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MASS RELATIONSHIPS IN CHEMICAL
REACTIONS
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CHAPTER 3 (80-107)
3.1 Atomic mass.
3.2 Avogadro’s number and the molar mass of an element.
3.3 Molecular mass.
3.5 Percent composition of compounds.
3.7 Chemical reactions and chemical equations.
3.8 Amounts of reactants and products.
3.9 Limiting reagents.
3.10 Reaction yield.
3.1 ATOMIC MASS.
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in
atomic mass units (amu)
One atomic mass unit is defined as a mass exactly equal to one –
twelfth the mass of one carbon-12 atom.
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H
= 1.008 amu
16O
= 16.00 amu
3.1
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x 6.015 + 92.58 x 7.016
= 6.941 amu
100
3.1
Average atomic mass (6.941)
3.2 AVOGADRO’S NUMBER
AND THE
MOLAR MASS OF AN ELEMENT.
The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C isotope
1 mol = NA = 6.0221367 x 1023
Avogadro’s number (NA)
3.2
eggs
shoes in grams
Molar mass M is the mass of 1 mole of marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
3.2
One Mole of:
S
C
32.07 g
12.01 g
Hg
Cu
63.55 g
200.6 g
Fe
55.85 g
10
3.2
How many amu are there in 8.4 g?
1 g = 6.022 x 1023 amu
8.4 g = 5.1 x 1024 amu
What is the mass in grams of 13.2 amu ?
1 amu = 1.66 x 10-24 g
13.2 amu = 2.2 x 10-23 g
11
M = molar mass in g/mol
NA = Avogadro’s number
m ‫كتلة العنصر‬
m/M
nM
‫عدد موالت‬
n ‫العنصر‬
nNA
‫عدد ذرات‬
N ‫العنصر‬
N/NA
3.2
m (g)
N (atoms)
n (mol ) 

M ( g / mol ) N A (atoms)
Do You Understand Molar Mass?
How many atoms are in 0.551 g of potassium (K) ?
n ( mol ) 
m (g)
N ( atoms)

M ( g / mol )
N A ( atoms)
N  mNA / M
N (atoms) = 0.551 (g)×6.022×1023/39.10 (g/mol)
= 8.49 x 1021 atoms K
3.2
m (g)
N (atoms)
n (mol ) 

M ( g / mol ) N A (atoms)
n (mol) = m (g) / M (g/mol)
n = 6.46 / 4.003
= 1.61 mol He
m (g)
N (atoms)
n (mol ) 

M ( g / mol ) N A (atoms)
m (g) = n (mol) × M (g/mol)
m = 0.356 × 65.39
= 23.28 g Zn
WORKED EXAMPLE 3.4
m (g)
N (atoms)
n (mol ) 

M ( g / mol ) N A (atoms)
N (atoms) = m (g) × NA (atoms) / M (g/mol)
N = 16.3 × 6.022×1023 / 32.07
= 3.06×1023 S atoms
3.3 MOLECULAR MASS.
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
3.3
a- SO2 = 32.07 amu + 2(16.00 amu)
= 64.07 amu
b- C8H10N4O2
= 8(12.01 amu) + 10(1.008 amu) + 4(14.01 amu) + 2(16.00 amu)
= 194.20 amu
WORKED EXAMPLE 3.6
m (g)
N (atoms)
n (mol ) 

M ( g / mol ) N A (atoms)
n (mol) = m (g) / M (g/mol)
= 6.07 / (12.01 + 4(1.008))
= 0.378 mol CH4
WORKED EXAMPLE 3.7
m (g)
N (atoms)
n (mol ) 

