent
composition: C = 60.00 %,
Find the empirical formula.
H = 4.48 % and
O = 35.53 % only.
Answer: C 9 H 8 O 4
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If subscripts are not whole numbers, multiply all the subscripts by a small
whole number to get whole number subscripts
Fractional subscript
Multiply by this number to get
Whole-number subscripts
__.10
10
__.20
5
__.33
3
__. 25
__.50
__.66
__.75
Practice Exercise 3.
4
2
3
4
A sample was found to contain 13.42 g of C, 2.25 g of H and 17.88 g of O only.
Determine the empirical formula for this compound.
Practice Exercise 4.
A sample has the following percentage compositions:
40% C, 6.71% H and 53.29% O
What is the empirical formula for this compound?
Practice Exercise 5
A 3.24 g sample of titanium reacts with oxygen to form 5.40 g of the metal
oxide. What is the formula of the oxide?
9
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2011
CALCULATING MOLECULAR FORMULAS FOR COMPOUNDS
To calculate the molecular formula, the following must be known:
(i)
(ii)
molar mass of the compound
empirical formula of the compound
Note: The formula calculated from percent composition is always the empirical formula
Molecular formula = Empirical formula × n, where n = 1,2,3……
Example:
Find the molecular formula for fructose (a sugar formed in fruit) from its empirical formula
CH 2 O and its molar mass, 180.2 g mol-1.
Solution:
Molar mass is a whole-number multiple of the empirical formula mass, the sum of the
masses of all atoms in the empirical formula.
Therefore molar mass = Empirical formula ×
and n =
n
Molar mass
Empirical formula molar mass
Empirical formula molar mass for fructose = 1(12.01) + 2(1.01) + 16 = 30.03 g mol-1
Therefore,
n=
180.2 g/mol
30.03 g/mol
= 6, hence molecular formula =
CH 2 O × 6= C 6 H12 O 6
Practice Exercise 1:
Given that the empirical formula of a compound is CH and the molar
mass is 104 g mol-1, determine the molecular formula.
10
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Solution:
Empirical formula mass = 12.01 g mol-1 + 1.01 g mol-1 = 13.01 g mol-1
Divide the molar mass of the compound by the empirical formula mass:
No of CH units =
104 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚−1
13.01 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚−1
= 8.00
Therefore molecular formula = CH × 8 = C 8 H 8
Practice Exercise 2.
Naphthalene , a compound containing carbon, and hydrogen only, is
often used in moth balls. Its empirical formula is C 5 H 4 and its molar
mass is 128.16 g mol-1. Find its molecular formula.
Solution: C 10 H 8
STOICHIOMETRY
Greek: Stoicheon = element metron = element measuring
Stoichiometry is the science of measuring the quantitative proportions or
mass ratios in which chemical elements stand to one another
Molar Ratios:
xA + yB → aA + zD
Stoichiometric relationship:
nA
x
=
nB
y
=
n𝐶𝐶
a
=
n𝐷𝐷
z
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Example 1:
Consider the following balanced equation and write the stoichiometric
relationship for all reactants and products
2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (l)
Solution:
nC2 H6 (g)
1.
2
=
nO2 (g)
7
=
nCO2 (g)
4
nH2 O(g)
=
6
Mole to Mole conversions: mol → mol
3 H 2 (g) + N 2 (g) → 2 NH 3 (g)
3 H 2 molecules ≡ 1 N 2 molecule ≡ 2 NH 3 molecules
3 mols H 2 ≡ 1 mol N 2 ≡ 2 mol NH 3
Example 1.
If 2.0 mol of N 2 (g) reacts with sufficient H 2 (g), how many mols of
NH 3 (g) will be produced?
Solution:
nN2
1
nNH
=
3
nNH3
2
⇒
2.0 mols
1
=
nNH3
= 2.0 mols x 2 = 4.0 mols
2
Example 2.
If 4.25 mols of H 2 (g) reacted with sufficient N 2 (g), calculate the number
of moles of NH 3 (g) that would be produced.
12
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Mole to mass conversions: mol → grams
Example 1:
How many grams of oxygen are produced when 1.50 mols of KClO 3 (s) are
decomposed according to the balanced equation?
∆
2 KClO 3 (s)
nKClO3 nO2
=
2
3
2 KCl(s) + 3 O 2 (g)
⇒
1.5 mols
2
Therefore nO 2 = 1.50 mols ×
nO
2
3.
