Transcript chapter 5
Chapter 4
The Mole and
Stoichiometry
Chemistry: The Molecular Nature
of Matter, 6E
Brady/Jespersen/Hyslop
β’ The molecular scale versus the
laboratory scale:
β’ Defining the mole:
β’ A number equal to the number of atoms
in exactly 12 gram of πππͺ atoms.
β’ 1 mole of element X= gram atomic mass of X
β’ Example:
β’ 1 mole of sulfur = 32.6 g
atomic mass of sulfur = 32.6 u
β’ The mole concept applied to compound:
β’ 1 mole of molecules X = gram molecular
mass of X
β’ Example:
β’ The molecular mass of water = 18.02 u
β’ The sum of atomic mass of two H atoms
and one O atmo.
β’ 1 mole of π―π O= 18.02 g
β’ 1 mole of ionic compound X = gram
formula mass of X
β’ Example:
β’ 1 mole of π¨π³π πΆπ =101.96 g
β’ π¨π³π πΆπ has 2AL with an atomic masses
26.98u and three oxygen with mass of
16 u.
β’ 1 mole of X = gram molar mass of X
β’ Converting from gram to moles:
β’ Example:
β’ In experiment to prepare of titanium(IV)
oxide we start with 23.5g sample of
titanium. How many moles of titanium do
we have?
β’ Solution:
β’ 1 mole of element X = gram atomic mass of
X
β’ 1 mole of Ti = 47.867g Ti
β’ 23 .5g Ti = ? Mole Ti
β’ Conversion factor:
π ππππ π»π
ππ. πππ ππ»π
β’ 23.5 gTi ×
π ππππ π»π
ππ,πππ ππ»π
= 0.491 mole Ti
β’ Conversion from mole to grams:
β’ Example:
β’ We need 0.254 moles of πππͺπ³π for
certain experiment. How many grams
would you need weight?
β’ Solution:
β’ 1 mole of πππͺπ³π = 162.204 g
β’ Molar mass πππͺπ³π = 55.845 g/mole +
( 3× 35.453) g/mole = 162.204 g/mole
β’ 0.254 mole πππͺπ³π = ? g πππͺπ³π
β’ Conversion factor:
πππ.πππ ππππͺπ³π
π ππππ πππͺπ³π
0.254 ×
πππ.πππ ππππͺπ³π
π ππππ πππͺπ³π
= 41.2 g πππͺπ³π
Learning check
* How many moles of aluminum are
there iN 3.47 gram sheet of aluminum foil.
- Atomic mass of AL=26.98u
β’ Avogadros number:
β’ The relationship between the atomic scale
and laboratory scale as
β’ 1 mole of X = π. πππ × ππππ unit of X
β’ The unit can be atoms , molecules , formula
unit.
β’ Example :
β’ 1 mole of Xe = π. πππ × ππππ atoms of Xe
β’ 1 mole of π΅πΆπ = π. πππ × ππππ molecules
of π΅πΆπ
β’ Converting from the laboratory scale to
the atomic scale.
β’ Example:
β’ In lightbulb the tungsten weight 0.653 g.
how many atoms of tungsten are there
in such sample.
β’ Solution:
β’ 0.632 g W = ? Atoms of W
β’ gram W to mole W to atom W
β’ 1 mole of W = 183.84 g W
β’ 1 mole W = π. πππ × ππππ atoms W
β’ Conversion factor :
β’ 1.
π ππππ πΎ
πππ.ππ π πΎ
β’ 2.
π.πππ×ππππ πππππ πΎ
π ππππ πΎ
β’ 0.635 ×
π ππππ πΎ
πππ.ππ π πΎ
×
π.πππ×ππππ πππππ πΎ
π ππππ πΎ
ππ
2.08 × ππ
=
atoms W
β’ Calculating the mass of molecules:
β’ Example :
β’ What is the average mass of one molecule of
carbon tetrachloride.
β’ Soluation:
β’ 1 molecule πͺπͺπ³π = ? g πͺπͺπ³π
β’ 1 mole of πͺπͺπ³π = π. πππ × ππππ molecules of
πͺπͺπ³π
β’ 1 mole of πͺπͺπ³π = 153.823 g πͺπͺπ³π
β’ Molar mass of πͺπͺπ³π = 12 + (4×
35.45)=153.823 g\mole
β’ Conversion factor :
β’ 1.
π ππππ πͺπͺπ³π
π.πππ×ππππ ππππππππ πͺπͺπ³π
β’ 2.
πππ.πππ ππͺπͺπ³π
π ππππ πͺπͺπ³π
β’ 1 molecule πͺπͺπ³π ×
π ππππ πͺπͺπ³π
πππ.πππ ππͺπͺπ³π
×
ππ
π.πππ×ππ ππππππππ πͺπͺπ³π
π ππππ πͺπͺπ³π
π. πππ × ππβππ g πͺπͺπ³π .
=
Learning check
β’ A sample of ethanol contains 1.6 mol C2H5OH,
Calculate the mass of C2H5OH molecules in
this sample. ( molar mass is , C = 12 , H=1 ,
O = 16 g/mol ).
β’ Calculate the mole number of 15.055 X 1023
CO molecules. ( Avogadro's number = 6.022 x
1023).
