Many problcms in gcncral chcmisrry involvc litttc morc than a changc in uniu-.- ih.y

CHAPTER 5

UNIT CONVERSIONS

Many problcms in gcncral chcmisrry involvc litttc morc than a changc

ih.y ".. rcidily solrcd by thc so-callcd "convcrsion factor
mcthod.'i Whiti thc'mathcmatics of rhis merhod is cxtrcmcly simplc, bc'
girining srulcnts ofidn fiil ro'appreciate its gcncral applicabilitp
T; illustratc thc'convcrsion factor approach. lct us apply it to a Par'
dcularly simflc irrob'lcm-: Supposc wc arc aslccd (o convcn a lcngth of 22
inchcs into [cct. To do rhii, wc makc usc of the convcrsion [actor:.
in uniu-.-

I ft =

l2in

(5-l)

Dividing both sidcs o[ this cquation by 12 in givcs a quoticnt which

is

cqual ro unity:

lfi
l2 in

I

--

lf wc now multiply 22 in by this quoticnt. wc do not changc rhc valuc o[
thc lcogrh, but wc do accomplish thc dcsircd convcrsion o[units:
22in x

lft
l-8 fr
l2tn =

Thc convcrsion factor givcn by Equation 5- l can bc used cqually wcll
wc
ro convcr( a lcngth givcn in [cet, lct us say 2.5 [t. to inchcs. In this casc.
dividcboth sidcs o[rhc cquation by I tt to obtain:
12

in

lft

CHAPTER 5

54
Muhiplying

2.5

ft by thc ratio

:

!!! ton""'

z.Sh- ,

l?Jn

th.

-

thc lcngth from fcet to inchcs:

30 in

(c'g'' 1 6 - 12 in) will always givc
Noticc that a singlc convcrsion factor
fi) which arc cqual to unity' In
in/l
q.",l.n,rlt f,7tZ in or 12

"r,*. .on,.rrion, r*c choosc rhc guoticnr which will cnablc us to canccl
-"fi.g "
out
-iil,unit that wc wish ro gct rid of'
--' thcpaniJubrly valuablc whcn wc arc dcaling with unapproach is

familiar units (ExamPte 5'l )'

Exomplc

5.1

Convcn 8'32

version factor I crg

-

2'39

x

x

lOi crgs ro calorics' using thc con'

l0-r cal'

cffcctivc usc ofthc convcrsion factor approach'
and
;;;.; " q,roti.n, in r"hicb cal'orics aPPcar in thc numcrator
is:
guoticnt
This
ienominaror'
crgs in thc

Solution To makc

:

1'3e

.

I-l9{cal = ,

Multiplying 8-32 x lOt crgs by this quoticnt:
E.32

x lo'os, *

4fl&!

-

l'eecal

by this mcthod' To
Multistcp convcrsions arc rcadily accomplishcd

factors:
convcrsio.n
thc
using
nna il. ,rr*b., of ,oonds in thrcc days.
I day

-

24

hrs;

I hr

-

60

min;

I miir = 60 scc

by the guoticnt 24
wc mighr 6rsl con,cn &ys to hours by mukiptying
hrs,/ t day:

3rravsx?f?="n"
convcrsion factor:
thcn convtrt ro minutcs' making usc ofrhc second

?2hcxH=432omin

55

UN'T CONYERSIONS.
a

and,6nally, cirangc from minutcs to scconds:
4320

tnh

^

60 scc

I mi{r

-

2-592@

x

lOt gcc

ofcoursc' ncccssary to
In carrying out a oultistcp convcrsion, it b not'
abovc cxamplc' Indccd, if

thc
in
wd
did
as
ans*crs,
ina*r,i.diatc
fo.
,ofr.
out with a slidc
;""..tti* ;rftiptications and divisions arc to bc carricd
linc' as in
a
singlc
on
problem
up
the
sctting
by

oftcn sar. rimc

.