=
mass of O 2
3
=
nO2
3
= 2.25 mols
2
⇒ mass of O 2 = 2.25 mols × 32.00 g mol-1
Molar mass of O 2
= 72.0 g
Mass to mole conversions
∆
2 KClO 3 (s)
Example:
2 KCl(s) + 3 O 2 (g)
If 80.0 g of O 2 (g) was produced in the above reaction, calculate the number of
moles of KClO 3 decomposed.
Solution:
Stoichiometry from balanced equation is:
nKClO3 nO2
2
=
3
⇒
nKClO3
=
nO
2
×
2
3
=
80 g
32.00 g/mol
×
2
3
= 1.67 mols of KClO 3 decomposed
13
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Practice Exercise:
Consider the following equation:
(a)
2H 2 (g) + O 2 (g) → 2H 2 O
How many grams of H 2 O are produced when 2.50 moles of
O 2 (g) is reacted?
(b)
If 3.00 moles of H 2 O is produced, calculate the mass of O 2 (g) that was
used.
4.
Mass to mass conversions
(c)
How many grams of H 2 (g) must be used, given the data in (b) above?
Example 1
How many grams of Cl 2 (g) can be liberated from the decomposition of
64.0 g of AuCl 3 in the following reaction:
2 AuCl 3 (aq) → 2 Au(s) + 3 Cl 2
Solution:
From stoichiometry
nAuCl3
2
nAuCl3
2
nCl
2
=
=
=
nAu nCl2
2
nCl2
3
nAuCl3
2
mass of Cl 2 (g)
=
3
x 3
=
64.0 g
303.32 g/mol
= 0.211 mols ×
= 22.40 g
×
3
2
= 0.211 mols
70.9 g mol-1
14
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Practice Exercises
1.
2.
Calculate the mass of AuCl 3 that can be produced from 100 g of Cl 2 in
the following reaction:
2 Au(s) + 3 Cl 2 → 2 AuCl 3 (aq)
Calculate the mass of AgCl(s) that can be produced by reacting 200.0 g
of AlCl 3 and sufficient AgNO 3 , using the following reaction:
3 AgNO 3 (aq) + AlCl 3 (aq) → 3 AgCl(s) + Al(NO 3 ) 3 (aq)
3.
4.
Using the following reaction:
2 KI(aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + 2 KNO 3 (aq)
Calculate the mass of PbI 2 (s) by reacting 30.0 g of KI with excess
Pb(NO 3 ) 2 .
How many grams of Na(s) are required to react completely with 75.0 g
of Cl 2 (g) using the following equation:
2 Na(s) + Cl 2 (g) → NaCl (unbalanced)
5.
A component of acid rain is sulfuric acid which forms when SO 2 (g), a
pollutant reacts with oxygen and rain water according to the following
reaction:
2 SO 2 (g) + O 2 (g) + H 2 O(ℓ) → H 2 SO 4 (aq)
Assuming that there is plenty of O 2 (g) and H 2 O(ℓ), how much H 2 SO 4 in
kilograms forms from 2.6 x 103 kg of SO 2 (g)?
15
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Limiting Reagents
The limiting reagent (or reactant) is the reactant that is completely consumed
in a chemical reaction.
The maximum amount of product formed depends on how much of this
(limiting) reactant was originally present.
Excess reagents are present in quantities greater than necessary to react with
the quantity of the limiting reagent.
Theoretical yield - the amount of product that can be made in a chemical
reaction based on the amount of limiting reagent.
Actual or (experimental) yield – the amount of product actually produced by a
chemical reaction.
Percentage Yield =
Actual Yield
Theoretical Yield
x 100
Consider a recipe to bake pancakes:
1 cup flour + 2 eggs + ½ tsp baking powder → 5 pancakes
Suppose we have:
3 cups flour + 10 eggs + 4 tsp baking powder → ? pancakes
We can make:
3 cups flour
→ 15 pancakes
10 eggs
→ 25 pancakes
4 tsp baking powder
→ 40 pancakes
Flour is the limiting reagent as it produces the least amount of pancakes
16
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Practice Exercise 1.
Consider the following reaction:
Ti(s) + 2Cl 2 (g) → TiCl 4 (s)
If we begin with 1.8 mol of Ti and 3.2 mol of Cl 2 , what is the limiting reagent
and calculate the theoretical yield of TiCl 4 in moles?
Solution:
Given:
Find: limiting reagent
1.8 mol Ti
3.2 mol Cl 2
Stoichiometry:
nTiCl4
1
nTiCl4
1
=
=
nCl2
2
nTi
1
=
Theoretical yield
3.2 mols
2
= 1.8 mols
= 1.6 mols
Since smaller amount of mols of TiCl 4 is produced from Cl 2 ,
Therefore Cl 2 is the limiting reagent while Ti is the excess reagent.
