chapter 3 notes

CHAPTER 3: Stoichiometry I: Equations, the Mole, and Chemical Formulas
3.1 Chemical Equations
C5H12 + O2 → H2O + CO2
reactants
products
Use subscripts (s), (l), (g), (aq) to indicate physical states
Balancing Chemical Equations
Al + H2SO4 → Al2(SO4)3 + H2
Be2C + H2O → Be(OH)2 + CH4
CH4 + Br2 → CBr4 + HBr
Cl2 + NaI → NaCl + I2
CO + O2 → CO2
CO2 + KOH → K2CO3 + H2O
Cu(NO3)2 → CuO + NO2 + O2
Fe2O3 + CO → Fe + CO2
Fe2O3 + CO → Fe3O4 + CO2
H2 + Br2 → HBr
H2O2 → H2O + O2
HCl + CaCO3 → CaCl2 + H2O + CO2
K + H2O → KOH + H2
KClO3 → KCl + O2
KClO3 → KClO4 + KCl
KNO3 → KNO2 + O2
KO2 + H2O + CO2 → KHCO3 + O2
Mg + O2 → MgO
N2 + H2 → NH3
(NH4)2Cr2O7 → Cr2O3 + H2O + N2
N2H4 + N2O4 → N2 + H2O
NaHCO3 → Na2CO3 + H2O + CO2
NH3 + CuO → Cu + N2 + H2O
NH3 + O2 → N2 + H2O
NH3 + O2 → NO + H2O
NH4NO2 → N2 + H2O
NH4NO3 → N2O + H2O
O3 → O2
P4O10 + H2O → H3PO4
PbS + O2 → PbO + SO2
S + HNO3 → H2SO4 + NO2 + H2O
S8 + O2 → SO2
XeF2 + H2O → Xe + HF + O2
Zn + AgCl → ZnCl2 + Ag
Types of Chemical Reactions
Combustion: reaction with oxygen
C + O2 → CO2
CH4 + O2 → CO2 + H2O
Fe + O2 → Fe2O3
C12H22O11 + O2 → CO2 + H2O
Neutralization: reaction of acid and base to yield water and salt
KOH + H3PO4 → K3PO4 + H2O
NaOH + H2SO4 → Na2SO4 + H2O
3.2 The Mole
Atomic Masses
Atomic masses originally defined as relative masses based on H. Changed to O to improve
12
accuracy, and to C (1961). Masses can be determined accurately with a mass
spectrometer.
For atoms which exist as isotopes, the atomic mass given is the weighted average of the
isotopoc masses.
12
C = 12 (by definition), 98.89%
C = 13.00335, 1.11%
To calculate weighted average, multiply each isotopic mass by its relative abundance.
C = (12.00000)(0.9889) + (13.00335)(0.0111) = 11.86680 + 0.14434=12.01114
however, relative abundance is only accurate to 4 significant figures, so the atomic mass of
C is given as 12.01
13
6
Li = 6.01512, 7.42%
Li = 7.01600, 92.58%
Li = (6.01512)(0.0742) + (7.01600)(0.9258) = 6.942
7
The Mole
A mole is a way to count atoms
1 mole = large number of atoms
since atoms are too small to count, we must count them indirectly by weighing then, so we
define the mole in terms of mass
1 mole = number of atoms in 12 g of
12
C
12
Because 12 is the atomic mass of C, and the atomic mass scale is a relative scale, any
time you have a mass in grams equal to the atomic mass of an element, you have one mole
of the element.
23
1 mole = 6.02214 x 10
23
atoms = atomic mass in grams
for Fe, 1 mole Fe = 6.022 x 10 atoms Fe = 55.85 g Fe
1 mole Fe
10.00 g Fe x 55.85 g Fe = 0.1791 mol Fe
55.85 g Fe
17.50 mol Fe x 1 mol Fe = 977.4 g Fe
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6.02214 x 10 atoms Fe
22
=1.078 x 10 atoms Fe
1.000 g Fe x
55.85 g Fe
Molecular Weight
molecular mass = sum of masses of all atoms in molecule
for C2H6, molecular mass = 2(12.01 amu) + 6(1.008 amu) = 30.07 amu
23
1 mol C2H6 = 6.022 x 10 molecules C2H6 = 30.07 g C2H6
molar mass - mass in grams of one mole of a substance
1 mol C2H6
10.00 g C2H6 x 30.07 g C H = 0.3324 mol C2H6
2 6
23
6.022 x 10 molecules C2H6
23
=
2.002
x
10
molecules C2H6
30.07 g C2H6
6 H atoms
23
24
2.002 x 10 molecules C2H6 x 1 molecule C H = 1.201 x 10 H atoms
2 6
10.00 g C2H6 x
3.3 Empirical Formulas
C2H4O2
C 2(12.01) = 24.02
H 4(1.008) = 4.032
O 2(16.00) = 32.00
60.05
24.02
%C = 60.05 x 100% = 40.00%
