수업 개요
Chapter 3
¾ 원자질량
¾ 몰
¾몰의 개념
Mass Relations in Chemistry:
Stoichiometry
(화학에서 질량관계 : 화학량론)
¾ 화학식에서 질량관계
¾ 화학식에서 원자(질량)비
¾ 반응에서의 질량관계
¾이론적 수율
3.1 Atomic and Formula Masses
Meaning of Atomic Masses
Meaning of Atomic Masses
• Relative masses of atoms of different elements are expressed in
Relative masses of different atoms and molecules
terms of their atomic masses
•Standard value is based on C-12 scale
Atomic Masses from Isotopic Composition
Atoms occur in nature as a mixture of two or more isotopes
Masses of Individual Atoms
Avogadro’s number represents the number of atoms of an
element which is equal to the atomic mass of the element.
•Most common isotope of carbon is assigned an atomic mass
of 12 amu.
Atomic Masses and Isotopic Abundance
The atomic masses of H, Cl, and Ni are
Atomic mass =
Isotope
Ne-20
Ne-21
Ne-22
Masses of Individual Atoms
H = 1.008 amu
Atomic Mass
20.00 amu
21.00 amu
22.00 amu
Percent
90.92
0.26
8.82
Cl = 35.45 amu
Ni = 58.69 amu
수소 1.008 g 중의 수소 원자의 수
= 염소 35.45 g 중의 염소 원자의 수
= 니켈 58.69g 중의 니켈 원자의 수
=NA
NA =
Masses of Individual Atoms
H 원자의 질량:
The Mole
Meaning : a collection of 6.0122 x 1023 items
Molar Mass : the mass of one mole of a substance
Mole-Mass Conversions
1g 니켈 중의 니켈원자의 개수:
Meaning
Mole-Gram Conversions
1 mol = 6.022 × 1023 items
m = MM x n
1 mol H = 6.022 × 1023 atoms; mass =
1 mol Cl = 6.022 × 1023 atoms; mass =
1 mol Cl2 = 6.022 × 1023 molecules; mass =
ex 1) Calculate mass in grams of 13.2 mol CaCl2
1 mol HCl = 6.022 × 1023 molecules; mass =
Molar Mass
ex 2) Calculate number of moles in 16.4 g C6H12O6
molar mass, MM = Σ atomic mass
3.3 Mass Relations in Chemical Formulas
Simplest Formula from Percent Composition
Percent Composition from Formula
• Simplest formula : simplest whole-number ratio of atoms
ex) Percent composition of K2CrO4?
molar mass = (78.20 + 52.00 + 64.00) g / mol
= 194.20 g / mol
i) Find mass of each element in sample of compound.
ii) Find numbers of moles of each element.
iii) Find mole ratio.
Simplest Formula from Percent Composition
ex) Simplest formula of compound containing
Simplest Formula from Percent Composition
ii) Find numbers of moles of each element.
26.6% K, 35.4% Cr, 38.0% O
i) Find mass of each element in sample of compound.
ii) Find numbers of moles of each element.
iii) Find mole ratio. ⇒
Simplest formula:
Simplest Formula from Chemical Analysis
Simplest Formula from Chemical Analysis
ex) A sample of acetic acid (C, H, O atoms) weighing 1.000 g
i) Find mass of C in sample (from CO2)
burns to give 1.446 g CO2 and 0.6001 g H2O.
Simplest formula?
Solution:
ii) Find mass of H in sample (from H2O)
i) Find mass of C in sample (from CO2)
ii) Find mass of H in sample (from H2O)
iii) Obtain mass of O by difference
iv) Find numbers of moles of each element
v) Find simplest mole ratio
iii) Obtain mass of O by difference
Simplest Formula from Chemical Analysis
Molecular Formula from Simplest Formula
iv) Find numbers of moles of each element
• Must know molar mass.
ex-cont) For acetic acid, simplest formula is CH2O
MM = 60 g/mol
∴ Molecular formula =
v) Find simplest mole ratio
Simplest formula is
3.4 Mass Relations in Reactions
Balancing : any calculation involving a reaction must be based
Writing and Balancing Chemical Equations
•
⇒
on the balanced equation for that reaction
•
The coefficients of a balanced equation represent numbers of
moles of reactants and products
•
Balancing:
Mass Relations in Reactions : the coefficients of a balanced
equation represent numbers of moles of reactants
and products
“Atoms are conserved!”
i) Adjust coefficients in front of formulas
ii) Start with an element that appears in only one species on
each side of the equation
iii) Use the simplest whole-number coefficients
Writing and Balancing Chemical Equations
ex) Combustion of propane in air
C3H8(g) + O2(g) → CO2(g) + H2O(l)
Balance C:
C3H8(g) + O2(g) → CO2(g) + H2O(l)
Balance H:
C3H8(g) + O2(g) → CO2(g) + H2O(l)
Balance O:
C3H8(g) + O2(g) → CO2(g) + H2O(l)
Mass Relations from Equations
ex 1) How many moles of CO2 are produced when 1.65 mol C3H8
burns?
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Use coefficients of balanced equation to obtain conversion fa
ctor:
ex 2) What mass of O2 is required to react with 12.0 g of C3H8?
Limiting Reactant and Theoretical Yield
Limiting Reactant : the least abundant reactant based on the
equation for a reaction
Theoretical Yield, Actual Yield, Percent Yield
Limiting Reactant and Theoretical Yield
• Ordinarily, reactants are not present in the exact ratio required for r
eaction.
• Instead, one reactant is in excess,
some of it is left when the reaction is over.
• The other,
is completely consumed to give the theoretical yield product.
Limiting Reactant, Theoretical Yield (cont.)
Limiting Reactant, Theoretical Yield (cont.)
• To calculate the theoretical yield and identify the limiting reactant:
ex) Calculate the theoretical yield of AgI and determine the limitin
g reactant starting with 1.00 g Ag and 1.00 g I2.
i) Calculate the yield expected if the first reactant is limiting
ii) Repeat this calculation for the second reactant
iii) The theoretical yield is the smaller of these two quantities
2Ag(s) + I2(s) → AgI(s)
Theoretical yield if Ag is limiting:
iv) The reactant that gives the smaller theoretical yield is the
limiting reactant
Theoretical yield if I2 is limiting:
Theoretical yield =
Experimental Yield; Percent Yield
• Experimental yield = actual yield
• Percent yield
% yield =
actual yield
theoretical yield
ex) Suppose actual yield of AgI were 1.50 g:
× 100
;
is limiting reactant