Everything you need to ace chemistry

C A L L M EC A L L M EC A L L M EC A L L M E S T A N N O U S S T A N N O U S S T A N N O U S S T A N N O U S C A L L M EC A L L M EC A L L M EC A L L M E S T A N N I C S T A N N I C S T A N N I C S T.C A L L M EC A L L M EC A L L M EC A L L M E S T A N N O U S S T A N N O U S S T A N N O U S S T A N N O U S C A L L M EC A L L M EC A L L M EC A L L M E S T A N N I C S T A N N I C S T A N N I C S T.

C A L L M E

S T A N N O U S

S T A N N I C

6 Wh at is t h e for mu la for h y poc h lorou s ac id?

247

plu mb ou s i on , t i n ( I V) i on OR s t an n i c ion, and

4

(ClO -) at t ac h e d t o a h y droge n at om

eac h individu al atom b e c au s e t h ey ar e s o s mall.

and six neutrons in its nucleus and is relatively easy to

measure accurately every time It was given the exact

atomic mass of 12 and is used as the standard for all

other atomic, molecular, and formulaic masses

MOLEThe SI base unit used to measure the

amount of a substance The number

of carbon atoms in exactly 12.000 g

of the carbon-12 isotope

249

Avogadro’s nu mb e r or 6.022 × 1 0 2 3 atoms (or molec u les)

A vogadro’ s n u mb e r w as n ot ac t u ally di s c ove r e d b y t h e

h e did propos e t h e h y pot h e s is t h at t w o s ample s of

gas of equ al volu me at t h e same t emperatu r e and

pr e s s u r e w ou ld c on t ain t h e s ame n u mb e r of mole c u le s

h y pot h e s is w as prove n Pe r r in n ame d t h e n u mb e r aft e r

Avogadro t o r e c ogn ize h im for h is s c ie n t ific c on t r ib u t ion

1 m o l o f p e n n i e s w o u l d

c o v e r t h e e n t i r e E a r t h f o r

a d e p t h o f 0 3 m i l e s

1 mol of c ar b on-12 h as exac t ly 6.022 × 1 0 2 3 atoms and

Avogadro’s n u mb e r w h e n c alc u lat in g, y ou mu s t s t ill r e por t

y ou r an s w e r u s in g on ly t w o s ign ific an t figu r e s

H o w d o y o u u s e t h i s i n f o r m a t i o n ?

Avogadro’s nu mb e r , you c an figu r e ou t t h e mas s of a s ingle

MOLAR MASSThe mass (in grams) of

1 mol of units of a substance

or grams/mole

ATOMIC MASSThe mass of one atom expressed

in atomic mass units

ATOMIC MASS UNITS

(AMU)The mass that is exactly equal

FOR EXAMPLE: Wh at is t h e mas s of a s in gle at om of

Th e molar mas s r elat ion s h ip c an b e u s e d as a c onve r s ion

Calculating Molecular Mass

If you know t h e atomic mass of

c alc u lat e mole c u lar mas s

To c alc u lat e t h e molec u lar mass of wat er (H

2O), follow

O has two hydrogen (H) atoms and one oxygen (O) atom

2. Ge t t h e at omic mas s for e ac h ele me n t from t h e pe r iodic

t ab le

A t omi c mas s of H = 1 0 0 7 8 amu

At omic mas s of O = 15 999 amu

3. Add t h e m t oge t h e r (You c an als o rou n d t o t h e h u n dr e dt h s

plac e for easier c alc u lat ion.)

present (in amu)

When adding numbers, therule for sig figs is to round

to the least number ofdecimal places

253

2 (atomic mass H) + 1 (atomic mass O) = molecular mass

6 (atomic mass of C) + 12 (atomic mass of H) + 6 (atomic mass

of O) = molec u lar mass of C

6H12O6

6 (12.009 amu ) + 12 (1.0078 amu ) + 6 (15 999 amu ) = 180.14 2 amu

Th e mole c u lar mas s of C

6H12O6

FOR EXAMPLE: If you h ave 15 g of magnesiu m (M g) in

fac t or on t h e left s o t h at t h e u n it s divide ou t (grams

of g M g in t h e de nominator , w h ic h w ill le ave mole s as t h e

u nit for t h e final answer

the periodic table

= 62

255

FOR EXAMPLE: If you h ave 6.5 8 g of c arb on, h ow many

atoms of c ar b on ar e t h e r e ?

1 mol C = 12.0 0 9 g

know h ow many atoms

of a single element are

1, 2, 3, 4, 5 It is not typical tohave a fraction of a mole present

in a chemical formula

257

2

S O

4

h as 2 mol H , 1 mol S , an d 4 mol O

Once you know how many moles of an element are present

may have

in 32.68 g of c affeine (C

8H

1 0N4O2)?

