D.H.S.G
Amount of Substance
. /\.
.., .. 0 c;c;, •
Amount of su6stance and tlie mole
~.&adro's constant
fl'\t
N.11- .r-o~ ~
numhtr of PClrtfCICS per moce Of PClftllleS
Moles andvolumes (Solutions a nd Gas~sl
Volume:
l va1ume
Volume -;;;-easurementsJcm• and
_
: GOl'lv
~~;IJ
Science Department
fleacting·Quantities,
Stolchlomet!'.}; (mole ratios!: rr.tt·io Of motCSin CQ\,l(IU0r')
CI:U JO :a:t f!:Ht~ I) IWOYl ti Ii( c;. r!QUlteU !0PYOHllt
CCYIQ i n ~U(IYllill/ Of proQlvct
•
e0111 CltS: 1. J cn111nr1r i ec; 1Y1t1r sno1,11C1 ..,,,
• .;, I 00 ' 0
\ m, lnumn er Of moc ec; .,. N-:7 lm 3 - -d l'Yl} <'-tm l
Q,uantitlesJrom.amounts aJ!l!.lqu~
f0Yrfl(C(
MolarMa.2_ \ q mo l ·•)
~ nn •11 WOH101.41 Pmo1.1nr fl'lmOl(SOI wnar 1,/0UOYC
u,e mo s ~ \Ql or one mo,c or HAbc;rance
Concentration: - qivtn
mo1t ~ l21 usetnteauahon1oworwouramoun1in mo1e,; or
-(QI on t molt c • 11 ·OCl ~tone ·- VO 11.im t
1.1n11novrn cauotrc,n
on, molt 11, 1- og Ill won101.1r un1.1nown •nformorr"on reQuireu
Example: Calculate the amount of NaCl, in Mol, In 30.0 cm• of a 2.00 mol
1
Determlnatlon of !ormula~ . dm·• solution Example Calculate the mass of aluminium oxide (Al 20 3), formed when 8.10g
Mole cu la rl=ormul ae - - -~o rm'= 0·03dm 1 of aluminium completely reacts with oxygen.
tl'\t ~ •type or ,ocn ,11mtn 1 in a compouncl molt r," LtAIC~l + ~OHQ> tl'll1011<.l moss , o -1s, 101
0-0~, 1-00: 0 ·0600mo1
rnou~ t'llll • 11-100 °" 1'1- ·o, o-;mo1 1 : IS · 3 I)
Example: Calculate the volume of a 0.160 mol dm·• solution that contains t : . o -1,;mo1 Alt 01
1.1 ••
Em[!lrlcal Formula . 3.25 x 10-3 mol of NaCl.
tnt c;impir<;t wnoir numbtt Lll!..LQ. Of atoms. 01 Example: Calculate the concentration of lithium hydroxide solution and
vo11.1mt,l~ s20-3tm~
ealn t1tmtn1 prrc;rnr in o compound ~-t<;'II o•\ volume of hydrogen gas produced when 0.552 g of lithium reacts with
water to form 125 cm' 1 l1· + lH10111 - ZLiOH1oq1 + Hzcq)
Finding Formulae: Chemical analysis of a compound gave the percentage Molar G~s Volyml!: \.it'YlOl.tS= o · <,<,t~ 6 · "1' 0·0900mOl
composition by mass C: 40.00%; H: 6.67%; 0: 55.33%. The relative formula vowm C<fWf J } e 1: .. o· 0>100 mo1 1-1,
mass of the compound Is 180. \ mo1ec,
' t'-l·O CcAYYl') '/IT n-P Li - H 1 ' I
·1. m mr lmOlfS ~c;mo11 rcuro cone (Li OH): o-OQOO = 0·6LIOmol c;1m· 1
Example: Calculatefhe amount (mol)o ffi yarogen (H ) in 480 cm' at RTP 2
O · 1 l ':>
C 1-10·1 . 11. ·o H~ I I
molts, ...!:Li.Q__. 0·01oomo1
H
0
(,-(,~'! I ·O
S<, ·V - 16 ·0 Vi',
f; · P' 1
I -
I
I
-tl'Ylpirico1: CH,0
'-'> m r: !O
II.I 000
Example: Calculate the volume (dm 3) of 0.150 mol of oxygen (0 2) at RTP.
vo1ume: o -,c,o, 1•1
t-11 lVOI):
.
O·Q'-1001< tLI
Cl60Gfm 3
0 o · q 6 cm 1
1&'0~'!,C, : {, : . 6,CHiO C1, H ,1 0, : i-GOc;lm 3
'
The Ideal gas equation:
\.< II : Pcnl0 3
llyllratell ~I\S~l
war n of cr y c;.1C111·~ 111ion, w111tr mo1ecu1t!> bonc:,1ec;1
Li! V, n RT[ Percentage_yield:
, ~OUQl!fltl~
: '1- 100 { imp1.1r1tres
into n,e cr11~1aunt ~1rucrute 01 o nvccratcGI compound p = I'll I!= S -31u rmo1·•w··
t ne Or ct i c II I I/ i t I cl <,itH rcoctron
1/, ml - ... mo11notnave cor IJ I ttect
Finding Formula of a hydrated salt: T = If. limiting reactant
~<;um,:itll>n-S] , , CAii 1"10 rer I<;
fl' mo ic,
•O~t th! YC<ltllnl tho t (S l'\OT in tHses<; .
2) no 1vrtntr GIHOmposi ti On Example: 0.1565 g of Xoccupies 80 cm' at lOlkPa and lOOOC, calculate the
Use the results below to determine the value of x and the formula of relative molecular mass of X. p, 1o I w Po , I O I , I O 3 Po
,I
PV ,.9..:.2Q!_ 11,tom.econ_om~:
CoCl2•xH 20 n' v: 9ocm 1 , s,-0110 -sm 1
M•ss of crucible/g
Mass of crucible + hvdrated salt/Ii'
Mass of crucible + anhvdrous salt/"
I17.265
I 18.438
I17.906
I
I
I
liT
motes, i - ~0110·"
mr = moss . o
5·1 (iLl·i'H
-,r,,s
(l. a 9 · '!, IU
t , 100°c • ~,;1<.
n,,
;um 0-1 mo1a r mcisHs or aesirre,I
<,um mo1eu rrit1sc;es Of prociutts
• I 00 I
moies - ~ o - u S_ustalnabll~
f (J,VOIAr~ rt(lltion~ Hln1 tl niqn Qtom economy;
mr, tiOI
Hwtr YOIW MC,IIC¥(CIIS (11ft IAHCI, ICSS W(l'i)t I<;
proctucea I
.,I
GIISO 1nc1uar<; fO<;~il f\lllS , Hriitt Yt',01,111(~
