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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter4.4c
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à 4.4cèEmpirical (Simplest) å Molecular Formulas
äèPlease fïd ê empirical formulas ç ê followïg compounds.
âèWhat is ê empirical formula ç a compound that is 46.5% Li å
53.5% O by mass?
In 100 g ç ê compound, êre are 46.5/6.941 = 6.70 mol Li å
53.5/16.00 = 3.34 mol O.èThe empirical formula gives ê relative
numbers ç aëms or moles ï ê compound usually as a whole number
ratio.èThe ratio ç Li ë O is 6.70:3.34 = 2.01:1.00.èThere are two
lithium aëms for each oxygen aëm.èThe formula is Li╖O.è
éSèThe empirical formula, also called ê simplest formula, is ê
simplest whole number ratio ç ê aëms ç each element ï one formula
unit ç ê compound.èIf you have one mole ç a compound, ên ê num-
ber ç aëms ç an element ï one formula unit will be ê same number ç
moles ç that element ï ê compound.èIn oêr words, ê mole ratio ç
ê elements ï ê compound is ê same as ê aëm ratio ï ê com-
pound.
In ê previous Section 4.3, we used ê aëm ratios ë fïd ê percent-
age composition.èNow we are essentially reversïg ê process ë fïd
ê empirical formula ç ê compound.èThe percentage composition gives
us ê mass ç each element ï 100 grams ç ê compound.èFrom this
mass, we can fïd ê number ç moles ç each element ï 100 grams ç ê
compound.èThe empirical formula only provides ê relative number ç
aëms ï a formula unit.èConsequently, we reduce ê ratios ç ê num-
bers ç moles ë simple whole number ratios.èThis procedure is illus-
trated by ê followïg example.
Fïd ê simplest formula ç a compound that is 41.10% K, 16.85% S, å
42.05% O.
1.èWe start by assumïg that we have 100 grams ç compound so that ê
percentage equals ê number ç grams ç each element.
2.èWe calculate ê number ç moles ç each element usïg ê aëmic
mass ç ê element.
èè? mol K = 41.10 g K x 1 mol K/39.10 g K = 1.051 mol K
èè? mol S = 16.85 g S x 1 mol S/32.06 g S = 0.5256 mol S
èè? mol O = 42.05 g O x 1 mol O/16.00 g O = 2.628 mol O
One mistake that some students make at this poït is ë round-çf ê
results ç êir calculations ëo soon.èWe are justified ï four signif-
icant figures so contïue ë carry four significant figures.
3.èNow we need ë get êse mole numbers as whole numbers.èThese mole
numbers are only relative numbers because êy happen ë be based on 100
grams.èWe could have started with 25 grams å all ê numbers would be
four times smaller.èWe want ê relative ratio ç ê aëms.èTake ê
smallest number ç moles å divide ê oêrs by ê smallest.
1.051/0.5256 = 2.000; 2.628/0.5256 = 5.000èThis reveals that êre are
twice as many K aëms as S aëms å five times as many O aëms as S
aëms.èThe relationship is
mol K : mol S : mol Oè=è1.051 : 0.5256 : 2.628è=è2.000 : 1 : 5.000
The formula ç ê compound is K╖SO║ (potassium persulfate).
We obtaïed this formula relatively easily.èSometimes ê first set ç
ratios are not all whole numbers.èIf whole numbers are not obtaïed, we
multiply by ïtegers until we get a set whole numbers.
Let's look at anoêr example.èWhat is ê empirical formula ç a com-
pound that is 68.42% Cr å 31.58% O?è(The aëmic masses are: Cr = 52.00
å O = 16.00)
1. Assumïg a 100 g sample, fïd number ç moles ç each element.
è ? mol Cr = 68.42 g Cr x 1 mol Cr/52.00 g Cr = 1.316 mol Cr
è ? mol O = 31.58 g O x 1 mol O/16.00 g O =è1.974 mol O
2. Fïd mole ratio.èDo not round-çf ëo soon.
è 1.974/1.316 = 1.500 mol O/mol Cr.èThe relative relationship is:
è mol Cr : mol O = 1.316 : 1.974 = 1 : 1.500
3. Multiply by ïtegers ë obtaï whole numbers.èHere we recognize .5
è as ê fraction 1/2.èIf we multiply through by two, we will get whole
è numbers: 2 x 1.5 = 3.èmol Cr : mol O = 1:1.5 = 2:3.èThe formula ç
è ê compound is Cr╖O╕.