;t,;:

Examplcs 5.2

I atm

-

ind

5-3-

I

760 mm Hg;

lit -

1000

ml;

I lit atm

-

24'22 cal

Hg'
Convcn-I.99 cal to thc cncrgy unit mlxmm

.Sototion. ln prgblcms of,this typc,

'

a.tt'".ir;;a';idiis
dicatcdsitir; cr;i

:.. -....,..:'.

1.99

bctore

fu

is hclpful to analyzc thc

p'tting i'iy

suc'

numbirs on papcr' Thc in-

2a'22cal)
.(l lit atm ml
(l
:
lir
)
aim :'
iZi;..Qgn-v9a'lit arm'to ml
"1000
760
mm Hg)
(l
atin
Hg
mlxmm
iri con.it, qnl3tm.to

(l)

Gonvcrt 1.99 cal to

cal

, ,;f6i

,

(l )

litcnatm

ml
lli(
(zl
-..x
1000

760 mm Hg

I et€
(l)

- 6.24 x l0'mlxmm Flg

Thc dcnsity o[ mcrcury (H-S)
is thc mass in pounds o[ 1.50 lit o[ mcrcury?

Eromple

5.3

it l3'6 g/ml' What

lo
Solulion Wc arc. in cffcct. askcd ro convcrt liters of mcrcury
pounds. A fcw momcnts rcflcction might suggcst thc lollowing
rhrcc-stcp path:

Hg; . (l lit - 1000ml)
(2) ml HS - g HS; (l ml Hg + t3'6 g)
(3) s Hs - lb Hs; (l lb -

(1) lit Hg

-

ml

-a5l'6s).

l.ioli( x 1a'

(t)

*

l;i-

(2)

x

IJJG
(l)

E tJ'u

GT'APIEN 5

s6

=, which-mians "cquivalcnt to"'
'Mathcmaricaily. convcrsion factors rclating cguiva'
;;Jmanncr'
guantitics arc handlcd in the ordinary
In this problcm, thc symdbl

*"t
lcnt

applied lo a rr'idc variety of
Ttrc convcrsion lacror approach can bc
considcr how onc might
cxamplc'
simplc
a
As
orobt.-, in chcmisrry'

;;;i;,i"

n,rmbcr

if

molcs of CaClz' givcn
Srams in 2'60

that:

- ltlg

ImolcCaClr

molc' wc obuin;
Dividing both ddc ot this cquation by I

lll g

r

I molc
Hcncc: nd.g'- 2.6omolcs

t

#L

-

289 g

. E*;Pi. i.4.iuustrat& lhc dpPlicaiion of thc cirnvcrsion factor.aP:
pr"""i-it ,rr. olcularion of gram cquivalcnt wcights of clements'

Eromplc

5-4

Thc gram cquivalcnt

w:tC\".f

an elcmcnt can bc

o:y.g:n'
Jln".a as thc .'=ighi rhat combincs with cighr grams- o[weighing
oxidc
mctal
ccnain
of
a
samplc
a
ihir
;;-t;;,6ods
i.fi;;J& l2l{ gof mctat rhcn rcduccd by hydrogen' Calculatc.d,-.

ir.-

cquiralcnt wcight of thc mcral'

metal which is
Solution Wc arc rcquircd to 6nd thc wcight of
Wc
nccd a con'
orygen'
of
gnms
crghr
ro
chcmirzlly cquirdcot
From
of
oxygen'
vcrsira Ectc rchring g;-* *t*td to granrs

ti. a.,a

wc dcduct

ilet ttr rrrtal oxidc samplc must

havc con-

taincd:

- l2l{g - 02ilgoxygcn
Hcnct: l2la 3 cd + A2i1 g oxygcn
t-/t6Eg

cEw@J - &tffig:crycs

.

ffi

= 38.2gmctats

'

37

UNII CONYERS'ONS

in solving problcms-dcal ing
cquation givc
with chcmical cquarions. Th. co.ffi.i.t ts of a balanccd
o[ diffcrcnr
molcs
o[
numbcrs
Ji.."tty tt. conr.rsion lactors rclaring thc
Convcrsion factors arc panicutarly uscful