8 (atomic mass of C) + 10 (atomic mass of H) + 4 (atomic mass

of N ) + 2 (atomic mass of O) = g/mol c affeine

1 0N4O2

1 mol C

8H

1 0N4O2

H

1 0N4O2

194 172 g C

8H

1 0N4O2

Using Molecular Mass

to Calculate Moles or Atoms

of a Compound from Grams

4) If you have

7.5 2 g of me t h an e , h ow man y mole s of CH

4

do you have?

1. You h ave 7.5 2 g b u t need to c onvert to moles You mu s t

fir s t c alc u lat e t h e mole c u lar mas s of CH

2 Wh at is t h e diffe r e nc e b e tw e e n molar mas s and

atomic mass?

6H12O6)?

5 Calc u lat e t h e mole c u lar mas s of M g

3N2

6 How many mole s of e t h ane (C

2H6) are in 5 6.2 g of e t h ane ?

261

is equal to Avogadro’s numb er or 6.022 × 1 0 2 3 atoms

2 M olar mas s is t h e mas s (in grams ) of 1 mol of u n it s of a

5 3 (24.31 g) + 2 (14.01 g) = 100.95 g/mol

6 5 6.2 g/30 0 8 g/mol = 1.87 mole s

To fin d t h e pu r it y of a c ompou n d or t h e amou n t of all of t h e

diffe r e n t ele me n t s , b y pe r c e n t age , s c ie n t is t s c alc u lat e t h e

To c alc u lat e t he perc ent age of b rown

eggs pr esent, divide t h e nu mb er of

b rown eggs (4 ) b y t h e tot al nu mb er of

eggs (12)

33% of eggs are b rown

263

1. D ivide t h e mas s of e ac h ele me nt in 1 mol of t h e c ompou n d

b y t h e molar mas s of t h e c ompou n d

2. M u l t i p ly b y 1 0 0 %

% Composition of element =

molar mas s of c ompou n d

a fe r t ilize r t o add n it roge n t o s oil Calc u lat e t h e pe r c e n t

N = 2 ( T h e t w o individu al N s [s u b s c r ipt s of 1] c omb ine to

mak e a t ot al of 2 mol.)

O = 3

2. A dd t h e molar mas s of N H

4

N O3

% Comp os i t i on of H =

4 (1 0 0 8 g)80.04 1 g

× 10 0 % = 5 0 37% H

% Comp os i t i on of O =

3 (15.999 g)80.04 1 g

% c ompos i t i on of N =

2 (14 0 0 6 g)80.04 1 g

Determining Percentage of a

Single Element in a Compound

S ome t i me s , s c i e n t i s t s n e e d t o k n ow t h e pe r c e n t age of a

FOR EXAMPLE: D e t e r min e t h e pe r c e n t age of oxy ge n in

15.999 g/mol O

18.0 15 g/mol H

2O

FOR EXAMPLE:

% F e =

molar mass of all F e

molar mas s of F e

2O3

2(5 5 85 g/mol Fe)

15 9.697 g/mol Fe

2O3

always c alc u lat ed b y

t ot al molar mas s of t h e ele me n t i n t h e c omp ou n d

molar mas s of t h e c ompou n d

EMPIRICAL FORMULA

for mu la i s s ome t i me s , b u t n ot alw ay s , t h e s ame as t h e

mole c u lar for mu la It ’s c alc u lat e d u s in g t h e pe r c e n t

c omp os i t i on of t h e c omp ou n d

EMPIRICAL FORMULA

A formula showing the proportions of

an element present in a compound

m o l a r m a s s f r o m

p e r i o d i c t a b l e

269

= K CN

2 Wh at is t h e proc e s s for fin din g pe r c e n t c ompos it ion ?

3 Wh at does t h e empir ic al for mu la mean?

percent by mass of each element in t hat compound.

S c i e n t i s t s u s e pe r c e n t c ompos i t i on t o de t e r mi n e t h e

pu r i t y of a c ompou n d or t h e amou n t of all of t h e

differ ent element s b y per c ent age

2

% Compos i t i on of

an element

=

molar mas s of c ompou n d

3 Th e e mpir ic al for mu la give s t h e s imple s t w h ole - n u mb e r

= 26.20%

Chapter 20

CHEMICAL

REACTIONS

CHEMICAL REACTIONS

Wh e n c h e mic als c omb in e , t h ey mak e a n ew s u b s t an c e Th is

on on e an ot h e r Th e b on ds

b e t w e e n t h e ir at oms ar e

b roken, and new b onds

are c reat ed, forming new

s u b s t an c e s

leave evidenc e of oc c u rring

CHEMICAL REACTION

A process during whichsubstances are changed intoone or more new substances

Evide nc e of a c h e mic al r e ac t ion c an b e :

t h e r ele as e of ligh t

t h e for mat i on of a gas

H ow did t h e s u b s t an c e s r e ac t ?