Ionic compounds do not possess distïguishable "molecules".èThe empiri-
cal formula is ê only formula that we write for an ionic compound.èOn
ê oêr hå, molecular substances do exist as molecules; å we can
write an empirical å a molecular formula.èThe molecular formula for
benzene is C╗H╗.èIts empirical formula just gives ê relative number ç
aëms å is just CH.
1èFïd ê empirical formula ç a compound that is 34.43% Fe å
èèèèè 65.57 % Cl.è(Aëmic Masses: Fe = 55.85; Cl = 35.45)
A) Fe╕Cl B) Fe╖Cl
C) FeCl╕ D) FeCl╖
ü The empirical formula is ê simplest whole number ratio ç ê
iron å chlorïe aëms ï ê compound.èIn 100 grams, we fïd
è ? mol Fe = 34.43 g Fe x 1 mol Fe/55.85 g Fe = 0.6165 mol Fe
è ? mol Cl = 65.57 g Cl x 1 mol Cl/35.45 g Cl = 1.850 mol Cl
Take ê larger number ç moles ç Cl å divide by ê smaller number
ç moles ç Fe.èThis gives ê followïg result:
mol Fe : mol Cl = .6165 : 1.850 = 1 : 3.001
The aëm ratio is 3 chlorïe aëms for each iron aëm.
The formula is FeCl╕.
Ç C
2èFïd ê empirical formula ç a compound that is 63.5% C,
12.2 % H, å 24.2% O.è(Aëmic Masses: C = 12.01, H = 1.008, O = 16.00)
A) C║H╢╖O╖ B) C║H╢╖O
C) C╝H╢╗O╖ D) C╗H╢╖O╖
ü The empirical formula is ê simplest whole number ratio ç ê
carbon, hydrogen, å oxygen aëms ï ê compound.èIn 100 g, êre are
è ? mol C = 63.5 g C x 1 mol C/12.01 g C = 5.29 mol C
è ? mol H = 12.2 g H x 1 mol H/1.008 g H = 12.1 mol H
è ? mol O = 24.2 g O x 1 mol O/16.00 g O = 1.51 mol O
Take ê larger number ç moles ç C å H, å divide by ê smaller
number ç moles ç O.èThis gives ê followïg result:
èè mol C : mol H : mol O = 3.50 : 8.01 : 1
We see that êre are 3î/╖ moles ç C ë one mole ç O.èIf we multiply
through by two, we will clear ê fraction å obtaï ê relation
mol C : mol H : mol O = 7.00 : 16.02 : 2.èThe 16.02 can be rounded ë
16.0 because ê number differs from a whole number ï ê uncertaï
digit.èThe aëm ratio is 7 carbon aëms ë 16 hydrogen aëms ë 2 oxygen
aëms.èThe formula is C╝H╢╗O╖.
Ç C
3èFïd ê empirical formula ç a compound that is 40.85% Cu,
18.01% N, å 41.14% O. (Aëmic Masses: Cu = 63.55, N = 14.01, O = 16.00)
A) Cu(NO╕)╖ B) Cu(NO╖)╖
C) Cu╖NO╖ D) CuNO╕
ü The empirical formula is ê simplest whole number ratio ç ê
copper, nitrogen, å oxygen aëms ï ê compound.èIn 100 g, êre are
è ? mol Cu = 40.85 g Cu x 1 mol Cu/63.55 g Cu = 0.6428 mol Cu
è ? mol Nè= 18.01 g N x 1 mol N/14.01 g N =èè1.286èmol N
è ? mol Oè= 41.14 g O x 1 mol O/16.00 g O =èè2.571èmol O
Take ê larger number ç moles ç N å O, å divide by ê smaller
number ç moles ç Cu.èThis gives ê followïg result:
èè mol Cu : mol N : mol O = 1 : 2.001 : 4.000
We see that êre are 2 moles ç N ë one mole ç Cu å 4 moles ç O ë
one mole ç Cu.èThe formula is CuN╖O╣.èThe nitrite ion, NO╖ú, has a
nitrogen ë oxygen ratio ç 1:2.èWe see that ê formula is consistent
with ê compound havïg two nitrite units.èThe compound is Cu(NO╖)╖.
