chemistry lecture slide week 2

LECTURE 2

Contents:

• Empirical & Molecular formula• Oxidation Number• Chemical Reaction• Stoichiometry calculation• Limiting Reagent

• Formula kimia bagi sebatian ditulis menggunakan simbol bagi menunjukkan semua unsur dalam sebatian mengikut nisbah tertentu.

• Formula empirik Formula empirik merupakan formula paling ringkas yang menunjukkan nisbah terkecil bilangan atom bagi semua unsur yang wujud dalam satu sebatian.

• Contoh: CH2O menunjukkan nisbah unsur C, H dan O dalam sebatian ialah 1:2:1.

• Formula empirik tidak menunjukkan bilangan sebenar tidak menunjukkan bilangan sebenar atom-atom yang bergabung dalam suatu molekul.

• Formula molekul ialah formula yang menunjukkan bilangan sebenar setiap atom yang wujud dalam satu molekul.

• Terdapat yang mempunyai formula empirik yang sama tetapi mempunyai formula molekul yang berbeza.

• Perkaitan antara formula empirik dan formula molekul diringkaskan seperti berikut:

yang mana n ialah integer

Apabila dianalisis, suatu sebatian didapati mengandungi 25.5% Mg dan 74.5% Cl mengikut jisim. Apakah formula empirik sebatian itu?

Andaikan jisim sampel sebatian = 100 gJisim Mg = 25.5 g dan Cl = 74.5 g

Unsur Mg ClJisim (g) 25.5 74.5Bil mol atom 25.5/24

= 1.06374.5/35.5

= 2.098Nisbah terkecil

1.063/1.063 = 1

2.098/1.063 = 2

Formula empirik sebatian = MgClMgCl22

Pembakaran 0.202 g suatu sampel bahan organik yang mengandungi unsur-unsur C, H dan O menghasilkan 0.361 g karbon dioksida dan 0.147 g air. Jika jisim molekul relatif sebatian ialah 148, tentukan formula molekulnya.

% O dalam sampel = 100 – (48.51 + 8.07) = 43.42%= 43.42%

Jisim C dalam CO2 = 1244

X 0.361 = 0.098 g

% C dalam sampel = 0.0980.202

X 100 = 48.51%

Jisim H dalam H2O = 218

X 0.147 = 0.0163 g

% H dalam sampel = 0.01630.202

X 100 = 8.07%

Pembakaran 0.202 g suatu sampel bahan organik yang mengandungi unsur-unsur C, H dan O menghasilkan 0.361 g karbon dioksida dan 0.147 g air. Jika jisim molekul relatif sebatian ialah 148, tentukan formula molekulnya.

Jisim formula molekul = n x jisim formula empirik 148 = n x [ (3x12) + (6x1) + (2x16) ] n = 2

Formula molekul sebatian Formula molekul sebatian = 2 (C3H6O2) = CC66HH1212OO44

Unsur C H O

Jisim (g) 48.51 8.07 43.42

Bil mol atom 48.51/12 = 4.04

8.07/1 = 8.07

43.42/16 = 2.71

Nisbah terkecil

4.04/2.71 = 1.5 @ 3

8.07/2.71 = 3 @ 6

2.71/2.71 = 1 @ 2

Formula empirik sebatian = CC33HH66OO22

• Amaun bahan yang wujud dalam bentuk gas biasanya diukur dalam unit isipadu.

• 1 mol gas pada suhu dan tekanan piawai, s.t.p. (T=273.15 K, P = 1 atm) menempati isipadu 22.4 dm3 dan dinamakan isipadu molarisipadu molar.

• Larutan adalah campuran bahan larut dalam sejumlah pelarut. Kepekatan larutan boleh dinyatakan dalam sebutan:

a.a. Kemolaran Kemolaran – bilangan mol bahan larut yang terlarut dalam 1 dm3 (1000 cm3) larutan (unit = mol dm-3).

b.b. KemolalanKemolalan – bilangan mol bahan larut yang terlarut dalam 1 kg pelarut (unit = mol kg-1).

Hitungkan kemolaran larutan yang disediakan dengan melarutkan 3.42 g sukrosa (C12H22O11) dalam 500 cm3 air.

Bilangan mol sukrosa = 3.42 = 0.01 mol 342

Mr C12H22O11 = (12 x 12) + (1 x 22) + (16 x 11)= 342

Kemolaran = 0.01500/1000

= 0.02 mol/dm3

Hitungkan kemolalan larutan yang mengandungi 6.50 g etilena glikol (C2H6O2) dalam 200 g air.

