atomic mass
DESCRIPTION
Atomic Mass. Atoms are so small, it is difficult to discuss how much they weigh in grams. Use atomic mass units. an atomic mass unit (amu) is one twelth the mass of a carbon-12 atom. The decimal numbers on the table are atomic masses in amu. They are not whole numbers . - PowerPoint PPT PresentationTRANSCRIPT
Atomic Mass Atoms are so small, it is difficult to
discuss how much they weigh in grams.
Use atomic mass units. an atomic mass unit (amu) is one
twelth the mass of a carbon-12 atom.
The decimal numbers on the table are atomic masses in amu.
kgmuamu C27
12 1066053892.112111
numberwhole
unpm
ukgm
ukgm
um
kgmuamu
C
n
p
C
C
12
27
27
12
2712
1067492735.1
1067262178.1
12
1066053892.112111
They are not whole numbers
Because they are based on averages of atoms and of isotopes.
mass
10000 Cu atom
6909 Cu-63
6909x62.93=434783.3
7 u
3091Cu-65
3091x64.93=200698.6
3 u
Total mass= 635482 uAverage mass= total mass/10000=63.55 u
93.64100
91.3193.62100
09.6955.63
......
...
...
3
2
1
isotope
isotope
isotope
MassAtAbunNat
MassAtAbunNat
MassAtAbunNatMassAtomicaverage
The Mole The mole is a number. A very large number, but still, just
a number. 6.022 x 1023 of anything is a mole A large dozen. The number of atoms in exactly 12
grams of carbon-12.
The Mole 1 mol of atoms of any element
weighs the number value of atomic weight in g.
1 mol Cu weighs 63.55 g 1 mol Ga weighs 69.723 g
Molar mass, M Mass of 1 mole of a substance. Often called molecular weight. To determine the molar mass of an
element, look on the table. To determine the molar mass of a
compound, add up the molar masses of the elements that make it up.
Find the molar mass of CH4 : CH4 → C + 4 H
mCH4= mC+ mH
MCH4= MC + 4×MH
Mg3P2 : M = 3×MMg + 2×MP
Ca(NO3)3 Al2(Cr2O7)3 CaSO4 · 2H2O
avNNn
Mmn
Number of moles, n
Avogadro’s number Nav=6.02x1023 mol-1
gmol
molgNNMm
Mm
NNn
avav
21123
1 1042.21002.6
6.243
Number of moles, number of atoms in 10 g Al?
2323
1
1023.21002.6371.0.
371.0.98.26
10
molmolNnN
molmolgg
Mmn
av
Number of moles, mass of 5.00x1020 atom Co?
gmolgmolMnm
molmolN
Nnav
414
4123
20
1089.4.93.581030.8.
1030.81002.6
1000.5
Number of molecules of isopentyl acetate in 1 ug?
Number of carbon atoms?
atomcarbonN
moleculeperatomscarbonofnomoleculesofnoN
moleculemolmolNnN
OHCmolmolgg
Mmn
molgMOHC
C
C
av
1615
15239
21479
1
6
2147
105.37105
..
1051002.6108.
108.18.130
101
/18.13000.162008.11401.127
Mass of CO32- in 4.86 mol CaCO3?
gmolgmolm
MnmMmn
molmol
COCaCO
CO
COCOCO
29216312186.4
86.486.4
23
23
23
23
233
Percent Composition Percent of each element a compound is
composed of.
What is the percent composition of C2H5OH?
%100% sample
X
mmX
%100%
%100%
%100%
52
52
52
OHHC
O
OHHC
H
OHHC
C
mmO
mmH
mmC
Suppose we have 1 mol C2H5OH
OHCOHHC MMMM
molmolmolmol
OHCOHHC
162
1621
62
52
52
%73.34%10007.46
00.161%100%100%
%13.13%10007.46008.16%1006%100%
%14.52%10007.46
01.122%1002%100%
5252
5252
5252
OHHC
O
OHHC
O
OHHC
H
OHHC
H
OHHC
C
OHHC
C
MM
mmO
MM
mmH
MM
mmC
• Sum of all percentages=100%• 100 g of C2H5OH contains 52.14 g C, 13.13 g H and
34.73 g O.