,ubr,"to

participating in a rcaction' Thc balanccd cquation:

+ 7 Ods) -; c COr(g) + 6HrO(l

z CiH.(g)

)

7 molcs of oxfgcnr
cquivalcnt

C:Hr
tclts us that in this rcaction. 2 molcs of cthenc'
i.not., of ."rUon dioxidc' and 6 molcs of watcr arc chcmically

'

lo cach othcr.
2 molcs C2H1

+

7 molcs 02

+

4 moles

COr

*

6 molcs HrO

of CrHe rcquircd to
I[ wc wcrc askcd to calculatc thc numbcr of molcs
writc:
*ith l6.O molcs of 01, wc would immcdiatcly

rc"cr

.

: l6.0rnolesqo

9f' '
*.2jl9Y
^' 7 rnolcsQ.

4'57 molcs CrHo

of any spccics is cqual
Funhcimorisincclthc mas' in grams of onc molc
to its granl fo.rmula wcight, wc havc: .. ...

.

.

- 2112.0 st + O ti'o sl = 30'0 g CrHr
- 32'0 g Or
I molc Or - 2(16.0 g)
I molc.COr - 12.0 g + 2(16'0 g) - 4a'0 g COz
' I molc HrO - 2(1.0 g) +'16'0 g - l8'0 g H1O
l.

molc

9rH..

with rhc
Using thcsc convcrsion factors in combination
varicry of problcms
work
a
wc
can
cquation.
batanccd
thc
g-ir;;Uy

molc rclationships

iot.-*ot

of

(Examplc 5'5)'
. gram-molc, or gram-gram typc

of
Eroraplc 5.5 Givcn thc balanccd cquation lor thc combustion
cthanc. calculatc:.

a. Thc

,,r*L.,

of grams of O1 rcquircd to react with 1.60

molcs of CrHe .
o[ molcs of COr formcd [rom l2'0 g of 01 '
Thc numbcr of grams of C1H1 rcquircd to form 2J2 g

b. The numbcr

c.

ol

HrO.
Solulion

a.

convcrsion' Wc 6rsl convcrr 1.60.o1., of CrH. to mol.s o[ O,. usinE tfic coeffWc can rcgard this as a two-stcP

thc

CHAPIER 5

58

cicnrs o[ rhe balanced cguation (2 molcs C2H3 h 7 molcs
Or)- Thcn wc conYcn to grirms of 01 (l molc 01 -

3Z0g Or).

r.6onpt*Gttr.
b.

r

r

ffi

- l?egor

ffi

Procecfing in I manncr cntirely analogous to that in srep
(a), wt coovcn

(t)

Sof

Or-molcsof

@) molcs Oz - molcs

r2.ose. ,.

; (t molcOr - 32.0gOr)
of COr; (7 molcs Or + 4 molcs COr)
01

,

ffi

-

ffi

o-2t4molccor

c- -Tbisprobltm may aPPcar to bc'somcwhat morc complcx,
ti.G il t oot cndrcty obvious what path to follow. Wc

r"!!rf .,gatf-- ttrc proftggljnjil-hg..9! two cquivalcnt

we)r
_
i- ntallii.g rhar thc cocfrcicnts of the balinccd cquation
- qivc us a 6-nriiisbo facror rclating molcs of CrHr ro molzr o[ HIO''.
-dtrortrh
moi.s as an intcrmcdiate, fotlowing the
L -,rfa rrod

.

path:

3

HrO --l;

Hcnce:2.92g-Hzo

moles

H2O

b

'

g'gC2H3

mol.s C2[I1

t motc*l*O 2 moler€rl{e
-ffi1ffi
'

(a)

'

x

(b)

30.0 s CrH
+
I motc€tlte

l.62gC1He

2. We look for a convcrsion factor relating g of HlO to g of
C2H1, rcaliiing that:

.

2 molcs C2H1

+

6 moles

HrO

.

