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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapter4.3c
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à 4.3cèPercentage Composition
äèPlease calculate ê percentage by mass ç ê specified element(s) or formula units ï ê
followïg compounds.
â Fïd ê percentage ç chlorïe ï phosphorous trichloride, PCl╕.
The percentage ç chlorïe is ê mass ç ê chlorïe ï 100 grams ç
PCl╕.èWe know that one mole ç PCl╕ contaïs three moles ç Cl.èOne mol
ç PCl╕ weighs 30.97 + 3(35.45) = 137.3 g.èThree moles ç Cl weigh
3 x 35.45 = 106.4 g.èThe mass ç chlorïe per 100 grams ç PCl╕ is
(106.4 g Cl/137.3 g PCl╕) x 100 = 77.49, so ê percentage ç chlorïe is
77.49%.èè
éSèPer cent means ê number ç parts out ç one hundred parts.
The percentage composition refers ë ê masses ç ê elements ï a com-
pound per one hundred grams ç ê compound.èWe can use ê formula ç
ê compound ë obtaï ê mass ç ê element ï one mole ç ê com-
pound.èWith that ïformation, we can easily calculate ê mass ç ê
element ï one hundred grams.è
Let's calculate ê percentage ç nitrogen ï ammonium nitrate, NH╣NO╕.
One mole ç ammonium nitrate contaïs two moles ç nitrogen, four moles
ç hydrogen, å three moles ç oxygen.èThe molar mass ç ammonium
nitrate is 2(14.01) + 4(1.008) + 3(16.00) = 80.05 g/mol.èThe mass ç
nitrogen ï 80.05 g ç ammonium nitrate is 2(14.01) or 28.02 g.èThe
percentage ç nitrogen is:
èèè grams ç nitrogen ï one mole ç compound
?% N = ───────────────────────────────────────── x 100
èèèèèèèèèmass ç one mole ç compound
èèèèèèèèè 2(14.01) g N
?% N = ─────────────────────────────────────── x 100 = 35.00% N
èèè 2(14.01) + 4(1.008) + 3(16.00) g NH╣NO╕
The ïformation ë fïd ê percentage can be presented ï oêr ways.
Consider ê followïg example.èWhen 2.000 grams ç tï reacted with
excess oxygen, 2.539 grams ç a tï oxide were formed.èWhat is ê
percentage ç tï ï ê compound?
The percentage ç tï is ê mass ç tï ï 100 grams ç ê oxide.
We can obtaï ê mass ç tï per gram ç compound from ê given masses.
Multiplyïg ê mass per gram by one hundred yields ê desired percent-
age.è
èèèèèèmass tï
?% Sn = ──────────────── x 100
èèèèèèèèmass ç compound
èèèèè 2.000 g Sn
?% Sn = ──────────────── x 100 = 78.77% Sn
èèèèèèèè2.539 g compound
So far we have only calculated ê percentage ç one element ï a com-
pound, but ê oêr percentages are calculated ï ê same manner.èOf
course, ê sum ç ê percentages ç ê elements ï a compound must
equal 100, but you already knew that!
Farmers are ïterested ï ê elemental percentage composition ç êir
fertilizers so that êy can use ê correct fertilizer ë encourage ê
correct growth ç êir crops.èSynêtic chemists use ê percentage
compositions ç êir products ë see if êy have prepared ê desired
compounds.
1èWhat is ê percentage ç uranium ï uranium hexafluoride?
(aëmic masses: U = 238.0, F = 18.99)
A) 92.02% B) 92.61%
C) 67.63% D) 47.87%
üèTo solve this problem, we must have ê correct formula ç ê
uranium hexafluoride, which is UF╗.èThe percentage ç uranium is ê
mass ç uranium ï 100 g ç uranium hexafluoride.èOne mole ç UF╗ has a
mass ç 238.0 + 6(18.99) = 351.9 g.èOne mole ç uranium hexafluoride
contaïs one mole ç uranium.èThe percentage is
èèè238.0 g U
%U = ─────────── x 100 = 67.63%.
èè 351.9 g UF╗
Ç C
2èWhat is ê percentage ç K ï K╖O?
(aëmic masses: K= 39.10, O = 16.00)
A) 83.0% B) 41.5%
C) 14.2% D) 14.2%
üèWe need ê mass ç potassium ï 100 g ç ê potassium oxide.
One mole ç K╖O weighs 94.2 g (2x39.1 + 16.00) å contaïs 2x39.1 g ç
potassium.èThe percentage is
èè 2x39.10 g K
%K = ─────────── x 100 = 83.01%.
èè 94.20 g K╖O
Ç A
3èWhat is ê percentage ç phosphorous ï calcium phosphate?