How did t h e r e ac t ant s c h ange ?

c h ange s t h at oc c u r w h e n c h e mic als r e ac t

r epr esent at ion of a c h emic al r eac t ion

periodic t ab le) to name eac h element

REACTANTS Ch e mic als on t h e r igh t of t h e y ield s ign ar e

Th e n u mb e r pr e c e din g t h e c h e mic al s y mb ol, t h e c oeffic ie n t ,

LAW OF CONSERVATION OF MASS

The mass of the products = the mass of the reactants

Th i s me an s t h at c h e mi c al e qu at i on s mu s t b e b alan c e d

BALANCING CHEMICAL

EQUATIONS

A b alanc e d c h e mic al e qu at ion h appe ns w h e n t h e nu mb e r

of t h e diffe r e n t at oms of ele me n t s on t h e r e ac t an t s s ide

i s e qu al t o t h e n u mb e r of at oms on t h e produ c t s s i de Th e

e qu at i on h as t h e s ame n u mb e r of e ac h t y pe of at om on

b o t h s i d e s

The number of moles of each element on the

left side of the equation must equal the number of

moles of the element on the right side of the equation

277

Steps for balancing

chemical equations:

1. Look for mu l t iple s of t h e c oeffic ie n t

You c an only c h ange t h e nu mb e r of mole c u le s , not t h e

molec u le it s elf Th is me ans c h ange only t h e c oeffic ie nt,

n ot t h e s u b s c r i pt

Forexample, hydrogen can react with oxygen to form water

If you h ave tw os , t r y to dou b le t h at nu mb e r to a fou r

H2+ O2

2)

2 mol H = 2 mol H b alanc ed

5. M u l t iply t h e ele me n t or c ompou n d, w h os e valu e is too

6. Ch e c k Cou n t t h e n u mb e r of mole s of t h e ele me n t s

one new produc t

2+ Br2

1 mol h ydroge n r e ac t s w it h 1 mol b romine and yields 2 mol

h ydroge n b romide

of a s y n t h e s i s r e ac t i on

AB ➜ A + B

2

2 mol mer c u r y oxide dec ompose to 2 mol mer c u r y and

1 mol molec u lar oxygen

HH

r e plac e d b y anot h e r ele me nt Ele me nt s t h at t e nd to for m

1 mol zin c c h lor ide c omb in e s w it h 1 mol c oppe r t o y ield 1 mol

c oppe r (II) c h lor ide an d 1 mol zin c

COMBUSTION: Oxy ge n r e ac t s w it h all ot h e r ele me n t s

in t h e or igin al c ompou n d, for min g oxide s

4+ 2O2

2+ 2H2O

(Th is u s u ally oc c u r s w h e n a h y droc ar b on r e ac t s w it h oxy ge n

to produ c e a c arb on dioxide and wat er.)

283

DOUBLE REPLACEMENT/METATHESIS:

react i n an aqu e ou s s olu t i on , an d t h e c at i on s an d an i on s of

t he two react ant s “ swit ch places” to form two new compounds

S O4

s olu t i on t o y i eld s i lve r c h lor i de

w a t e r

REACTANTS AND PRODUCTS

S ome t i me s , i t is h elpfu l t o k n ow t h e ph y s ic al s t at e of a

s u b s t an c e in an e qu at ion Ar e t h e r e ac t an t s s olids ? Liqu ids ?

It is h elpfu l t o k n ow t h e s t at e s of t h e r e ac t an t s , b e c au s e

It is h elpfu l t o k n ow t h e s t at e s of t h e r e ac t an t s , b e c au s e

You mu s t pic k t h e r igh t s t at e t o ge t t h e produ c t t h at y ou

+ O

2(g)2(g) gggg(g ))

2(g)2(g) gggg(g ))

22 (g ggg

g a s gg

The type of subscript that

is used to show the physicalstate is different from thesubscript in a compound

This subscript is a letter inparentheses, not a number

285

FOR EXAMPLE:

KBr(aq )+ A gN O

3( a q )

3( a q )+ AgBr

3 Wh at is t h e diffe r e nc e b e tw e e n a c h e mic al r e ac t ion and

a c h emic al equ at ion?

4 Wh at ar e t h e main t ype s of c h e mic al r e ac t ions ?

3

2CO3+ H2

2O

3PO4

7 Wh y is i t i mp or t an t t o k n ow t h e p h y s i c al s t at e of t h e

r e ac t an t s an d t h e produ c t s ?

