Download BMC seminar

BMC seminar
Week 2
2012
Chemical formulas,
Naming chemical compounds,
Chemical equations,
Avogadro’s number and the mole,
Atomic, molecular and molar mass,
Yields of chemical reactions,
Empirical and molecular formulas
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C=gray, H=ivory, N=blue, O=red
CxHyNzON
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Naming Binary Ionic Compounds
• Identify the positive ion
• Identify the negative ion
• + -ide
Examples:
LiF
Lithium-fluoride
CaBr2
Calcium-bromide
AlCl3
Aluminium-chloride
In any neutral compound, the total number of + charges must equal
the total number of - charges.
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Naming Binary Ionic Compounds
• Some metals form more than one kind of cation.
– Distinguished by using Roman numeral in parantheses
• Fe2+[Iron(II) ion], Fe3+[Iron(III) ion],
• Sn2+ [Tin(II) ion], Sn4+ [Tin(IV) ion]
• Pb2+ [Lead(II) ion], Pb4+ [Lead(IV) ion]
• Cr2+ [Chromium(II) ion], Cr3+ [Chromium(III) ion]
Examples:
CrCl3
Chromium(III) chloride
PbS
Lead(II) sulfide
Fe2O3
Iron(III) oxide
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Naming Binary Molecular Compounds
•
•
•
Identify the more cationlike element (with numerical prefix)
Identify the more anionlike element (with numerical prefix)
+ -ide
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Naming Compounds with Polyatomic Ions
• Identify the positive ion
• Identify the negative ion
• + different endings
– most polyatomic anions end in – ite or – ate
– Hidroxide (OH-), cyanide (CN-) and peroxide (O22-) have – ide
ending
– Oxoanions:
hypo … ite
… ite
more
… ate
oxygen
per … ate
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Naming Acids
•
Most of the acids are oxoacids
– Names are related to the nemes of oxoanions
• … ite ion Æ … ous acid
• … ate ion Æ … ic acid
•
Some common acids don’t contain oxygen
– Hydro … ic acid (HCl - hydrochloric acid)
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12
(amu)
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Numerically, molecular mass (or more generally formula mass)
equals the sum of the atomic masses of all atoms in the molecule.
MOLECULAR MASS [amu]
Sum of atomic masses of all atoms in a molecule.
FORMULA MASS [amu]
Sum of atomic masses of all atoms in a formula
unit of any compound, molecular or ionic.
amu: atomic mass unit
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n=
m
mass of potassium ( K )
1g
=
=
= 0.026 mol
M molar mass of potassium 39.1 g / mol
n=
m
mass of gold ( Au )
1g
=
=
= 0.005 mol
M molar mass of gold 196.97 g / mol
m = n ⋅ M = number of moles of glu cos e ⋅ molar mass of glu cos e
m = 0.25 mol ⋅180.1 g / mol = 45,025 g
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1 g = 1000 mg ⇒ 125 mg = 0,125 g
m
0,125 g
n=
=
= 6.44 ⋅10 − 4 mol
M 194,2 g / mol
6.02 ⋅10 23 molecules
6.44 ⋅10 mol ⋅
= 3.87 ⋅10 20 molecules
1 mol
−4
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23
1
6
2
2
2
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n = m /Methylene = 4,2 / 28 = 0,15 mol
1 mol
0,15
1 mol
0,15
1 mol
0,15
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0,624 mol
0.624
11.2
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28
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C
40% Æ 0,4 x 90,08g = 36,03 g
MC= 12 g/mol
36,03g / 12g => 3,002 mol ~ 3 mol
H
Molecular
formula is
6,71% Æ 0,0671 x 90,08g = 6,044 g MH = 1 g/mol
6,044g / 1g => 6,044 mol ~ 6 mol
O
53,3 % Æ 0,533 x 90,08g = 48,01 g MO = 16 g/mol
48,01g/ 16g => 3,001 mol ~ 3 mol
Divide by the smallest number (3) Æ
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