M ( g / mol ) N A (atoms)
N (atoms) = m (g) × NA (atoms) / M (g/mol)
= 25.6 × 6.022×1023 × 4 / 60.06
= 1.03 ×1024 H atoms
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
N (atoms) = m (g) × NA (atoms) / M (g/mol)
= 72.5 × 6.022×1023 × 8 / 60
= 5.82 × 1024 H atoms
3.3
3.5 PERCENT COMPOSITION OF
COMPOUNDS.
Percent composition of an element in a compound =
is the percent by mass of each element in a compound
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
3.5
H3PO4 = 3(1.008) + 1(30.97) + 4(16.00)
= 97.99 g
3 x (1.008 g)
%H =
x 100% = 3.086%
97.99 g
%P =
%O =
30.97 g)
97.99 g
x 100% = 31.61%
4 x (16.00 g)
x 100% = 65.31%
97.99 g
Percent Composition and Empirical
Formulas
‫أوالا‪ :‬حساب عدد الموالت بناء على النسبة المئوية لكل عنصر‬
‫‪percent‬‬
‫‪M‬‬
‫‪40.92‬‬
‫‪12.01 g‬‬
‫= ‪nC‬‬
‫‪= 4.54 mol H‬‬
‫‪4.58‬‬
‫‪1.008 g‬‬
‫= ‪nH‬‬
‫‪= 3.406 mol C‬‬
‫‪54.50‬‬
‫‪16.00 g‬‬
‫= ‪nO‬‬
‫‪= 3.407 mol C‬‬
‫=‪n‬‬
‫ثانيا ا‪ :‬تتم قسمة نواتج الموالت المحسوبة على أصغر عدد للموالت‬
‫‪3.406 = 1‬‬
‫‪3.406‬‬
‫=‪O‬‬
‫‪4.54 = 1.33‬‬
‫=‪H‬‬
‫‪3.406‬‬
‫‪3.407 ≈ 1‬‬
‫‪3.406‬‬
‫ثالثا ا‪ :‬الضرب في عدد صحيح للحصول على أرقام صحيحه لعدد الموالت‬
‫‪1.33 × 2 = 2.66‬‬
‫‪1.33 × 3 = 3.99 ≈ 4‬‬
‫‪1.33 × 4 = 5.32‬‬
‫رابعا ا‪ :‬ضرب موالت الذرات في ‪ 3‬للحصول على الصيغة األولية‬
‫‪empirical formula‬‬
‫وتصبح كالتالي‬
‫‪C3H4O3‬‬
‫=‪C‬‬
Determine the empirical formula of a compound that has
the following percent composition by mass: K 24.75,
Mn 34.77, O 40.51 percent.
nK =
percent
M
24.75
39.10 g
nMn =
34.77
54.94 g
= 0.6329 mol Mn
nO =
40.51
16.00 g
= 2.532 mol O
n=
= 0.6330 mol K
38
3.5
‫ثانيا ا‪ :‬تتم قسمة نواتج الموالت المحسوبة على أصغر عدد للموالت‬
‫‪2.532 = 4‬‬
‫‪0.6329‬‬
‫=‪O‬‬
‫‪0.6329 = 1‬‬
‫=‪Mn‬‬
‫‪0.6329‬‬
‫‪0.6330 ≈ 1‬‬
‫‪0.6329‬‬
‫ثالثا ا‪ :‬بما أننا حصلنا على أرقام صحيحه في الخطوة الثانية فإن الصيغة‬
‫األولية ‪ empirical formula‬تصبح كالتالي‬
‫‪KMnO4‬‬
‫=‪K‬‬
3.7 CHEMICAL REACTIONS AND
CHEMICAL EQUATION.
A process in which one or more substances is changed into one
or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
reactants
products
3.7
HOW TO “READ” CHEMICAL EQUATIONS
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on
the left side and the correct formula(s) for the
product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
2. Change the numbers in front of the formulas
(coefficients) to make the number of atoms of
each element the same on both sides of the
equation. Do not change the subscripts.
2C2H6
NOT
C4H12
3.7
Balancing Chemical Equations
3. Start by balancing those elements that appear in
only one reactant and one product.
C2H6 + O2
2 carbon
on left
C2H6 + O2
6 hydrogen
on left
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
multiply CO2 by 2
2CO2 + H2O
2 hydrogen
on right
2CO2 + 3H2O
multiply H2O by 3
3.7
Balancing Chemical Equations
4. Balance those elements that appear in two or
more reactants or products.
C2H6 + O2
2 oxygen
on left
2CO2 + 3H2O
multiply O2 by 7
2
4 oxygen + 3 oxygen = 7 oxygen
(3x1)
on right
(2x2)
C2H6 + 7 O2
2
2CO2 + 3H2O
2C2H6 + 7O2
4CO2 + 6H2O
remove fraction
multiply both sides by 2
3.7
Balancing Chemical Equations
5. Check to make sure that you have the same
number of each type of atom on both sides of the
equation.
2C2H6 + 7O2
4CO2 + 6H2O
4 C (2 x 2)
4C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
