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à 4.2cèMolar Mass, Moles, Molecules, å Ions
äèPlease determïe ê molar mass ç ê followïg compounds.
âèThe molar mass ç ammonium sulfate,è N: 2x14.01 =è28.02
(NH╣)╖SO╣ is found by addïg ê aëmicèèèèH: 8x1.008 =è 8.064
masses ç ê elements ï ê compound.èèèèS: 1x32.07 =è32.06
Ammonium sulfate has 2 nitrogen aëms,èèèè O: 4x16.00 =è64.00
8 hydrogen aëms,one sulfur aëm, åèèèèèèèèèèè ───────
4 oxygen aëms.èCalculation ç ê molarèèèmolar mass = 132.14 g/mol
mass is shown ë ê right:è
éS The molar mass is ê mass ï grams ç one mole ç ê substance.
Various labels are used for ê molar mass ç different classes ç com-
pounds.
Let's consider NaCl.èSodium chloride is an ionic substance.èWe know it
is ionic because it conducts electricity ï ê molten (liquid) state.
This reveals that NaCl actually exists as Naó å Clú ions.èThere is no
"NaCl" molecule ï sodium chloride.èThe formula ç an ionic substance
merely gives ê relative numbers ç ions ï ê compound.èIn this case,
êre is one Naó ion for each Clú ion.èNaCl is called ê formula unit.
The one molar mass ç NaCl contaïs one molar mass ç sodium plus one
molar mass ç chlorïe.èThis ëtal mass is 22.99 + 35.45 = 58.44 g/mol.
This molar mass is also called ê "gram formula mass" or just ê
"formula mass".
Now let's consider aceëne, C╕H╗O.èAceëne is a molecular substance.
This compound does not conduct electricity.èThis tells us that êre are
no ions ï aceëne or that it exists as ïdividual molecules.èEach
molecule has three carbon aëms, six hydrogen aëms, å one oxygen aëm.
Therefore, one mole ç aceëne contaïs three moles ç carbon, six moles
ç hydrogen, å one mole ç oxygen.èThe molar mass ç aceëne is found
by addïg ê masses ç carbon, hydrogen, å oxygen.
C:è3 x 12.01è= 36.03
H:è6 xè1.008 =è6.048
O:è1 x 16.00è= 16.00
èèèèèèèèèèèè ─────
èèèèmolar massèè = 58.08 g/moleè(rounded ë hundredths)
The molar mass ç a molecular substance may be called ê gram molecular
mass or simply ê molecular mass.è
Regardless ç wheêr ê substance is ionic or molecular, ê molar mass
is obtaïed by addïg ê aëmic masses ç ê elements ï ê substance.
1èWhat is ê molar mass ç Ba(NO╕)╖?
A) 247.3 g/mol B) 261.3 g/mol
C) 167.3 g/mol D) 118.0 g/mol
üèThe molar mass is ê mass ç one mole ç barium nitrate.èOne
mole contaïs one mole ç Ba, two moles ç N, å six moles ç O.
The molar mass is:
èèè Ba:è 1 x 137.3è=è137.3
N:èè2 x 14.01è=è 28.02
O:èè6 x 16.00è=è 96.00
èèèèèèèèèèèèèè──────
èèèèèèèmolar mass =è261.3 g/mol
Ç B
2èWhat is ê molar mass ç ethyl alcohol, C╖H║OH ?
A) 45.06 g/mol B) 26.00 g/mol
C) 46.07 g/mol D) 29.02 g/mol
üèThe molar mass is ê mass ç one mole ç ethyl alcohol.èOne
mole contaïs two moles ç carbon, six moles ç hydrogen, å one mole
ç oxygen.èThe molar mass is:
èèè C:è 2 x 12.01è =è24.02
H:è 6 xè1.008è=è 6.048
O:è 1 x 16.00è=è 16.00
èèèèèèèèèèèèèè──────
èèèèèèèmolar mass =è46.07 g/mol
Ç C
3èWhat is ê molar mass ç calcium phosphate, Ca╕(PO╣)╖?
A) 310.2 g/mol B) 279.2 g/mol
C) 87.05 g/mol D) 522.3 g/mol
üèThe molar mass is ê mass ç one mole ç calcium phosphate.èOne
mole ç calcium phosphate contaïs three moles ç Ca, two moles ç P, å
eight moles ç O.èThe molar masses ç ê elements are êir aëmic
masses ï grams.èThe molar mass is:
èèè Ca:è 3 x 40.08è=è120.24
P:èè2 x 30.97è=è 61.94
O:èè8 x 16.00è=è128.00
èèèèèèèèèèèèèè──────
èèèèèèèmolar mass =è310.18 g/mol
We are only justified ï four significant figures so ê mass should be
reported as 310.2 g/mol.