Kemolalan C2H6O2 = 0.105200/1000

= 0.524 mol/kg

Bilangan mol C2H6O2 = 6.5

2 (12) + 1 (6) + 2 (16)= 0.105 mol

How to prepare 1 dm3 of 0.1 M CuSO4 starting with 2 M CuSO4 ?

50 g of benzene (C6H6) is dissolved in 80 g of toluene (C6H5CH3). Calculate the molality of the solution.

n C6H6 = 50

12 (6) + 1 (6)= 0.640 mol

Using M1V1 = M2V2

= 0.1 M x 1 dm3

2 M= 0.05 dm3 = 50 cm3V1 = M2V2

M1

Molality C6H5CH3 = = n C6H6

80/1000 kg= 8.01 mol/kg0.640 mol

80/1000

c.c. Pecahan mol (X) Pecahan mol (X) Bilangan mol bahan larut dibahagikan dengan jumlah

bilangan mol semua komponen yang terdapat dalam larutan.

Contoh: Larutan yang mengandungi sebatian A, B dan C. Pecahan mol bagi sebatian A dalam larutan ialah:

nA, nB dan nC adalah bilangan mol A, B dan C masing-masing

Pecahan mol tidak mempunyai unit dan jumlah pecahan mol semua komponen bernilai 1 iaitu:

d. Peratusan mengikut jisim atau isipadud. Peratusan mengikut jisim atau isipadu Kepekatan larutan dalam sebutan peratusan boleh

dinyatakan sebagai jisim bahan larut/jisim larutan (w/w)% atau isipadu bahan larut/isipadu larutan (V/V)%

Contohnya asid sulfurik, H2SO4 berkepekatan 36% mengikut jisim mengandungi 36 g H2SO4 dalam 64 g air. Bagi larutan 5% NaCl mengikut isipadu, terdapat 5 g NaCl dalam 100 cm3 larutan.

• In solution prepared by dissolving 24 g of NaCl in 152 g of water,

• 20% w/w HCl → 20 g of HCl dissolved in 80 g of water.

Calculate the percentage by mass of NaBr in an aqueous soluting containing 8 g of NaBr in 80 g of water.

% w/w = 8

(8 + 80)x 100%

% w/w NaCl = 24 g x 100%24 g + 152 g

= 14%

= 9.09%

e. ppme. ppm Kepekatan larutan yang sangat cair dinyatakan

dalam bahagian persejuta (ppm).

Calculate the volume of K3Fe(CN)6 used to prepare a 125 ml 85% v/v solution.

% v/v = Volume of K3(CN)6

(125)x 100%

= 106.25 mlVolume of K3(CN)6 = 85 x 125

100

The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.

The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.

1.1. Free elements Free elements (uncombined state) have an oxidation number of zero.

Na, Be, K, Pb, H2, O2, P4 = 0

2. In monatomic ionsmonatomic ions, the oxidation number is equal to the charge on the ion.

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2

3. The oxidation number of oxygen is usually –2. In H2O2 and O2

2- it is –1.

4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1 (hydride eg: H2S).

6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.

5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.

HCO3-

O = -2 H = +13x(-2) + 1 + ? = -1

C = +4

Oxidation numbers of C in HCO3

- ?

Determine the oxidation number of the following underlined elements:a. Cu2O b. SO4

2- c. ICl3 d. OF2

a. Oxidation no of O = - 2 Let oxidation no of Cu be x 2x - 2 = 0 x = +1

b. Oxidation no of O = -2 Let oxidation no of S be y y - 8 = - 2 y = +6

c. Cl more electronegative than I Oxidation no of Cl = - 1 Let oxidation no of I be q q - 3 = 0 q = +3

d. F more electronegative than O Oxidation no of F = - 1 Let oxidation no of O be r r - 2 = 0 r = +2

• Persamaan kimia menggunakan simbol dan formula bagi menggambarkan secara ringkas perubahan yang berlaku dalam suatu tindak balas kimia.

• Tindak balas am:

A dan B adalah reaktan manakala C dan D adalah hasil tindak balas.

Tanda ‘+’ = memisahkan setiap reaktan dan hasil tindak balas.

Anak panah = bermakna ‘untuk menghasilkan’

Example: 3 ways of representing the reaction of H2 with O2 to form H2O

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

IS NOT

2 grams Mg + 1 gram O2 makes 2 g MgO

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

2C2H6 NOT C4H12

C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbonon left

1 carbonon right

multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right

multiply H2O by 3

3. Start by balancing those elements that appear in only one reactant and one product.

4. Balance those elements that appear in two or more reactants or products.

multiply O2 by 72

2 oxygenon left

4 oxygen(2x2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3x1)

= 7 oxygenon right

C2H6 + O2 2CO2 + 3H2O72

remove fractionmultiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O2C2H6 + 7O2 4CO2 + 6H2O

5. Check to make sure that you have the same number of each type of atom on both sides of the equation.

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products

4 C12 H14 O

4 C12 H14 O

4 C (2 x 2) 4 C12 H (2 x 6) 12 H (6 x 2)14 O (7 x 2) 14 O (4 x 2 + 6)

Ion-electron methodIon-electron method

• The ion-electron method is mainly used for balancing the redox (reduction-oxidation) equations.

