Chapter 09 I Counting A toms: The Mole 135
- --- -
What <:Jre the Different Kinds of
9.4 Chemical Formulae?
Chemists commonly use .
. . three kinds of formula e f o r compounds They are·
molecu Iar f ormu Ia, empmcal formula • and structura Iformu 1a. · ·
;!, I
I I )I ~ I
ball-and-stick model
To illustrate the difference
. . between the three k,.nds of f ormu Iae, cons,·d er the
compound ethane. This 1s a compound found in crude oil. H H
I I
• The m olecular formul a of ethane is C2H6 . That is, a molecule of ethane has two. H-C-C-H
carbon atoms and six hydrogen atoms. I I
H H
• The :~tio of the numbers of carbon and hydrogen atoms is 1 : 3. Thus, the structural formula
em pirical form ula of ethane is C1 H3 or simply CH 3 .
Figure 9.3 The ethane molecule
• To show the structural form ula of ethane, we need to make a ball-and-stick
model of the molecule showing the bonds between the atoms. Figure 9.3 shows
a ball-and-stick model of ethane and how its structural formula is represented
on paper.
Are these Formulae Always Different?
Sometimes, the empirical formula and molecular formula of a compound are the
same . An example is methane . The molecular formula of methane is CH 4 . The ratio
of the atoms present is C : H = 1 : 4 . As this is already the simplest ratio of the atoms
present. the empirical formula is identical to the molecular formula.
In the next section , we focus our attention on empirical formu lae. And in Section 7,
Organic Chemistry, you will learn more about structural formulae .
Calculating the Ernpirical Form ula of a C ompound
The empirical formula of a compound can be found from the masses of the. eleme~ts
that combine together. For example, the empirica l formula of magn_es1um oxide
(Figure 9.4) can be found by burning a known mass of magnesium 1n air.
Typical results for such an experiment are as follows:
Mass of magnesium burnt = 0.48 g
Mass of magnesium oxide produced = 0.80 g
~herefore, the mass of oxygen that combined with 0.48 g of magnesium
- O.Bo - 0.48 = 0.32 g
The numb f . nd oxygen atoms that combine together can Figure 9.4 Magnesium oxide formed
er o moles of magnesium a f atoms of each element 1 · I I d after burning magnesium metal in air
then be f d . . s ca cu ate as
sh 1
oun . Tlie ratio of the mo es o
own in Table 9.2. .
b
136 Sec tion 3
1 Che rni cc1I Co k.:ulo tions
' . .. .
I Magnesium (Mg) oxygen (0)
Step 1
' Writ e dow n th e mass or eAch
0.48 g 0.32 g
elem ent .
Step 2
Writ e dow n th e molar mass of
24 g/mol 16 g/mol
each elem ent.
Step 3 l-
Divid e each mass by its molar 1
mass to obtain the 11un1ber of 0 48
· = O02 mol 2:E. = 0.02 mol
moles.
211 . 16 I
I
Step 4
Divide each num ber of moles by 0.02 =
the smallest number. 1 ~= 1
0.02 0.02
Table 9.2
The final num bers are 1 and 1. These
numbers give the ratio of the atom s in
com poun d . Therefor e, the empirical form the
ula is Mg, 0 1. This is normally writt en as
MgO .
The following example produces .,in empirical formula which does not haye a 1 : 1
• t •
.. .
' • '
'
· ratio.
k 'Wor~
--., PFT
e •
·
A sample of an oxide of copper contains ' '
8 g of copper combined with i1 g of oxyg
Find the empirical formula of the compoun en.
d. 1
I
,- -- -- -
Solution - - - - - -. ---
-- I
Cop per (Cu) - -. --
Oxygen (0)
Step 1
Write dow n the mass of each element. 8g
1g
Step 2
Write dow n the molar mass of each elem 64 g/mol
ent.
16 g/mo l
Step 3
Divide each mass by its molar mass to
obta in the number of moles.
8
64 = 0.125 mol
,
16 == 0.0625 mol
I Step 4
Divid e each number of moles by the 0.125
I s1 na lles t number.
--- --- -
~5 == 2 I 0.0625
-
The final numbers are 2 and l These give
,.._
the ratio of the atoms i th
I.. O:Os25 == 1
the empirical formula is Cu20 ,. This is norm • ~
ally written as Cu n 8 com poun d Ther
. 2O. f
• e ore,
Chapter 09 1 Counting Atoms: The M o 1e 137
Calculating the Empiric a l Form ula 1from Percent age
Composition
Th e empirical formula of a compound can also be found from its percentage :
composition . In the ca lculations, we assume that 100 g of the compound is
analysed . Each percentage then gives the mass of the element in grams in 100 g of_
th e compo und . ·
r.