4.032
%H = 60.05 x 100% = 6.714%
32.00
%O = 60.05 x 100% = 53.29%
Determining the Formula of a Compound
1 mol C
93.70% C x 12.01 g C = 7.80 mol C
1 mol H
6.30% H x 1.008 g H = 6.24 mol H
reduce to a simple ratio
7.80 mol C 1.25 mol C 5 mol C
= 1 mol H = 4 mol H
6.24 mol H
empirical formula is C5H 4
empirical formula mass = 5(12.01) + 4(1.008) = 64.09
molecular mass (determined by experiment) = 128.18
molecular mass 128.18
= 64.09 = 2 so molecular formula = 2(empirical formula) = C10H8
formula mass
Combustion Train
Oxygen
carbon dioxide
water vapor out
Oxygen in
Sample
H2O absorber [Mg(ClO4)2]
CO2 absorber [NaOH]
10.000 g of a compound containing C, H, and O → 21.193 g CO2 and 3.253 g H2O
12.01 g C
21.193 g CO2 x 44.01 g CO = 5.783 g C
2
2.016 g H
3.253 g H2O x
18.016 g H2O = 0.364 g H
10.000 g compound - 5.783 g C - 0.364 g H = 3.853 g O
1 mol C
5.783 g C x 12.01 g C = 0.4815 mol C ÷ 0.361 = 1.33 mol C
1 mol H
0.364 g H x 1.008 g H = 0.361 mol H ÷ 0.361 = 1.00 mol H
1 mol O
3.853 g O x 16.00 g O = 0.2408 mol O ÷ 0.361 = 0.667 mol O
multiply by 3 to convert fractions → C4H3O2 empirical formula mass = 83.06
from experiment, molecular mass = 166.13 so molecular formula = C8H6O4
3.4 Mass Relationships in Equations
Ca3N2 + 6H2O → 3Ca(OH)2 + 2NH3
What mass of calcium hydroxide can be produced by the reaction of 100.00 g calcium
nitride with excess water?
1) convert mass of calcium nitride to moles
100.00 g Ca3N2 x \f(148.26 g Ca3N2) = 0.6745 mol Ca3N2
2) convert moles of calcium nitride to moles of calcium hydroxide
3 mol Ca(OH)2
0.6745 mol Ca3N2 x
1 mol Ca3N2 = 2.024 mol Ca(OH)2
3) convert moles of calcium hydroxide to mass of calcium hydroxide
74.10 g Ca(OH)2
2.024 mol Ca(OH)2 x
1 mol Ca(OH)2 = 149.93 g Ca(OH)2
H2O + Cl2O → 2HOCl
How many grams of Cl2O are required to make 15.000 g HOCl?
1 mol Cl2O 86.90 g Cl2O
1 mol HOCl
15.000 g HOCl x 52.46 g HOCl x 2 mol HOCl x 1 mol Cl O = 12.42 g Cl2O
2
3.5 Limiting Reactants
Ca3(PO4)2 + 8C → Ca3P2 + 8CO
What mass of Ca3P2 can be produced from 50.00 g Ca3(PO4)2 and 25.00 g C?
1 mol Ca3(PO4)2
50.00 g Ca3(PO4)2 x 310.18 g Ca (PO ) = 0.1612 mol Ca3(PO4)2
3
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1 mol C
25.00 g C x 12.01 g C = 2.082 mol C
2.082 mol C
12.9 8
=
0.1612 mol Ca3(PO4)2
1 > 1 soC is in excess and Ca3(PO4)2 will run out first
1 mol Ca3P2
182.18 g Ca3P2
0.1612 mol Ca3(PO4)2 x 1 mol Ca (PO ) x 1 mol Ca P = 29.37 g Ca3P2
3
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3 2
Zn + 2AgNO3 → 2Ag + Zn(NO3)2
What mass of Ag can be fromed from 3.22 g Zn and 4.35 g AgNO3?
1 mol Zn
3.22 g Zn x 65.39 g Zn = 0.0492 mol Zn
1 mol AgNO3
4.35 g AgNO3 x
169.9 g AgNO3 = 0.0256 mol AgNO3
Zn is in excess
2 mol Ag
107.9 g Ag
0.0256 mol AgNO3 x 2 mol AgNO x 1 mol Ag = 2.76 g Ag
3
4LiH + AlCl3 → LiAlH4 + 3LiCl
What mass of LiAlH4 can be formed from 40.00 g LiH and 100.00 g AlCl3?
1 mol LiH
40.00 g LiH x 7.949 g LiH = 5.032 mol LiH
1 mol AlCl3
100.00 g AlCl3 x
133.33 g AlCl3 = 0.7500 mol AlCl3
6.7
5.032 mol LiH
= 1 so LiH is in excess
0.7500 mol AlCl3
1 mol LiAlH4 37.95 g LiAlH4
0.7500 mol AlCl3 x 1 mol AlCl x 1 mol LiAlH = 28.46 g LiAlH4
3
4
Percent Yield
actual yield
percent yield = theoretical yield x 100%
example: if 24.04 g LiAlH4 is actually obtained,
24.04 g
% yield = 28.46 g x 100% = 84.5%

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