• Reduction is a process whereby the oxidation number of an atom decreases.

• Conversely, oxidation involves an increase in the oxidation number.

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2-

2Mg (s) + O2 (g) 2MgO (s)0 0 2+ 2-

Oxidation half-reaction - Lose e-. - Also a reducing agent by giving up its electrons causing the other substance to be reduced.

Reduction half-reaction- Gain e-. - Also a oxidizing agent removes electrons from another substance by acquiring them itself; thus the agent is itself reduced.

Example: Balance the equation showing oxidation of Fe2+ to Fe3+ by Cr2O7

2- in acid solution?

Using ion-electron method (reaction divided into two half reactions)

1. Write the unbalanced equation for the reaction ion ionic form. Fe2+ + Cr2O7

2- Fe3+ + Cr3+

2. Separate the equation into two half-reactions.

(reducing agent) Oxidation:

Cr2O72- Cr3+

+6 +3(oxidizing agent) Reduction:

Fe2+ Fe3++2 +3

3. Balance the atoms other than O and H in each half-reaction.

Cr2O72- 2Cr3+

4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms.

Cr2O72- 2Cr3+ + 7H2O

14H+ + Cr2O72- 2Cr3+ + 7H2O

6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O

6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients.

6Fe2+ 6Fe3+ + 6e-

5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction.

6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2OFe2+ Fe3+ + 1e-

7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides The number of electrons on both sides must cancel.must cancel.

6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O

6Fe2+ 6Fe3+ + 6e-Oxidation:

Reduction:

14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O

8. Verify that the number of atoms and the charges are balanced.

14x1 – 2 + 6x2 = 24 = 6x3 + 2x3

Stoichiometrically Equivalent Molar ratiosStoichiometrically Equivalent Molar ratios

• Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

• Use moles to calculate the amount of product formed in a reaction – mole methodmole method.

• Therefore, the stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance.

• Example: N2 (g) + 3H2 (g) 2NH3(g)

• The equation can be read as “ 1 mole of N2 gas combines with 3 moles of H2 gas to form 2 moles of NH3 gas”.

• In stoichiometric calculation stoichiometric calculation – 3 moles of H2 are stoichiometrically equivalent to 2 moles of NH3 – 1 mole N2 are stoichiometrically equivalent to 2 mol NH3 – 1 mole N2 are stoichiometrically equivalent to 3 mol H2

• The conversion factors conversion factors from this equivalence:

1 molecule 3 molecule 2 molecule

3 mol H2 2 mol NH3

2 mol NH3 3 mol H2

and

The general approach for solving stoichiometry problems:

i. Write a balanced equation for the reaction.

ii. Convert the given amount of the reactant (gram @ other unit) to number of moles.

iii. Use the mole ratio from the balanced equation to calculate the number of moles of product formed.

iv. Convert the moles of product to gram ( or other units) of product.

• Example: 16.0 g of H2 react completely with N2 to form NH3. How many grams of NH3 will be formed?

grams H2 moles of H2 moles of NH3 grams of NH3

16 g H21 mol H22.016 g H2

x 2 mol NH33 mol H2

x 17.03 g NH31 mol NH3

x

Grams of NHGrams of NH33 = =

= 90.61 g NH= 90.61 g NH33

N2 (g) + 3H2 (g) 2NH3(g)

Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH 1 mol CH3OH32.0 g CH3OH

x 4 mol H2O2 mol CH3OH

x 18.0 g H2O1 mol H2O

x

= 235 g H2O

• Goal of reaction is to produce the maximum quantity from the starting material.

• Consequently, some reactant will be left over at the end of reaction.

• The reactant used up first in a reaction is called the limiting reagentlimiting reagent - the maximum amount of product formed depends on how much of this reactant was originally present.

• Excess reagents Excess reagents are the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent.

6 green used up6 red left over

Sebanyak 12 g zink dan 6.5 g sulfur bertindak balas mengikut persamaan:

a.Apakah bahan yang merupakan reaktan penghad di dalam tindak balas?

Daripada persamaan,

Zn(p) + S(p) ZnS(p)

Bilangan mol Zn = = 0.1836 mol1265.37

Bilangan mol S = = 0.2031 mol6.532

1 mol Zn bertindak balas dengan 1 mol SOleh itu 0.1836 mol Zn memerlukan 0.1836 mol S

# Reaktan penghad ialah Zn# Reaktan penghad ialah Zn


b. Berapa gramkah ZnS yang terhasil?