Ç A
äèPlease fïd ê number ç moles ç ê given compounds or ions.
â How many moles ç acetic acid are ï 15.0 g ç acetic acid,
HC╖H╕O╖?èThe molar mass is ê mass ç one mole ç ê compound å
provides ê lïk between mass å moles.èThe molar mass ç acetic acid
is 2(12.01) + 4(1.008) + 2(16.00) = 60.05 g/mol.èThe number ç moles
ï 15.0 g is:
èèèèèèèèèèèèèèè 1 mol HC╖H╕O╖
èè ? mol = 15.0 g HC╖H╕O╖ x ─────────────── = 0.250 mol HC╖H╕O╖
èèèèèèèèèèèèèèè60.05 g HC╖H╕O╖
éSèThe molar mass is ê mass ï grams ç one mole ç ê sub-
stance.èThe molar mass allows us ë ïterrelate ê moles å grams ç a
sample ç ê substance.èIf you are not provided with ê molar mass ç
ê compound, ên you must calculate ê mass.èWe çten calculate moles
from grams å vice versa.
To fïd ê number ç moles from ê mass ç ê substance, we divide ê
mass by ê molar mass.èThe number ç moles ç NaCl ï 250. g ç NaCl is
èèèèèèè 1 mol NaCl
èèèè250. g NaCl x ──────────── =è4.28 mol NaCl.
èèèèèèèèèèè58.44 g NaCl
The molar mass ç NaCl is 22.99 + 35.45 = 58.44 g/mol.èNotice that ê
gram units cancel, as ï ê oêr problems that we treated via dimen-
sional analysis.
To fïd ê mass ç a given number ç moles, we multiple ê number ç
moles by ê molar mass.èThe mass ç 0.800 mol ç NaCl is
èèèèèèèè 58.44 g NaCl
èèèè0.800 mol NaCl x ──────────── = 46.8 g NaCl.
èèèèèèèèèèèèè1 mol NaCl
For an ionic substance, we can fïd ê number ç moles ç an ion from
ê number ç moles ç ê compound via its formula.èFor example, ê
formula ç magnesium nitrate is Mg(NO╕)╖.èOne mole ç Mg(NO╕)╖ contaïs
one mole ç Mgìó ions å two moles ç NO╕ú ions.èIf we had 0.300 mol
ç magnesium nitrate, ên êre would be 0.300 mol ç Mgìó ions å
2 x 0.300 = 0.600 mol ç nitrate ions ï ê sample.
4èHow many moles ç propane, C╕H╜, are ï 500. g ç propane?
A) 2.20x10Å mol B) 0.0882 mol
C) 0.114 mol D) 11.3 mol
üèTo fïd ê number ç moles, we must first know ê molar mass ç
propane.èPropane has 3 carbons å 8 hydrogens so ê molar mass is
3(12.01) + 8(1.008) = 44.09 g/mol.èThe number ç moles is
èèèèèèèèèèèèèèèèè 1 mol C╕H╜
? mol C╕H╜ = 500. g C╕H╜ x ──────────── = 11.3 mol
èèèèèèèèèèèèèèèèè 44.09 g C╕H╜
Ç D
5èHow many moles ç ammonium sulfate, (NH╣)╖SO╣, are ï 350. g
ç ammonium sulfate?
A) 2.65 mol B) 0.378 mol
C) 4.63 mol D) 7.68x10Ä mol
üèTo fïd ê number ç moles, we must first know ê molar mass ç
ammonium sulfate.èAmmonium sulfate has 2 nitrogen aëms, 8 hydrogen
aëms, one sulfur aëm, å 4 oxygen aëms.èThe molar mass is
2(14.01) + 8(1.008) + 32.06 + 4(16.00) = 132.14 g/mol.èThe number ç
moles is
èèèèèèèèèèèèèèèèèè 1 mol (NH╣)╖SO╣
? mol (NH╣)╖SO╣ = 350. g (NH╣)╖SO╣x ────────────────── = 2.65 mol
èèèèèèèèèèèèèèèèèè132.14 g (NH╣)╖SO╣
Ç A
6èWhat is ê mass ç 0.662 mol BaCl╖?