Ç B
4èWhen 0.224 g ç tï, Sn, reacted with an excess ç chlorïe,
0.492 g ç a tï chloride salt was obtaïed.èWhat is ê empirical
formula ç ê salt?è (aëmic masses: Sn = 118.7, Cl = 35.45)
A) Sn║Cl╢╢ B) Sn║Cl╗
C) SnCl D) SnCl╣
üèThe empirical formula is ê smallest whole number ratio ç ê
tï å chlorïe aëms ï ê compound.èThis ratio also equals ê mole
ratio ç ê elements ï ê compound.èWe could convert ê masses ïë
percentages å solve as before.èThis is not necessary, because we only
need ê mole ratio.èThe mass ç tï is 0.224 g, å ê mass ç
chlorïe is 0.492 g - 0.224 g = 0.268 g Cl.èThe numbers ç moles are
? mol Sn = 0.224 g Sn/118.7 g Sn/mol Sn = 0.00189 mol Sn,
? mol Cl = 0.268 g Cl/35.45 g Cl/mol Cl = 0.00756 mol Cl.
Dividïg ê larger Cl by ê smaller Sn yields 0.00756/0.00189 = 4.00
There are 4 moles ç chlorïe for each mole ç tï.èThe formula ç ê
compound is SnCl╣.
Ç D
5èFïd ê empirical formula ç a compound that is 26.95% S,
13.45% O, å 59.60% Cl.(aëmic masses: S = 32.06, O = 16.00, Cl = 35.45)
A) SOCl╖ B) S╖OCl╣
C) SOCl D) S╖O╖Cl
üèThe empirical formula is ê smallest whole number ratio ç ê
sulfur, oxygen, å chlorïe aëms ï ê compound.èThis ratio also
equals ê mole ratio ç ê elements ï ê compound.èIn 100 g, ê
numbers ç moles are
? mol Sè= 26.95 g S x 1 mol S/32.06 g Sèè= 0.8406 mol S
? mol Oè= 13.45 g O x 1 mol O/16.00 g Oèè= 0.8406 mol O
? mol Cl = 59.60 g Cl x 1 mol Cl/35.45 g Cl = 1.681èmol Cl.
Dividïg ê larger Cl eiêr by ê smaller S or O yields
1.681/0.8406 = 2.000.èThe mole relationship is
mol S : mol O : mol Cl = 1 : 1 : 2.è
The formula ç ê compound is SOCl╖.
Ç D
6èWhen 0.445 g ç potassium, K, reacted with excess oxygen,
0.809 g ç a yellow oxide was obtaïed.èWhat is ê empirical formula ç
ê oxide?è (aëmic masses: K = 39.10, O = 16.00)
A) K╖O╖ B) K╖O
C) KO D) KO╖
üèThe empirical formula is ê smallest whole number ratio ç ê
potassium å oxygen aëms ï ê compound.èThis ratio also equals ê
mole ratio ç ê elements ï ê compound.èWe could convert ê masses
ïë percentages å solve as before.èThis is not necessary, because we
only need ê mole ratio.èThe mass ç potassium is 0.445 g, å ê mass
ç oxygen is 0.809 g - 0.445 g = 0.364 g O.èThe numbers ç moles are
? mol K = 0.445 g K x 1 mol K/39.10 g K = 0.0114 mol K
? mol O = 0.364 g O x 1 mol O/16.00 g O = 0.0228 mol O
Dividïg ê larger moles ç O by ê smaller moles ç K yields
0.0228/0.0114 = 2.00.èThere are 2 moles ç oxygen for each mole ç
potassium.èThe formula ç ê compound is KO╖.
Ç D
äèPlease fïd ê molecular formula ç ê followïg compounds.