Daripada persamaan,

Zn(p) + S(p) ZnS(p)

Bilangan mol ZnS terbentuk = 0.1836 mol

Jisim ZnS terbentuk = 0.1836 x 97.37 = 17.88 g

Oleh kerana Zn ialah reaktan penghad maka amaun hasil ditentukan oleh amaun Zn

1 mol Zn menghasilkan 1 mol ZnS


c. Namakan bahan yang berlebihan dan kirakan bilangan gram bahan tersebut yang masih belum bertindak balas.

Bahan berlebihan ialah S

Zn(p) + S(p) ZnS(p)

Bilangan mol S berlebihan = 0.2031 - 0.1836 = 0.0195 mol

Jisim Sulfur berlebihan = 0.0195 x 32 = 0.624 g

• Example: Urea is prepares by reacting ammonia with CO2

2NH3 (g) + CO2 (g) (NH2) 2CO (aq) + H2O (l)

637.2 g of NH3 are treated with 1142 g CO2. Which of the two reactant is limiting reagent?

grams NH3 moles of NH3 moles of (NH2) 2CO

637.2 g NH31 mol NH3

17.03 g NH3

x 1 mol (NH2) 2CO 2 mol NH3

x

moles of (NH2) 2CO ==

= 18.71 mol (NH= 18.71 mol (NH22)) 2 2COCO



637.2 g of NH3 are treated with 1142 g CO2. Which of the two reactant is limiting reagent?

grams CO2 moles of CO2 moles of (NH2) 2CO

1142 g CO21 mol CO2

44.01 g CO2

x 1 mol (NH2) 2CO 1 mol CO2

x

moles of (NH2) 2CO ==

= 25.95 mol (NH= 25.95 mol (NH22)) 2 2COCO

Therefore, NH3 must be the limiting reagent because it produces a smaller amount of (NH2) 2CO



Calculate the mass of (NH(NH22)) 2 2CO CO formed.

18.71 mol (NH2) 2CO 60.06 g (NH2) 2CO 1 mol (NH2) 2CO

x

Mass of (NH2) 2CO ==

= 1124 g (NH= 1124 g (NH22)) 2 2COCO

The molar mass of (NH2) 2CO is 60.06 g. Use the conversion conversion factorfactor to convert from moles of (NH2) 2CO to gram of (NH2) 2CO.

Do You Understand Limiting Reagents?

In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

org Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al

27.0 g Alx

1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent

Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3

x

= 234 g Al= 234 g Al22OO33

Do You Understand Limiting Reagents?

In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

• Theoretical Yield is the amount of product that would result if all the limiting reagent reacted (from calculation using the balanced equation).

• Actual Yield is the amount of product actually obtained from a reaction (almost always less than the theoretical yield).

• Percent yield - to determine reaction efficiency, which describe the proportion of the actual yield to the theoretical yield.

% Yield = Actual Yield

Theoretical Yieldx 100

• Example:

TiCl4 (g) + 2Mg (l) Ti (s) + 2MgCl2 (l)

3.54 x 107 g of TiCl4 are reacted with 1.13 x 107 g of Mg. Calculate the theoretical yield of Ti in grams.Calculate the theoretical yield of Ti in grams.

g of TiCl4 Mol of TiCl4 Mol of Ti

First: to find limiting reagent, calculate the no. of moles of Ti that could produces if all the TiCl4 reacted

= 1.87 x 105 mol Ti

Next: calculate the no. of moles of Ti formed from 1.13 x 107 g of Mg.

g of Mg Mol of Mg Mol of Ti = 2.32 x 105 mol Ti

• Example:


3.54 x 107 g of TiCl4 are reacted with 1.13 x 107 g of Mg. Calculate the theoretical yield of Ti in grams.Calculate the theoretical yield of Ti in grams.

Therefore, TiCl4 is the limiting reagent because it produces a smaller amount of Ti

= 8.95 x 106 g Ti1.87 x 105 mol Ti47.88 g Ti

1 mol Tix

Mass of Ti = Mass of Ti =

• Example:


3.54 x 107 g of TiCl4 are reacted with 1.13 x 107 g of Mg. Calculate the percent yield if 7.91 x 10Calculate the percent yield if 7.91 x 1066 g of Ti are actually g of Ti are actually obtained. obtained.

The percent yield is given by:

% Yield = Actual YieldTheoretical Yield

x 100

% Yield = 7.91 x 106 g8.95 x 106 g

x 100

= 88.4%

chemistry lecture slide week 2

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