A) 315 g B) 138 g
C) 3.18x10úÄ g D) 1.84x10úì g
üèTo fïd ê number ç grams from ê number ç moles, we must
first fïd ê molar mass ç BaCl╖.èIts molar mass is 137.3 + 2(35.45)
which equals 208.2 g/mol.èThe number ç grams is
208.2 g BaCl╖
? g BaCl╖ = 0.662 mol x ───────────── = 138 g BaCl╖
èèèèèèèèèèèèèèèè 1 mol BaCl╖
Ç B
7èWhat is ê mass ç 0.0848 mol HC╖H╕O╖, acetic acid?
A) 708 g B) 1.46 g
C) 1.41x10úÄ g D) 5.09 g
üèTo fïd ê number ç grams from ê number ç moles, we must
first fïd ê molar mass ç acetic acid.èThe molar mass ç acetic acid
is 4(1.008) + 2(12.01) + 2(16.00) which equals 60.05 g/mol.èThe number
ç grams is
è 60.05 g HC╖H╕O╖
? g HC╖H╕O╖ = 0.0848 mol x ─────────────── = 5.09 g HC╖H╕O╖
èèèèèèèèèèèèèèèèèè1 mol HC╖H╕O╖
Ç D
8èHow many moles ç chloride ions are ï 55.6 g ç AlCl╕?
A) 0.523 mol Clú B) 4.71 mol Clú
C) 1.57 mol Clú D) 1.25 mol Clú
üèThe formula ç this ionic compound shows that êre are three
moles ç chloride ions ï one mole ç alumïum chloride.èYou need ë
know how many moles ç alumïum chloride êre are ï ê 55.6 grams.
The molar mass provides ê connection between ê mass å moles.
The molar mass ç AlCl╕ is 26.98 + 3(35.45) = 133.33 g/mol.èThe pathway
for solvïg for ê moles ç Clú is g AlCl╕ ──¥ mol AlCl╕ ──¥ mol Clú.
èè1 mol AlCl╕èèè3 mol Clú
? mol Clú = 55.6 g AlCl╕ x ────────────── x ─────────── = 1.25 mol Clú
èèèèèèèèèèèèè 133.33 g AlCl╕è 1 mol AlCl╕
Ç D
9èHow many moles ç nitrate ions are ï 45.0 g ç Mg(NO╕)╖?
A) 0.607 mol B) 6.59 mol
C) 0.363 molèè D) 0.303 mol
üèThe formula ç this ionic compound shows that êre are two moles
ç nitrate ions ï one mole ç magnesium nitrate.èWe need ë know how
many moles ç magnesium nitrate êre are ï ê 45.0 grams.èThe molar
mass provides ê connection between ê mass å moles.èThe molar mass
ç Mg(NO╕)╖ is 24.31 + 2(14.01) + 6(16.00) = 148.33 g/mol.èThe pathway
for solvïg for ê moles ç NO╕ú is g Mg(NO╕)╖ ──¥ mol Mg(NO╕)╖ ──¥
mol NO╕ú.
èèèè1 mol Mg(NO╕)╖èèè 2 mol NO╕úè
? mol NO╕ú = 45.0 g Mg(NO╕)╖ x ───────────────── x ──────────────
èèèèèèèèèèèèèèè 148.33 g Mg(NO╕)╖è 1 mol Mg(NO╕)╖
? mol NO╕ú = 0.607 mol
Ç A
äèPlease fïd ê number ç molecules, aëms, or ions ï ê followïg samples.
âèFïd ê number ç molecules ï 250.0 g ç water.èWe know that
êre are 6.022x10ìÄ molecules ï one mole ç a molecular substance like
water.èWe must fïd ê number ç moles ç water ï 250 g ç water.èThe
molar mass ç water is 2(1.008) + 16.00 = 18.02 g/mol.èThe number ç
molecules isèèèèè1 molèè 6.022x10ìÄ molec
èèèèèèè250 g x ─────── x ──────────────── = 8.35x10ìÅ molec H╖O
èèèèèèèèèèè18.02 gèèèè1 mol
éSèIn chemical calculations, we frequently convert between grams
å moles, å sometimes we are ïterested ï ê number ç molecules,
aëms, or ions.èAs we mentioned ï ê previous Section 4.1, we perform
this pattern ç conversions agaï å agaï.
è ┌─────────┐ ┌─────┐
┌─────┐ ÷ molar mass ┌─────┐ x 6.022x10ìÄ │molecules│ chemical │aëms│
|grams| ============ |moles| ============ |èèorè │ ======== │ orè|
└─────┘ x molar mass └─────┘ ÷ 6.022x10ìÄ │ formula │ formulaè│ions │
è │èunitsè│ └─────┘
è └─────────┘
The ëp lïe ïdicates ê pattern from left ë right, å ê botëm
lïe shows ê pattern from right ë left.èWhen you know ê mass ç ê
substance å want ë know ê number ç aëms or ions, you would divide
ê mass by ê molar mass, multiply by Avogadro's number, å fïally
multiply by ê number ç aëms or ion ï a formula unit.