I
, ' I
An oxide of sulfur contains 40% sulfur and 60% oxygen by mass. Find th_e empirical form ~la.
S~ufion /
Step 1
J____
Write down the mass of each element.
---=~-r---
The percentage composition means that 100 g ·of the compound contains 40 g of!sulfur tjnd
60 g of oxygen. Therefore, we will take 100 g of the compound for ou 6 calculatio1.
.-
I
.
1
Sulfur (S)
40 g
___ __~-
Oxyg~n (0)
60 g
1
Step 2
, : ~ :: -o_'!:'~t_!I~ ~olar __m~ss ~ f each element. ,_ _
32
g/mol --+- __ I
1~ g/mol
Divide each mass by its molar mass to
11
.iQ_ = 1.25 mo/ ~ = 3.:75 mo!
,-
1 f.
obtain the number of moles. I 32 16 .'
--· ·-· .. -· --·- - - . -· ·--------- ·-··· - -- -- -- - - - - -- ..
Step 4 I 3 75
1 25 =-= 3
Divide each number of moles by the - = 1 -
srnalles~~um_ber::_ ____ __ ______ -"--
1 25
-
12
- 5 ___ __ J
The final numbers are 1 and 3. These give the ratio of the atoms in the empirical formula~'
The empirical formula is S1 0 3 . This is normally written as S03.
From Empirica l Formula to Molecu lar Formula
The molecular formula can be calculated from the empirical formula and the relative:
molecular mass.
Propene has the empirical formula CH 2 • The relative molecular mass of ~---'
propene is 42. Find the molecular formula of propene . !
Solution
Let the molecular formula be C11H211·
The relative molecular mass= 12n + 2n
= 14n
14n = 42
42
n = 14
:3
.-. The molecu lar formula is C3H5.
8 ~2L. tion 3 , Chen,ical Calc ulations
Uses or Empirica l Formulae
An irnportant use of empirical forrnula calculations is in Organic Chemistry (Section 7).
Aln,ost every day, a new organic con,pound is either discovered or made in the labora~ory.
To find the formula of the new substance, a sample is analysed to obtain the mass or
percentage con,position of each element in the compound. From the data, the empjrical
r:, ractical
LIi Workboolt
forn,ula is then worked out. The relative molecular mass can be determined, follow~d by • Experiment 9.1
the n-iolecular formula.
II f I I ' '
••
1. Find the empirical lormuloo of the following: 3. (a) 17.4 g of a compound of chlorine and oxygen contui11s
(a) a compound consisting of 3.5 g of nitrogen combined 3.2 g of oxygon. Fi11d the empirical formula of the
with 8.0 g at oxygen, compound. ; ,
(b) a cotr1pound consisting of 2.8 g oi iron combined with (b) 66.8 9 of a c?mpound of iodine ancl oxygon contains 16.0 U
1.2 g at oxygen , · of ox~gen. Find th~ empirical formula of the compouncl.
(c) a compound consisting 9f 2.4 g of carbon, 6.4 g of [ calculBttng]
oxygen and 0.2 g of hydrogen, and
(d) a compound consisting of 9.2 g of sodium, 12.8 g of For O.uestions 4 to 6, use relative ~tomic rnassos in Table 8.1 tu
sulfur and 9.6 fl of oxygen. holp you answer the que~tions. . •
{1:alc11/nting\ 4. The o:1,piricnl fo~mula of a compound is C0H -1, Its ro lativu
2. Fllld t11e 1m,pirical 1ormu1Au ot the fallowing: molecule~ mtiss is 62. Find tlHi moleculi:ir f on~uln
(a) n compound with co111positin.n 75% curl>on und 25% [ c11/c11/ot111gl ·
llydrogon by mass, .. 5. Tht:1 ontpirical tonnulu.of u corn .
{b) u 1:ompound with composition 4o.7% s1llco11 and 53.3% molecular mass is Fini , pound is H:,CO. ll s rnl1111v11
90 th
oxygen by mass, lcolculatingl · ' c o rnolecu la1 lor1111t111 .
(c) u compound with co111position 43.4% sorlium, 11 .3% r:arhon
r1n d 45.3% oxygen by mass, nnd 6. Tha empirical formula of u ,
, molecular muss is 46 F' I compound IS H~co~. lls I o l11ll\ll
1
(ct) " compound with composition 48.6% corho11, 43.3%
lct1/c:11/atingl . · · ll1C th o molHculnr fo11rn 1L 1
oxygcrn unll 8.1% hyclronon by muss. 1
I cn/c:11/Rtingl