(aëmic masses: Ca = 40.08, P = 30.97, O = 16.00)
A) 35.58% B) 71.15%
C) 19.97% D) 11.09%
üèTo solve this problem, we must have ê correct formula ç ê
calcium phosphate, which if Ca╕(PO╣)╖.èThe percentage ç phosphorous is
ê mass ç phosphorous ï 100 g ç calcium phosphate.èOne mole ç
calcium phosphate has a mass ç 310.2 g (3(40.08) + 2(30.97) + 8(16.00))
One mole ç calcium phosphate contaïs two moles ç phosphorous.
The percentage is
èèè 2x30.97 g P
%P = ───────────────── x 100 = 19.97%.
èè 310.2 g Ca╕(PO╣)╖
Ç C
4èFïd ê percentage ç iodïe ï a metal iodide when 0.530 g
ç ê metal reacts with excess iodïe ë form 2.796 g ç ê iodide.
(aëmic mass ç I = 126.9)
A) 19.0% B) 81.0%
C) 23.4% D) 52.8%
üèThe percentage ç iodïe is ê mass ç iodïe ï 100 g ç ê
compound.èWe were given ê mass ç ê metal å ê mass ç ê com-
pound.èThe mass ç iodïe is ê difference ï those masses.èThe mass
ç iodïe equals 2.796 g - 0.530 g = 2.266 g.èThe percentage is
èèèmass ç iodïe
%I = ──────────────── x 100
èè mass ç compound
èèèè2.266 g I
%I = ──────────────── x 100 = 81.04%
èè 2.796 g compound
Ç B
5èThe percentage ç water ï iron(III) nitrate hexahydrate,
Fe(NO╕)╕∙6H╖O, is ...è(aëmic masses: Fe = 55.85, N = 14.01, O = 16.00,
H= 1.008)
A) 5.15% B) 30.89%
C) 44.70% D) 19.58%
üèThe percentage ç water is ê mass ç water ï 100 grams ç ê
compound.èOne mole ç ê iron(III) nitrate hexahydrate contaïs six
moles ç water.èWe can calculate ê mass ç one mole ç ê salt å ç
ê water.èDividïg six times ê molar mass ç water by ê molar mass
ç ê entire salt gives ê mass ç water per gram ç salt.èMultiplyïg
this fraction by 100 yields ê percentage.èThe molar mass ç ê
nitrate salt is 55.85 + 3(14.01) + 15(16.00) + 12(1.008) =è349.98 g/mol.
The molar mass ç water is 2(1.008) + 16.00 = 18.02 g/mol.
The percentage ç water is
èèèèè 6 x 18.02 g H╖O
% H╖O = ────────────────────── x 100 = 30.89%
èèèèèèèè349.98 g Fe(NO╕)╕∙6H╖Oè
Ç B
6èWhat is ê percentage ç carbon monoxide ï nickel carbonyl,
Ni(CO)╣?è(aëmic masses: Ni = 58.71, C = 12.01, O = 16.00)
A) 56.42% B) 89.10%
C) 36.51% D) 65.62%
üèThe percentage ç carbon monoxide is ê mass ç carbon monoxide
ï 100 grams ç ê nickel carbonyl.èOne mole ç Ni(CO)╣ contaïs four
moles ç carbon monoxides.èWe can calculate ê mass ç one mole ç
Ni(CO)╣ å ç CO.èDividïg four times ê molar mass ç carbon monoxide
by ê molar mass ç ê entire compound gives ê grams ç CO per gram
ç Ni(CO)╣.èMultiplyïg this fraction by 100 yields ê percentage.èThe
molar mass ç Ni(CO)╣ is 58.71 + 4(12.01) + 4(16.00) =è170.75 g/mol.
The molar mass ç CO is 12.01 + 16.00 = 28.01 g/mol.
The percentage ç carbon monoxide is
èèèè4 x 28.01 g CO
% CO = ──────────────── x 100 = 65.62%
èèèèèèè 170.75 g Ni(CO)╣
Ç D
7èWhat is ê percentage ç water ï Na╖CO╕∙10H╖O?
(aëmic masses: Na = 22.99, C = 12.01, O = 16.00, H = 1.008)
A) 62.96% B) 26.09%
C) 6.30% D) 45.29%
üèThe percentage ç water is ê mass ç water ï 100 grams ç ê
Na╖CO╕∙10H╖O.èOne mole ç Na╖CO╕∙10H╖O contaïs ten moles ç water.èWe
can calculate ê mass ç one mole ç H╖O å ç Na╖CO╕∙10H╖O.èDividïg
ten times ê molar mass ç water by ê molar mass ç ê entire com-
pound gives ê grams ç H╖O per gram ç Na╖CO╕∙10H╖O.èMultiplyïg this
fraction by 100 yields ê percentage.èThe molar mass ç Na╖CO╕∙10H╖O is
2(22.99) + 12.01 + 13(16.00) + 20(1.008) =è286.2 g/mol.