If you are given ê number ç aëms or ions ï a substance å want ë
fïd ê number ç grams ç ê substance, you would divide ê number ç
aëms or ions by ê number ç aëms or ions ï one formula unit, divide
by Avogadro's number å multiply by ê molar mass.
How many bromide ions are ï 5.00 grams ç FeBr╕?èThe molar mass ç
FeBr╕ is 55.85 + 3(79.90) = 296.5 g/mol.
èèèèèèèèèèèèè1 molèè 6.022x10ìÄ FeBr╕ unitsè 3 Brú
? Brú ions = 5.00g FeBr╕x ─────── x ────────────────────── x ───────
èèèèèèèèèèèèè296.5 gèè 1 molèèèèèèèèè1 FeBr╕
? Brú ions = 3.05x10ìì ions.
Of course, we can reverse this process.èWhat is ê mass ç 3.3x10îô
molecules ç CO╖?èThe molar mass ç CO╖ is 44.01 g/mol.
èèèèèèèèèèèèèèèè 1 mol CO╖èèèèèè44.01 g CO╖
? g CO╖ = 3.3x10îô molec CO╖ x ──────────────────── x ───────────
èèèèèèèèèèèèèèè 6.022x10ìÄ molec CO╖èè1 mol CO╖
? g CO╖ = 2.4x10úÅ g CO╖è (or 0.24 mg)
10 How many molecules ç sugar are ï 5.00 g ç sugar, C╢╖H╖╖O╢╢?
The molar mass ç sugar is 342 g/mol.
A) 4.12x10ìÉ molec B) 8.80x10ìî molec
C) 2.83x10ìÉ molec D) 1.14x10ìÅ molec
üèFrom ê mass ç ê sugar we can calculate ê number ç moles.
Usïg Avogadro's number with ê number ç moles, permits calculation ç
ê number ç molecules that are ï ê sample.
èèèèèèèèèèèè 1 mol sugarè 6.022x10ìÄ molec sugar
? molec = 5.00 g sugar x ─────────── x ──────────────────────
èèèèèèèèèèèè 342 g sugarèèè 1 mol sugar
? molec sugar = 8.80x10ìî molec
Ç B
11èHow many ammonium ions are ï 75.0 g (NH╣)╖SO╣?èThe formula
mass ç ammonium sulfate is 132.14 g/mol.
A) 3.42x10ìÄ ions B) 1.06x10ìÅ ions
C) 2.12x10ìÅ ions D) 6.84x10ìÄ ions
üèFrom ê mass ç ê ammonium sulfate, we can calculate ê
number ç moles.èThe molar mass states that one mole is 132.15 grams.
Usïg Avogadro's number with ê number ç moles enables us ë fïd ê
number ç formula units ï ê molecules that are ï ê 75.0 g sample.
Each formula units contaïs two ammonium ions.
èèèèèèèèèèèèèè 1 mol (NH╣)╖SO╣èè6.022x10ìÄ (NH╣)╖SO╣'s
? ions = 75.0g (NH╣)╖SO╣ x ────────────────── x ────────────────────── x
èèèèèèèèèèèèèè132.14 g (NH╣)╖SO╣èè 1 mol (NH╣)╖SO╣
2 NH╣ó ions
────────────────
1 (NH╣)╖SO╣ unit
? NH╣ó ions = 6.84x10ìÄ ions
Ç D
12èHow many oxygen molecules are ï 200. g ç O╖?
A) 7.71x10ìÆ molec B) 7.53x10ìÅ molec
C) 3.76x10ìÅ molec D) 9.64x10ìì molec
üèAvogadro's number is ê number ç molecules ï one mole ç a
molecular compound like oxygen.èWe need ë determïe ê number ç moles
ç oxygen ï ê 200. grams.èThe molar mass ç oxygen is 2(16.00) =
32.00 g O╖/mol O╖.èThe number ç molecules is
èèèèèèèèèèèèè1 mol O╖èè6.022x10ìÄ molec O╖
? molec O╖ = 200. g O╖ x ────────── x ───────────────────
èèèèèèèèèèèè 32.00 g O╖èèèè1 mol O╖
? molec O╖ = 3.76x10ìÅ molec
Ç C