287

t h e c h an ge s t h at oc c u r b e t w e e n c h e mic als w h e n

t hey reac t

2 F or t h e c h e mi c al e qu at i on t o b e b alan c e d, t h e n u mb e r of

e qu al t o t h e n u mb e r of mole s of t h e ele me n t on t h e r igh t

s i de of t h e e qu at i on

c h ange d in t o produ c t s A c h e mic al e qu at ion u s e s c h e mic al

s ymb ols to r e pr e s e nt t h at proc e s s

4 Th e main t y pe s of c h e mic al r e ac t ion s ar e s y n t h e s is ,

single replac ement, dec omposit ion, dou b le replac ement,

2O

3PO4

r e ac t an t s an d t h e produ c t s b e c au s e t h is le t s y ou k n ow

h ow t h e e xpe r ime n t w as ac t u ally c ar r ie d ou t

8 Aqu e o u s me an s t h at a s u b s t an c e h as b e e n dis s olve d

in wat er

289

Mole-to-Mole Stoichiometry

I n mole - t o- mole s t oi c h i ome t r y , w e s t ar t w i t h mole s an d e n d

are represent ed b y moles

2+ 3H2

3 mol of hydrogen gas to form 2 mol of ammon ia gas

(meaning “element”) and

metron (meaning “measure”)

291

FOR EXAMPLE: I n t h e e qu at i on N

2+ 3H2

3, how

man y mole s of N H

3

w ill b e produ c e d w h e n 4 0 mol H

2react s

c omple t ely w i t h N

2

?

a rat io from t h e c h emic al r eac t ion

A problem-solving method that uses the idea that any number, compound,

or element can be multiplied by a ratio from the chemical equation

without changing its value; used to convert units of measurement

Grams in 1 mol N H

3: 14 0 1 g N + 3 (1.0 0 8 g H ) = 17.0 3 g

2.7 mol NH

3

×17.0 3 g

1 mol N H

3

= 4 6 g of N H

3produ c ed

4. Conver t t h e moles of produ c t to grams of produ c t

You don’t ac tu ally c alc u lat e eac h sec t ion of t h e prob lem

individu ally ; in s t e ad, y ou c alc u lat e t h e e n t ir e prob le m as

a w h ole

293

mas s A ➜ mole A ➜ mole B ➜ mas s B

4H

1 0), undergoes

c omb u s t ion t o produ c e c ar b on dioxide an d w at e r

Wit h 236.5 g b u t an e , h ow man y grams of c ar b on dioxide ar e

1. Wr it e a b alan c e d e qu at ion

means t hat oxygen (O

2) is a reac t ant and c arb on dioxide

(CO

2

) and w at e r (H

2O) ar e t h e produ c t s

m o l a r m a s s

m o l a r m a s s( p er i o d i c t a b le)

2. Conve r t t h e amou n t of b u t an e t o mole s

Calc u lat e t h e molar mas s of b u t an e

Ge t t h e mole rat io from t h e e qu at ion 2 mol b u t an e = 8 mol

4. Convert moles to grams of produ c t

Calc u lat e molar mas s of c ar b on dioxide

To c alc u lat e t h e atoms of CO

2, mu l t iply t h e n u mb e r of mole s

FOR EXAMPLE: Hydroge n s u lfide gas b u r ns in oxygen to

produ c e su lfu r dioxide and wat er Wit h 5 6.2 g of oxygen for

t h i s c omb u s t i on r e ac t i on , h ow man y grams of w at e r w i ll t h e

reac t ion produc e?

Th e r e ar e 5 6.2 g O

2(informat ion given)

2

×

2 mol H

2O

3 mol O

2

×18.0 2 g H

2O

1 mol H

2O

= 21.1 g H

2O

produced

==

mol O

==

Mass-to-Volume Stoichiometry

S ome t i me s in gas r e ac t an t s or produ c t s , t h e me as u r e me n t s

S t an dard t e mp e rat u r e an d p r e s s u r e ( STP) =

273

273 K elvin (K ) an d 1 at mos ph e r e (at m)

1 mol any gas

A theoretical gas that consistently obeys the gas laws

The molar volume of a gas at STP is equal to 22.4 L

for 1 mol of any ideal gas at a temperature equal to

273.15 K and a pressure equal to 1.00 atm

297

To solve mass-to-volume

1. Wr it e a b alan c e d c h e mic al e qu at ion for t h e r e ac t ion

2. Conve r t amou n t of r e ac t an t t o n u mb e r of mole s

3. U s e t h e mole rat io from t h e b alan c e d e qu at ion t o

c onve r t t h e amou nt (in mole s ) of produ c t c r e at e d

4. Conve r t t h e mole s of produ c t t o lit e r s of produ c t u s in g

t h e STP c onve r s ion fac t or for gas e s

Moles of B Volume of B

m o l a r m a s s( p er i o d i c t a b le)