Therefore theoretical yield of TiCl 4 (s) = 1.6 mols
Practice Exercise 1.
Consider the following reaction:
2 Al(s) + 3Cl 2 (g) → 2 AlCl 3 (s)
If we begin with 0.552 mol of aluminium and 0.887 mol of chlorine,
what is the limiting reagent and the theoretical yield?
17
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Limiting Reagent, Theoretical Yield, and Percent Yield from Initial masses
of Reactants
Consider the following reaction:
2 Na(s) + Cl 2 (g) → 2 NaCl(s)
If we begin with 53.2 g of Na and 65.8 g of Cl 2 , what is the limiting reactant
and theoretical yield?
53.2g
nNa = 22.99g/mol
= 2.31 mols,
nNaCl nNa
2
nNaCl
2
=
=
2
nCl2
1
nCl
= 2.31 mols
⇒
2
=
65.8g
70.90g/mol
= 0.928 mols
nNaCl = 0.928 mols x 2 = 1.856 mols
Therefore the limiting reagent is Cl 2
Theoretical yield (calculated from Cl 2 )
Mass of NaCl = 1.856 mols x 58.44 g mol-1 = 108 g NaCl
Suppose when the synthesis was carried out, the actual yield of NaCl was
found to be 86.4 g. What is the percent yield?
Percentage Yield =
=
=
Actual Yield
Theoretical Yield
86.4 g
108 g
x 100
80.0 %
x 100
18
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Practice Exercise
1.
2.
Ammonia can be synthesized by the Haber Process according to the
following reaction:
3 H 2 (g) + N 2 (g) → 2 NH 3 (g)
(a)
What is the maximum amount of ammonia in grams that can be
synthesized from 25.2g of N 2 (g) and 8.42g of H 2 (g)?
(b)
What is the maximum amount of ammonia in grams that can be
synthesized from 5.22g of H 2 (g) and 31.5 of N 2 (g)?
Consider the following reaction:
Cu 2 O(s) + C(s) → 2 Cu(s) + CO 2 (g)
When 11.5 g of C are allowed to react with 114.5 g of Cu 2 O(s), 87.4 g of
Cu are obtained. Find the limiting reagent , theoretical yield and percent
yield.
19
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COMBUSTION REACTIONS
A combustion reaction is one in which the elements in a compound react with
molecular oxygen to form the oxides of those elements. For example C in a
carbon-containing compound will be converted to CO 2 and if there is
hydrogen it will be converted to H 2 O
Notice that by accurately measuring the mass of CO 2 obtained by combustion
of the carbon-containing compound, the mass of carbon in the original sample
can be calculated.
Similarly, by measuring the mass of H 2 O formed in the reaction, the mass of
hydrogen in the original sample can be calculated. These calculations assume
that all the carbon in the sample is captured in the CO 2 and that all the
hydrogen is captured in the H 2 O
EXAMPLE 1: Combustion reaction involving C,H and O only
Vitamin C is a compound that contains the elements C. H and O. Complete
combustion of a sample of mass 0.2000 g of vitamin C produced 0.2998 g of
CO 2 and 0.08185 g of H 2 O. Determine the empirical formula of vitamin C.
Molar masses (in g mol-1) :
CO 2 = 44.01
H 2 O = 18.02 ; C = 12.01 H = 1.008 O = 16.00
CxH yOz +O2 → x CO2 +
y
2
H2O
All the C is converted to CO 2 and all the H is converted to H 2 O.
20
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Solution:
Step1:
Determine the mass of C in CO 2
12.01 g mol −1
Step 2:
44.01 g mol −1
x 0.2998 g CO 2 = 0.08181 g of C
Calculate the mass of water hydrogen in water
2 x 1.008 g mol −1
18.02 g mol −1
Step 3:
(note: there are 2 mols of hydrogen in water)
Calculate the mass of O which is obtained by difference
Mass of O
Step 4:
Moles
x 0.08185 g H 2 O = 0.009157g of H
= Mass of sample – (mass of C + mass of H)
= 0.2000 g – (0.08181 g + 0.00915 g)
= 0.1090 g of O
mass
Convert to moles
(Atomic mass )
0.006812
0.009084
0.006814
0.009084
0.006814
1.33
1
C
H
÷ by smallest number of moles
0.006812
0.006812
1
0.006812
O
0.006812
convert to whole numbers by multiplying by 3
1x3
3
1.33 x 3
4
1x3
3
Therefore the empirical formula of vitamin C is C 3 H 4 O 3
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EXAMPLE 2.