âèWhat is ê molecular formula ç a compound that is 30.4% N å
69.6% O by mass å has a molar mass ç 92.02 g/mol.è
One mole ç ê compound contaïs
92.02 g x 0.304 g N/g compound x 1 mol N/14.01 g N = 2.00 mol N, å
92.02 g x 0.696 g O/g compound x 1 mol O/16.00 g O = 4.00 mol O.
One mole ç ê compound contaïs 2 mol N å 4 mol O.èTherefore, one
molecule contaïs 2 aëms ç N å 4 aëms ç O.èThe formula is N╖O╣.
éSèThe empirical formula gives ê relative number ç aëms ï ê
compound.èThe buildïg blocks ç an ionic substance are ê cations å
anions.èThe numbers ç cations å anions are those needed ë make ê
substance electrically neutral.èThe formula ç ê ionic compound is ê
empirical formula.
In contrast ë ê ionic substance, ê buildïg block ç a molecular
substance is a molecule.èThe molecular formula specifies ê numbers ç
each kïd ç aëm ï ê molecule.èThus, ê molecular formula provides
more ïformation than ê empirical formula.èBoth acetylene å benzene
have ê empirical formula CH.èThe molecular formulas are C╖H╖ for
acetylene å C╗H╗ for benzene.èWe can tell immediately that benzene is
a slightly larger å more complex molecule than acetylene.
To determïe ê molecular formula ç a compound, we must know its molar
mass.èIf we know ê empirical formula, we can calculate ê mass ç one
empirical formula unit.èThe molar mass ç ê molecule must be evenly
divisible by ê empirical formula mass, because êre is a whole number
ç empirical formula units ï one molecule.
The mass ç ê CH unit is 12.01 + 1.008 = 13.02.èThe molar mass ç
benzene is 78.11 å that ç acetylene is 26.04.èDividïg ê molar mass
ç benzene by ê unit mass gives 78.11/13.02 = 5.999, which may be
rounded ë 6.èThere are six empirical formula units ï one benzene mole-
cule (6 CH ─¥ C╗H╗).èSimilarly for acetylene, 26.04/13.02 = 2.000.
There are two empirical formula units ï one acetylene molecule (2 CH ─¥
C╖H╖).
We do not need ë know ê empirical formula ë fïd ê molecular form-
ula, if we know ê percentage composition å ê molar mass.èConsider
benzene, C╗H╗, agaï.èOne molecule ç benzene contaïs 6 carbon aëms
å 6 hydrogen aëms.èWe can also say that one mole ç benzene contaïs
6 moles ç carbon å 6 moles ç hydrogen.èIf we fïd ê numbers ç
moles ç each element ï one molar mass ç ê compound, ên those num-
bers will be ê subscripts ï ê molecular formula.
The molar mass ç benzene is 78.11 g/mol, å its composition is 92.25% C
å 7.75% H by mass.èWe fïd ê number ç moles ç each element ï one
mole ç ê compound by usïg ê percentages with ê molar mass.
Let Bz represent benzene.
èèèèèèèè 78.11 g Bzè 0.9225 g Cè 1 mol C
? mol C/mol Bz = ────────── x ────────── x ───────── = 6.000 mol C/mol Bz
èèèèèèèèè1 mol Bzèè 1 g Bzèèè12.01 g C
èèèèèèèè 78.11 g Bzè 0.0775 g Hè 1 mol H
? mol H/mol Bz = ────────── x ────────── x ───────── = 6.00 mol H/mol Bz
èèèèèèèèè1 mol Bzèè 1 g Bzèèè1.008 g H
These numbers show that one molecule ç benzene contaïs 6 carbon aëms
å 6 hydrogen aëms.èThe molecular formula is C╗H╗.
7èThe empirical formula ç cyclohexane is CH╖, å its molar
mass is 84.16 g/mol.èWhat is ê molecular formula ç cyclohexane?
A) CH╖ B) C║H╖╣
C) C╢╡H╖╡ D) C╗H╢╖
üèThe molecular formula must contaï a whole number ç empirical
formula units.èThis means that ê molar mass must be evenly divisible
by ê empirical formula mass.èThe mass ç a CH╖ unit is 12.01+2(1.008)
or 14.03.èDividïg ê molar mass by ê unit mass produces
84.16/14.03 ≈ 6.èThere are six CH╖ units ï one molecule.èThe molec-
ular formula is C╗H╢╖.