Find the percent compositions of the following compounds:
Al2(Cr2O7)3 CaSO4 · 2H2O
%100% 22
sample
OH
mm
OH
From percent composition, you can determine the empirical formula.
Empirical Formula the lowest ratio of atoms in a molecule.
A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O, what is its empirical formula?
C H N Osample mass=100 g 59.53 g 5.38 g 10.68 g 24.4 g
n=m/M 59.53/12.01
5.38/1.008
10.68/14.00
24.4/16.00
n 4.96 5.34 0.76 1.53Divide by smallest
value6.53 7.03 1 2.01E.F.
C13H14N2O4
Empirical Formula and Molecular Formula
H2O2 EF: HO Such compound doesn’t exist! MF = n × EF H2O2 = 2 (HO) MMF=n × MEF 34 = 2 × 17
2×MH + 2×MO = 2 (MH+MO) Alkenes: C2H4 , C3H8 , C4H8 , C5H10 … EF : CH2
Cl C Hsample mass=100 g 71.65 g 24.27 g 4.07 g
n=m/M 71.65/35.45
24.27/12.01
4.07/1.008
n 2.021 2.021 4.04Divide by smallest
value1 1 2
E.F. CH2Cl
2
)48.49(96.98
)45.351008.1201.121(96.98
n
n
n
MnM EFMF
M.F. C2H4Cl2
Elemental Analysis
A substance is known to be composed of C, H and nitrogen. When 0.1156 g of this substance is reacted with oxygen, 0.1638 g CO2 and 0.1676 g H2O. Find the empirical formula.
Assumptions:1) All C in sample converted into
CO2.
ggmm
mmC
MMC
mm
COC
CO
CCO
CO
CCO
sampleCCOC
04470.02729.01638.0%100%29.27
%29.27%100%
%29.27%10001.4401.12%100%
?
2
2
2
2
2
2
22
22
2%100%100%
CO
C
CO
C
CO
C
CO
CCO
mm
MM
mm
MMC
2) All H in sample converted into H2O.
ggmm
mmH
MMH
mm
OHH
OH
HOH
OH
HOH
sampleHOHH
01875.01119.01676.0%100%19.11
%19.11%100%
%19.11%100016.18008.12%1002%
?
2
2
2
2
2
2
OH
H
OH
H
OH
H
OH
HOH
mm
MM
mm
MMH
22
22
2
2
%100%1002%
gm
m
mmmm
N
N
NHCsample
0521.0
01875.004470.01156.0
C H N
mass 0.04470 g 0.01875 g 0.0521 g
n=m/M 0.04470/12.01
0.01875/1.008
0.0521/14.00
n 0.003722 0.01860 0.003721
Divide by smallest value 1 5 1
E.F. CH5N
Chemical Equations Are sentences that describe what
happens in a chemical reaction. Reactants Products
CH4 + O2 CO2 + H2O Redistribution of bonds, atoms don’t
change! Equations should be balanced: Have the
same number of each kind of atoms on both sides.
Law of conservation of mass. CH4 + 2O2 CO2 + 2 H2O Balancing by inspection.
Abbreviations (s) ¯ (g) l (aq) heat D catalyst
Practice Ca(OH)2 + H3PO4 H2O +
Ca3(PO4)2 Cr + S8 Cr2S3 KClO3(s) Cl2(g) + O2(g) Solid iron(III) sulfide reacts with
gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas.
Fe2O3(s) + Al(s) Fe(s) + Al2O3(s)
Stoichiometric CalculationsWhat mass of oxygen will react completely
with 96.1 g of propane?C3H8 + O2 CO2 + H2O
Balance the equation!C3H8 + 5 O2 3 CO2 + 4 H2O
Language of moles, not grams!
molmolgg
Mm
nHC
HCHC 18.2
/1.441.96
83
83
83
C3H8 + 5 O2 3 CO2 + 4 H2O
1 52.18 ?
molnO 90.101
518.22
2349/3290.10
.222
2
2
2
Ogmolgmol
Mnm
Mm
n
OOO
O
OO
Grams of CO2 produced?C3H8 + 5 O2 3 CO2 + 4 H2O
1 3
2.18 ?
molnCO 54.61
318.22
2288/01.4454.6
.222
2
2
2
COgmolgmol
Mnm
Mm
n
COCOCO
CO
COCO
Mass of CO2 absorbed by 1.00 kg LiOH(s)?