Reactants
4C
12 H
14 O
Products
4C
12 H
14 O
3.7
WORKED EXAMPLE 3.12
3.8 AMOUNTS OF REACTANTS AND
PRODUCTS.
Amounts of Reactants and Products
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
4. Convert moles of sought quantity into desired units
3.8
Methanol burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
64 g of 2CH3OH
209 g
72 g of H2O
xg
x = 72×209/64
= 235 g H2O
3.8
WORKED EXAMPLE 3.13A
WORKED EXAMPLE 3.13B
n (mol) = m (g) / M (g/mol)
n (mol) = 856/180.2
= 4.75 mol
1 mol of C6H12O6
4.75 mol
6 mol of CO2
x mol
x = 6×4.75 / 1
= 28.5 mol CO2
m (g) = n (mol) × M (g/mol)
= 28.5 × 44.01
= 1.25 × 103 g CO2
13.882 g of Li
2.016 g of H2
xg
9.89 g
x = 13.882×9.89 / 2.016
= 68.1 g Li
3.9 LIMITING REAGENTS
Limiting Reagents is the
reactant used up first
In a reaction
Excess reagents are the
Reactants present in quantities
Greater than necessary to react
With the present quantity of
The limiting reagent.
2NO + 2O2
2NO2
NO is the limiting reagent
O2 is the excess reagent
3.9
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
n (mol) = m (g) / M (g/mol)
nAl = 124 / 26.98
= 4.596 mol
nFe2O3 = 601 / 160
3.756 mol
‫نسبة كل مركب‬
Al = 4.596/2 = 2.298
Fe2O3 = 3.756/1 = 3.756
limiting reagent ‫األصغر عدديا هو الكاشف المحدد‬
‫ هو الكاشف المحدد‬Al ‫وبناء على النتائج فإن‬
3.9
‫لحساب كمية ‪ Fe2O3‬نستخدم الكاشف المحدد‬
‫‪53.96 g of Al‬‬
‫‪101.96 g of Al2O3‬‬
‫‪124 g‬‬
‫‪xg‬‬
‫‪x = 124×101.96/53.96‬‬
‫‪= 234.3 g Al2O3‬‬
WORKED EXAMPLE 3.15A
3.10 REACTION YIELD
Reaction Yield
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
The percent Yield is the proportion of the actually
yield to the theoretical yield which can be obtained
from the following relation:
% Yield =
Actual Yield
x 100
Theoretical Yield
3.10
‫أوالً يتم تحديد الكاشف المحدد‬
‫)‪n (mol) = m (g) / M (g/mol‬‬
‫‪nTiCl4 = 3.54×107 / 189.7‬‬
‫‪= 1.87×105 mol‬‬
‫‪nMg = 1.13×107 / 24.31‬‬
‫‪4.65×105 mol‬‬
‫نسبة كل مركب‬
‫‪TiCl4= 1.87×105 /1 = 1.87×105‬‬
‫‪Mg = 2.32×105 /2 = 2.32×105‬‬
‫األصغر عدديا هو الكاشف المحدد ‪limiting reagent‬‬
‫وبناء على النتائج فإن ‪ TiCl4‬هو الكاشف المحدد‬
‫ نستخدم الكاشف المحدد‬Ti ‫لحساب الكمية النظرية لـ‬
189.7 g of TiCl4
47.88 g of Ti
3.54×107 g
xg
x = 47.88 × 3.54×107 / 189.7
x (theoretical yield) = = 8.95×106 g Ti
%yield = actual yield/theoretical yield × 100
6
6
= (7.91×10 g / 8.95×10 g) × 100
= 88.4%
35
37
17
1.
What information would you need to calculate the average
atomic mass of an element?
A) The number of neutrons in the element.
B) The atomic number of the element.
C) The mass and abundance of each isotope of the element.
D) The position in the periodic table of the element.
2.
The atomic masses of Cl (75.53 percent) and Cl (24.47 percent)
are 34.968 amu and 36.956 amu, respectively. Calculate the
average atomic mass of chlorine. The percentages in parentheses
denote the relative abundances.
A) 35.96 amu
B) 35.45 amu
C) 36.47 amu
D) 71.92 amu
3.
How many amu are there in 8.4 g?
A) 8.4 x1023 amu
B) 1.4 x 10-23 amu
C) 8.4 amu
D) 5.1 x 1024 amu
4.
How many atoms are there in 5.10 moles of sulfur (S)?
A) 3.07 x 1024
B) 9.59 x 1022
C) 6.02 x 1023
D) 9.82 x 1025
5.
6.
What is the mass in grams of a single atom of As?
A) 1.244 x 10-22 g
B)
2.217 x 10-26 g
C)
8.039 x 1021 g
D) 4.510 x1025 g
How many atoms are present in 3.14 g of copper (Cu)?