Ç D
8èAdipic acid, used ï some types ç nylon, has ê empirical
formula C╕H║O.èIts molar mass is 114.14 g/mol.èWhat is ê molecular
formula ç adipic acid?
A) C╗H╢╡O╖ B) C╝H╢╣O
C) C╖╢H╕║O╝ D) C╕H║O
üèThe molecular formula must contaï a whole number ç empirical
formula units.èThis means that ê molar mass must be evenly divisible
by ê empirical formula mass.èThe mass ç a C╕H║O unit is
12.01 + 5(1.008) + 16.00 = 57.07.èDividïg ê molar mass by ê unit
mass yields 114.14/57.07 = 2.èThere are two C╕H║O units ï one molecule.
The molecular formula is C╗H╢╡O╖.
Ç A
9èA sugar with a molar mass ç 150.13 g/mol was found ë be
40.00% C, 6.71% H, å 53.29% O.èFïd ê molecular formula ç ê
sugar.
A) C╣H╖╖O║ B) C║H╢╡O║
C) C╣H╗O╗ D) C╗H╢╣O╣
üèOne mole ç ê sugar contaïs ê same number ç moles ç each
element as ê number ç aëms ï ê molecular formula.èOne mole ç ê
sugar weighs 150.13 g.èWe can fïd ê moles ç each element ï one mole
ç ê sugar as follows.èLet SG represent ê sugar.
èèèèè150.11 g SGè 0.4000 g Cèè1 mol C
? mol C = ─────────── x ────────── x ───────── = 5.000 mol C/mol SG
èèèèè 1 mol SGèèè1 g SGèèè12.01 g C
èèèèè150.11 g SGè 0.0671 g Hèè1 mol H
? mol H = ─────────── x ────────── x ───────── = 9.99 ≈ 10 mol H/mol SG
èèèèè 1 mol SGèèè1 g SGèèè1.008 g H
èèèèè150.11 g SGè 0.5329 g Oèè1 mol O
? mol O = ─────────── x ────────── x ───────── = 5.000 mol O/mol SG
èèèèè1 mol SGèèè 1 g SGèèè16.00 g O
One mole ç ê sugar contaïs 5 mol C, 10 mol H, å 5 mol O.
The molecular formula is C║H╢╡O║.
Ç B
10èThe molar mass ç nicotïe is 162.2 g/mol å its percentage
composition is 74.04% C, 8.70% H, å 17.27% N.èFïd ê molecular
formula ç nicotïe.
A) C╢╡H╢╣N╖ B) C╛H╢╖N╕
C) C╢╡H╢╣N╖ D) C╜H╢╡N╜
üèOne mole ç nicotïe contaïs ê same number ç moles ç each
element as ê number ç aëms ï ê molecular formula.èOne mole ç
nicotïe weighs 162.2 g.èWe can fïd ê moles ç each element ï one
mole ï nicotïe as follows.èLet Nic represent nicotïe.
èèèèè162.2 g Nicè 0.7404 g Cèè1 mol C
? mol C = ─────────── x ────────── x ───────── = 9.999 ≈ 10 mol C/mol Nic
èèèèè 1 mol Nicèè 1 g Nicèè 12.01 g C
èèèèè162.2 g Nicè 0.0870 g Hèè1 mol H
? mol H = ─────────── x ────────── x ───────── = 14.0è≈ 14 mol H/mol Nic
èèèèè 1 mol Nicèè 1 g Nicèè 1.008 g H
èèèèè162.2 g Nicè 0.1727 g Nèè1 mol N
? mol N = ─────────── x ────────── x ───────── = 1.999 ≈è2 mol N/mol Nic
èèèèè 1 mol Nicèèè1 g Nicèè14.01 g N
One mole ç nicotïe contaïs 10 mol C, 14 mol H, å 2 mol N.
The molecular formula is C╢╡H╢╣N╖.
Ç C