2011
Combustion reaction involving C,H,O and N only
The compound caffeine contains the elements C, H, N and O. Combustion
analysis of a 1.500 g sample of caffeine produces 2.737g of CO 2 and 0.6814 g
of H 2 O. A separate further analysis of another sample. of mass 2.500 g of
caffeine produces 0.8677 g of NH 3 .
Determine the empirical formula of caffeine.
Molar masses(in g mol-1) : H = 1.008 C = 12.01 ; N = 14.01 ; O = 16.00
CO 2 = 44.01 ; H 2 O = 18.02 ; NH 3 = 17.04.
Important features of this problem are that:
1.
2.
You are analysing for 4 elements – C, H, N and O.
The analysis is performed on two samples (for example sample 1 and 2)
which have different masses.
NOTE:
In the one sample you analyse for carbon and hydrogen.
In the other you analyse for nitrogen. Remember that the oxygen is always
obtained by difference.
22
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Solution:
In sample 1 - 1.500g of caffeine
Step1:
Determine the mass of C in CO 2
12.01 g mol −1
Step 2:
44.01 g mol −1
x 2.737 g CO 2 = 0.7469 g of C
Calculate the mass of water hydrogen in water
2 x 1.008 g mol −1
18.02 g mol −1
In sample 2 - 2.500 g of caffeine
Step 3.
x 0.06814 g H 2 O = 0.07623 g of H
Determine the mass of N in NH 3
14.01 g mol −1
Step 4:
Note:
17.04 g mol −1
x 0.8677 g NH 3 = 0.7134 g of N
Calculate % N in sample 2 = % N in sample 1
The % composition of a pure compound is constant. If you know the percentage of N in
sample 1 then the percentage of nitrogen in sample 2 is exactly the same.
Therefore % N in sample 2 =
Step 5.
0.7134 g
2.500 g
x 100 = 28.54 %
You must determine the masses of all the elements in the
compound in a common sample therefore find the mass of N in
sample 1
i.e.
28.54 g
100 g
X 1.500 g = 0.4280 g mass of N
23
Chemistry for Engineers 1A (CHEM 181) – UKZN – School of Chemistry , Howard College Campus,
compiled by A.Bissessur
Step 6.
2011
The mass of O in sample 1 is obtained by difference
Mass of O = Mass of sample 1 – (mass of C + mass of H + mass of N)
= 1.500 g – (0.7469 g + 0.07623 g + 0.4280 g)
= 0.2489 g
Determine the empirical formula
C
H
N
O
Mass (g)
0.7469
0.07623
0.4280
0.2489g
moles:
0.06219
0.07563
0.03057
0.01555
Atomic mass (g mol-1)
12.01
1.008
14.01
16.00
Divide by the smallest number of mol = 0.01555
4
4.9
~5
∴ Empirical formula is
2
1
=
C4H5N2O
24
Chemistry for Engineers 1A (CHEM 181) – UKZN – School of Chemistry , Howard College Campus,
compiled by A.Bissessur
2011
PRACTICE EXERCISES
1.
An 0.1888g sample of a hydrocarbon produces 0.6260g of CO 2 and
0.1602g of H 2 O in combustion analysis. Its molecular weight is found to
be 106 amu.
For this hydrocarbon determine
(a)
its mass percent composition
(c )
its molecular formula
(b)
2.
its empirical formula
Ans. 90.47% C.
9.50% H
Ans. C 4 H 5
Ans. C 8 H 10
Dimethylhydrazine is a carbon–hydrogen-nitrogen containing
compound. Combustion analysis of a 0.312g sample of the compound
produces 0.458 g of CO 2 . From a separate 0.525 g sample the nitrogen
content is converted to 0.244 g N 2 .
(a)
(b)
3.
What is the empirical formula of dimethylhydrazine? Ans. CH 4 N
If the molecular weight was found to be 60.02 what is the
molecular formula of the compound?
Ans. C 2 H 8 N 2
A 1.35 g sample of a substance containing C. H. N and O is burned in air
to produce 0.810 g of H 2 O and 1.32 g of CO 2 . In a separate analysis of
the same substance all of the N in a sample of mass 0.735 g is converted
to 0.284g NH 3 . Determine the empirical formula of the substance.
Ans. CH 3 NO
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