The molar mass ç water is 2(1.008) + 16.00 = 18.02 g/mol.
The percentage ç water is
èèèè10 x 18.02 g H╖O
% H╖O = ──────────────────── x 100 = 62.96%
èèèèèèèè286.2 g Na╖CO╕∙10H╖O
Ç A
äèPlease calculate ê amount ç a compound required ë supply an amount
ç an element or a subunit ï ê compound.
âèHow many grams ç PCl╕ are required ë furnish 15.0 g ç Cl?èWe
know that one mole ç PCl╕ contaïs three moles ç chlorïe, so we also
know that 137.3 g ç PCl╕ contaïs 106.4 g ç Cl.èWe want 15.0 g ç Cl
from PCl╕.èTherefore, we must make a conversion facër that shows ê
grams ç PCl╕ per gram ç Cl ï ê PCl╕.èèè
èè? g PCl╕ = 15.0 g Cl x (137.3 g PCl╕ / 106.4 g Cl) = 19.4 g PCl╕
Notice that êè"g Cl" unit cancels.
éSèTo fïd ê mass ç a substance that will supply a subunit ç
ê substance, we essentially reverse what we have been doïg å divide
by ê percentage.èWe also can view this as a dimensional analysis type
ç problem.è
Many ïorganic salts form hydrates, å we frequently want ë know what
ê dry mass ç ê compound would be.èFor example, Epsom salt is ê
heptahydrate ç magnesium sulfate, MgSO╣∙7H╖O.èThis colorless crystal-
lïe salt has seven water molecules ïcluded ï ê crystal with each
MgSO╣ unit.èWhat mass ç MgSO╣∙7H╖O would you take ï order ë obtaï
50.0 g ç MgSO╣?èIn one mole ç ê compound contaïs one MgSO╣ unit.
The molar mass ç MgSO╣ is 24.31 + 32.07 + 4(16.00) = 120.37 grams.èThe
molar mass ç ê whole compound is 120.37 + 14(1.008) + 7(16.00) =
246.48 grams.èOur conversion facër should have ê units ç
g MgSO╣∙7H╖O/g MgSO╣ because we know how many grams ç MgSO╣ we want.
èèè 246.48 g MgSO╣∙7H╖O
?g MgSO╣∙7H╖O = 50.0 g MgSO╣ x ─────────────────── = 102 g MgSO╣∙7H╖O
èèèè 120.37 g MgSO╣
102 g MgSO╣∙7H╖O will furnish 50.0 g MgSO╣.
8èHow many grams ç K╖O would supply 15.0 g ç K?
(aëmic masses: K = 39.10, O = 16.00)
A) 12.5 g B) 18.1 g
C) 21.1g D) 10.6 g
üèOne mole ç K╖O contaïs two moles ç potassium.èIn terms ç ê
mass,è94.2 g ç K╖Oècontaïs 78.2 g ç K.èWe need ë convert ê 15.0
g ç K ïë grams ç K╖O.
èè94.2 g K╖O
?g K╖O = 15.0 g K x ────────── = 18.1 g K╖O
èè78.2 g K
Ç B
9èHow many grams ç CoCl╖∙6H╖O are required ë obtaï 22.0 g ç
CoCl╖?è(aëmic masses: Co = 58.93, Cl = 35.45, H = 1.008, O = 16.00)
A) 12.0 g B) 25.1 g
C) 26.0 g D) 40.3 g
ü We know that one mole ç CoCl╖∙6H╖O contaïs one mole ç CoCl╖.
Usïg ê molar masses, we can say 237.93 g CoCl╖∙6H╖O contaïs 129.83 g
CoCl╖.èWe start with ê 22.0 grams ç CoCl╖ å convert ë ê mass ç
CoCl╖∙6H╖O.
237.93 g CoCl╖∙6H╖O
? g CoCl╖∙6H╖O = 22.0 g CoCl╖ x ─────────────────── = 40.3 g CoCl╖∙6H╖O
è129.83 g CoCl╖
Ç D
10èWhat mass P╖O║ contaïs 150. g ç phosphorous?
(aëmic masses: P = 30.97, O = 16.00)
A) 228 g B) 181 g
C) 344 g D) 688 g
üèWe must convert from grams ç phosphorous ïë grams ç P╖O║.
The formula tells us that one mole ç phosphorous penëxide contaïs two
mole ç phosphorous.èThe molar mass ç P╖O║ is 2(30.97) + 5(16.00) =
141.94 grams.èOne molar mass ç P╖O║ contaïs 2(30.97) = 61.94 grams ç
phosphorous.èThe conversion from P ë P╖O║ is
èèè141.94 g P╖O║
? g P╖O║ = 150. g P x ───────────── = 344 g P╖O║
èèèè61.94 g P
Ç C