A) 2.98 x 1022
B) 1.92 x 1023
C) 1.89 x 1024
D) 6.02 x 1023
7.
Calculate the molar mass of Li2CO3.
A) 73.89 g
B) 66.95 g
C) 41.89 g
D) 96.02 g
8.
How many molecules of ethane (C2H6) are present in 0.334 g of
C2H6?
A) 2.01 x 1023
B)
6.69 x 1021
C)
4.96 x 1022
D) 8.89 x 1020
9.
10.
Allicin is the compound responsible for the characteristic smell
of garlic. An analysis of the compound gives the following
percent composition by mass: C: 44.4 percent; H: 6.21 percent;
S: 39.5 percent; O: 9.86 percent. What is its molecular formula
given that its molar mass is about 162 g?
A) C12H20S4O2
B) C7H14SO
C) C6H10S2O
D) C5H12S2O2
The formula for rust can be represented by Fe2O3. How many
moles of Fe are present in 24.6 g of the compound?
A) 2.13 mol
B)
0.456 mol
C)
0.154 mol
D) 0.308 mol
11.
12.
13.
14.
How many grams of sulfur (S) are needed to react completely
with 246 g of mercury (Hg) to form HgS?
A) 39.3 g
B) 24.6 g
C) 9.66 103 g
D) 201 g
Tin(II) fluoride (SnF2) is often added to toothpaste as an
ingredient to prevent tooth decay. What is the mass of F in
grams in 24.6 g of the compound?
A) 18.6 g
B) 24.3 g
C) 5.97 g
D) 75.7 g
What is the empirical formula of the compound with the
following composition?
2.1 percent H, 65.3 percent O, 32.6 percent S.
A) H2SO4
B) H2SO3
C) H2S2O3
D) HSO3
Which of the following equations is balanced?
A) 2C + O2
CO
B)
2CO + O2
2CO2
C)
H2 + Br2
HBr
D) 2K + H2O
2KOH + H2
15.
Consider the combustion of carbon monoxide (CO) in oxygen gas:
2CO(g) + O2(g) 2CO2(g)
Starting with 3.60 moles of CO, calculate the number of moles
of CO2 produced if there is enough oxygen gas to react with all
of the CO.
A) 7.20 mol
B) 44.0 mol
C) 3.60 mol
D) 1.80 mol
16.
Nitrous oxide (N2O) is also called “laughing gas.” It can be
prepared by the thermal decomposition of ammonium nitrate
(NH4NO3). The other product is H2O. The balanced equation for
this reaction is:
NH4NO3 N2O + 2H2O
How many grams of N2O are formed if 0.46 mole of NH4NO3 is
used in the reaction?
A) 2.0 g
B)
3.7 101 g
C)
2.0 101 g
D) 4.6 10-1 g
17.
The fertilizer ammonium sulfate [(NH4)2SO4] is prepared by the
reaction between ammonia (NH3) and sulfuric acid:
2NH3(g) + H2SO4(aq) (NH4)2SO4(aq)
How many kilograms of NH3 are needed to produce 1.00 105 kg
of (NH4)2SO4?
A) 1.70 104 kg
B) 3.22 103 kg
C) 2.58 104 kg
D) 7.42 104 kg
18.
19.
Nitric oxide (NO) reacts with oxygen gas to form nitrogen
dioxide (NO2), a dark-brown gas:
2NO(g) + O2(g) 2NO2(g)
In one experiment 0.886 mole of NO is mixed with 0.503 mole of
O2. Calculate the number of moles of NO2 produced (note: first
determine which is the limiting reagent).
A) 0.886 mol
B)
0.503 mol
C)
1.01 mol
D) 1.77 mol
Hydrogen fluoride is used in the manufacture of Freons (which
destroy ozone in the stratosphere) and in the production of
aluminum metal. It is prepared by the reaction
CaF2 + H2SO4 CaSO4 + 2HF
In one process 6.00 kg of CaF2 are treated with an excess of
H2SO4 and yield 2.86 kg of HF. Calculate the percent yield of
HF.
A) 93.0 %
B)
95.3 %
C)
47.6 %
D) 62.5 %
ANSWER KEY
1-C 2-B 3-D 4-A 5-A 6-A 7-A 8-B 9-C
10-D 11-A 12-C 13-A 14-B 15-C 16-C 17-C
18-A
19-A
Problems
 3.5 – 3.6 – 3.7 – 3.8
 3.14 – 3.16 – 3.18 – 3.20 – 3.22
 3.24 – 3.26 – 3.28 – 3.40 – 3.42
 3.44 – 3.46 – 3.48 – 3.50 – 3.52 – 3.60
 3.84 – 3.86