LiOHmolmolgg
Mmn
LiOH
LiOHLiOH 8.41
/95.231000
2 1
41.8 ?
molnCO 9.202
18.412
2920/01.449.20
.222
COgmolgmol
Mnm COCOCO
Antiacids
molmolgg
Mm
n
molmolgg
Mm
n
OHMg
OHMgOHMg
NaHCO
NaHCONaHCO
0171.0/32.58
00.1
0119.0/01.84
00.1
2
2
2
3
3
3
)(
)()(
1 1
0.0119 ?molnHCl 0119.0
1 2
0.0172 ?molnHCl 0344.0
120172.0
Examples One way of producing O2(g)
involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O2(g) are produced?
How many grams of potassium chloride?
How many grams of oxygen?
Examples A piece of aluminum foil 5.11 in x
3.23 in x 0.0381 in is dissolved in excess HCl(aq). How many grams of H2(g) are produced?
How many grams of each reactant are needed to produce 15 grams of iron form the following reaction?
Fe2O3(s) + Al(s) Fe(s) + Al2O3(s)
Examples K2PtCl4(aq) + NH3(aq)
Pt(NH3)2Cl2 (s)+ KCl(aq)
what mass of Pt(NH3)2Cl2 can be produced from 65 g of K2PtCl4 ?
How much KCl will be produced? How much from 65 grams of NH3?
Limiting Reactant
31:Re toratiotricstoichiomeinareagents
mixturetricstoichiome
1 25 ?=10 mol NH3
3 29 ?=6 mol NH3
15 (not available)
H2 is LR.
3 mol nitrogen12 mol hydrogen
1 23 ?=6 mol NH3
3 212 ?=8 mol NH3
14 (not available)
N2 is LR.
Limiting Reactant Reactant that determines the
amount of product formed. The one you run out of first. Makes the least product.
N2 + H2 NH3
What mass of ammonia can be produced from a mixture of 100. g N2 and 500. g H2 ?
How much unreacted material remains?
molmolg
gMm
n
molmolg
gMm
n
H
HH
N
NN
01.248/016.2
500
57.3/00.28
100
2
2
2
2
2
2
1 23.57 ?=7.14 mol NH3
3 2248 ?=165.3 mol NH3
356.121/024.1714.7
.333
NHgmolgmol
Mnm NHNHNH
Reaction Yield The theoretical yield is the amount
you would make if everything went perfect.
The actual yield is what you make in the lab.
Percent Yield%100%
yieldltheoreticayieldactualyield
68.5 kg CO is reacted with 8.60 kg H2. Calculate the percent yield if the actual yield was 3.57x104
g CH3OH. %100% yieldltheoretica
yieldactualyield
)(3)(2)( 2 lgg OHCHHCO
molmolgg
Mm
n
molmolgg
Mmn
H
HH
CO
COCO
9.4265/016.2
106.8
6.2445/01.28
105.68
3
3
2
2
2
)(3)(2)( 2 lgg OHCHHCO
1 12445.6 ?=2445.6 mol CH3OH
2 14265.9 ?=2132.9 mol CH3OH
OHCHgmolgmol
Mnm OHCHOHCHOHCH
36.68343/042.329.2132
.333
%2.52%1006.683431057.3%100%
4
gyieldltheoretica
yieldactualyield
Examples Aluminum burns in bromine
producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield.
2 Al + 3 Br2 2 AlBr3
Examples Years of experience have proven
that the percent yield for the following reaction is 74.3%
Hg + Br2 HgBr2
If 10.0 g of Hg and 9.00 g of Br2 are reacted, how much HgBr2 will be produced?
Examples Commercial brass is an alloy of Cu
and Zn. It reacts with HCl by the following reaction Zn(s) + 2HCl(aq) ZnCl2 (aq) +
H2(g)
Cu does not react. When 0.5065 g of brass is reacted with excess HCl, 0.0985 g of ZnCl2 are eventually isolated. What is the composition of the brass (i.e. Zn% and Cu%)?