β’ Percentage composition:β’ Called percentage by mass of element:
is the number of grams of the element
present in 100g of the compound.
β’ Percentage by mass of element =
ππππ ππ πππππππ
ππππ ππ πππππ ππππππ
× πππ%
β’ Example:
β’ A sample of liquid with a mass of 8.657 g
was decomposed into its element and gave
5.217g of carbon, 0.9620 g of hydrogen, and
2.478 g of oxygen. What is the percentage
composition of this compound?
β’ solution:
β’ For C :
β’ For H :
π.πππ π
× πππ% = ππ. ππ%
π.πππ π
π.πππ π
× πππ% = ππ. ππ %
π.πππ π
C
β’ For O :
π.πππ π
π.πππ π
× πππ % = ππ. ππ %
β’ Sum of percentage: 99.99%
Learning check
* An organic compound weighing 0.6672 g
is decomposed , giving 0.3481 g of carbon
0,087 g hydrogen. What is the percentage
of hydrogen and carbon in this
compound?
Lreaning check
β’ Calculate the percentage composition of each
element in KNO3 (Atomic masses ,K = 39 ,
N = 14 , O = 16 g/mol).
β’ Determining empirical and molecular
formulas:
β’ Calculating an empirical formula from
mass data:
β’ Example:
β’ 2.57 g sample of compound composed on only tin
and chlorine was found to contain 1.17 g of tin.
What the compound empirical formula?
β’
β’ Solution:
β’ Mass of CL = 2.57 g compound β 1.17 g Sn =
1.40 g CL
β’ 1 mol Sn = 118.7 g Sn
β’ 1 mol CL = 35.45 g CL
β’ C.V :
β’ 1.
1 πππ ππ
118.7 π ππ
β’ 2.
1 πππ πΆπΏ
35.45 π πΆπΏ
β’ 1.17g ×
1 πππ ππ
118.7 π ππ
β’ 1.40 g ×
= 0.00986 mol Sn
1 πππ πΆπΏ
35.45 π πΆπΏ
= 0.0395 mol CL
β’ Formula: ππ0.00986 πΆπΏ0.0395
β’ To convert the decimal subscripts to integer by
dividing each by the smallest number in the
set .
β’ ππ 0.00986 πΆπΏ 0.0395 = ππ1.00 πΆπΏ4.01
0.00986
0.00986
β’ The empirical formula is SnπΆπ4
Learning check
A 1.525g sample of a compound between
nitrogen and oxygen contain 0.712g of
nitrogen. Calculate its empirical formula.
Determining a molecular formula from an
empirical formula and a molecular mass:πππππππππ ππππ ππ ππππππππ
πππππππππ πππππππ ππππ ππ ππππππππ
= integer
Example:Styrene has an empirical formula of CH it molecular
mass is 104. what is its molecular formula?
Solution:
The formula mass is
12.01 + 1.008 = 13.02
πππ
ππ.ππ
= 7.99 = 8
Molecular formula of styrene is πͺπ π―π
learning check
The empirical formula of hydrazine is
Nπ―π and its molecular mass is 32.0
what is its molecular formula?
β’ The mole and chemical reaction:β’ Writing and balancing equations:
β’ Always the balancing of an equation as a
two-step process:
β’ Step 1: write the unbalanced equation.
β’ Step 2: adjust the coefficient to get equal
number of each kind of atoms on both side.
β’ Some guideline for balancing equation:
β’ 1. start balancing with the most complicated
formula first . Element, particularly
π―π πππ
πΆπ .
β’ 2. balance atoms that appear in only two
formula.
β’ 3. balance as a group those polyatomic ions
that appear un changed on both side of the
arrow.
β’ Example:β’ Sodium hydroxide and phosphoric acid
π―π π·πΆπ React to give sodium phosphate and
water. The sodium posphate remain in
solution. Write the balanced equation for this
reaction?
β’ Solution:
β’ NaOH + π―π π·πΆπ
π΅ππ π·πΆπ + π―π O
(unblanced)
β’
β’
β’
β’
1. balance element ( Na and P)
3 NaOH + π―π π·πΆπ
π΅ππ π·πΆπ + π―π O
2. balance particular ( π―π β πΆπ )
3 NaOH + π―π π·πΆπ
π΅ππ π·πΆπ + π π―π O
LEARNING CHECK
Balance the following equation:
1. Caπͺπ³π + π²π π·πΆπ
πͺππ (π·πΆπ )2 + KCL
2. π²π πΊπΆπ + NaOH
π΅ππ πΊπΆπ + KOH
β’ Example :
β’ How many moles of sodium phosphate can be
made from 0.24 mol of sodium hydroxide by the
following equation:
β’ 3NaOH (aq) + H3PO4(aq) β Na3PO4(aq) + 3 H2O
sol:
3 mol of NaOH β 1mol of Na3PO4
0.24 mol NaOH β ? mol of Na3PO4
C.V :
1mol of Na3PO4
3 mol of NaOπ»
β’ 0.24 mol NaOH × 1mol of Na3PO4 =
3 mol of NaOπ»
0.08 mol Na3PO4
Learning chick
β’ How many moles of O2 are needed to
produced 6.76 moles of SO3 from the
following equation:
2SO2 + O2 β 2SO3