Sincc onc molc o[ CzHr wcighs 30-0 g and
wcighs t 8.0 g, wc havc:
2(30.0 g) CrHe.a 6(18.0 g)HrO

60.0gC2H1+l08gHrO

I

molc of HrO

UNII

59

CONYERS'ONs.

Hcncc: 2.929H2o x

- l.62gcrHr

ffi

This approach can' o[ coursc, bc uscd lor any multistcp convcrsion- It.involvcs combining onc or morc convcrsion lactors ro
havc
6nd thc panicular factor that i,c nccd. [n pan (b)' wc could

that:

rcasoocd

7 molcs

-

but, sinci I moleofOr

Or

^

4 molcs C'Ol

32-0gofO1

:

a 4 ruolcs COr
224 gOt + 4 molcs COr

7(32.0 g

l2.o

g-or :.

O:)

-

##=

o-2r4 morc

cor

Convcrsioo'faciors'can i[s6 bc usid in working problcms dcaling
cncrgy rclationships in thcrnical rcactions (Examplc 5.6)..

Eromple

5.6

'r

'--"

:::::

u'ith

For thc rcaction:

CrHrk) + 5Or(g)

-

3COr(S) + a HrO(t)

AH = -525 kcal. lnothcr words, 525 kcal ofhcat is cvolvcd whcn
onc molc of C1H1 burns. Calculatc:
a. Thc amount o[ hcat cvolvcd whcn onc gram of C1 H. burns'
"'. b. Atlwhcn onc molc otCOris formcdSolution

a. Our convcrsion

factor is:

-525 kcal ^ I molc CjHr = 44.0

g

ClHj

- r.oosQ". , ffiHil -

-n.e

For onc gram of

An
i.c..

I

l

9 kcal

C3 H3 ,

ofhcat

wc havc:

is cvolvcd.

kcar

t0

CHAPTEN, S

'

b.
l.

LH-

+ I molc CrHr
I rirolc ClHj a 3 molcs COr
-525 kcal

I nrohCo,

*

t molcGrl{r
3

-525 kcal

" I muH*Ir
rnolcs@r

Thc principal advantagc o[ thc convcrsion lactor approach is that it
forccs us to analyzc a problcm to dctcrminc whar path wc arc going ro fol-

ger from rhc information wc arc givcn to thc quantity which is rcadv.antagc of sctting up the arithmetic in a
form u,hich is rcady-madc for slidc rulc calcularions.

torr, ro

quircd. tr has thc additional

:";;:,

.:""

i.:-:i:'.ra1:::::l

f:'.ii:."

-:,:.;??;?":{:IiF-:;*tEtriiiSf:i'EiIiF;EIEt

PROBLEMS
.

Solve cach o[thc follou'ing pr6blcms.using rhc convcrsion facior'aiproach. lf rhc appropriatc. convcrsion factgrs.arc not givcn,'look them uP
in

yourtixt

5.1

or in a [and!-oo!!.-

A strip of rin foil

- a.mm'...:'
b-

b.

5.3

o[a

-

c.. iriches
d. Angstroms
.

sodium atom is calcutatcd to bc 26.9

141'.

Calcu-

c. in'

c*!
litcrs

A ccnain gas €xcns a prcssurc of 6t8 mm

in:

.

cm. Exprcss its lcngrh in:

meters

5.2 The volumc
late its volumc in:
".

:

has a length of 3.28

Hg.

Exprcss its prcssurc

a. atm

b. dyncs/cmr

5.4
1.00

Thc dcnshy of u'arer.is 62.i

lb/fi!.

Show that

it

has a dcnsity of

g/cm'.

5.5 An oxygcn molcculc ar 25'C has an avcragc vclochy of 4.82
cm/scc. \{har is its velocity in milcs/hr?
5.6

x l0'

Thc gram cquivalcnt wcight of a mctat can bc takcn to bc thc wcight
that combines with t6.0 g of sulfur.' lt is found that whcn 1.200 g o[ a cer'