3.1 Atomic mass.

Mass Relationships in Chemical Reactions

Chapter 3

Chapter 3 (80-107)

3.1 Atomic mass.

3.2 Avogadro’s number and the molar mass of an element.

3.3 Molecular mass.

3.5 Percent composition of compounds.

3.7 Chemical reactions and chemical equations.

3.8 Amounts of reactants and products.

3.9 Limiting reagents.

3.10 Reaction yield.

3.1 Atomic mass.

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro World

atoms & molecules

Macro World grams

3.1

The unit of atomic mass is amu

By definition:

1 atom ¹²C “weighs” 12 amu

On this scale

1

H = 1.008 amu

16

O = 16.00 amu

3.1

Average atomic Mass

• Each atom have more than one isotope with different abundance

• Average atomic Mass: the average mass of all of the isotopes of an element, each one weighted by its proportionate

abundance

• Science each atom have more than one isotope with different abundance

Average atomic mass=Atomic mass × abundance percent + Atomic mass × abundance percent

100

Isotope 1 Isotope 2

Natural lithium is:

7.42% ⁶Li (6.015 amu) 92.58% ⁷Li (7.016 amu)

7.42 x 6.015 + 92.58 x 7.016

100 = 6.941 amu

3.1

Average atomic mass of lithium:

Atomic number

Average atomic mass

Example

Average atomic mass (6.941)

The atomic masses of (75.53 percent) and (24.47 percent) are 34.968 amu and 36.956 amu, respectively. Calculate the average atomic mass of chlorine.

Average atomic mass⁼ Atomic mass × abundance percent + Atomic mass × abundance percent

100

Isotope 1 Isotope 2

Solution:

Isotope I

Atomic mass = 34.698 amu Abundance percent = 75.53 %

Isotope II

Atomic mass = 36.956 amu Abundance percent = 24.47

%

Average atomic mass= 34.698 × 75.53+ 36.956 × 24.47

100

Isotope 1 Isotope 2

35.45 amu

=

35Cl

17 ₁₇³⁷Cl

Iodine has two isotopes

¹²⁶

I and

¹²⁷

I, with the equal abundance.

Calculate the average atomic mass of Iodine (

₅₃

I).

 

(a) 126.5 amu

 

(b) 35.45 amu

 

(c) 1.265 amu

 

(d) 71.92 amu

equal abundance MEAN each atom has abundance 50% .

What information would you need to calculate the average atomic mass of an element?

(a) The number of neutrons in the element.

(b) The atomic number of the element.

(c) The mass and abundance of each isotope of the element.

(d) The position in the periodic table of the

element.

3.2 Avogadro’s number and the

molar mass of an element.

Pair = 2

Dozen = 12

The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00

grams of ¹²C isotope

3.2

1 mol = N_A = 6.0221367 x 10²³ Avogadro’s number (N_A )

ةقلح وه ورداجوفأ ددع تاميسجلا ملاع نيب لصولا

)

تايئزجلا ةيئرملا ريغ (

تايئرملا ملاعو تارذلاو حلملاو ءاملاك

نويئايميكلا همدختسيو تانويلاو تارذلاو تائيزجلا دعل لوملا مدختسي ةنيعلا يف تاميسجلا ددع ةفرعمل ةمئلم ةقيرط رفوي هنل

Molar mass ^(M) is the mass of 1 mole of in grams

shoeseggs Marbles atoms

1 mole ¹²C atoms = 6.022 x 10²³ atoms = 12.00 g 1 ¹²C atom = 12.00 amu

1 mole lithium atoms = 6.941 g Li 1 lithium atom = 6.941 amu

For any element

atomic mass (amu) = molar mass (g/mol)

Molar mass

Atomic mass

Mass of 1 mole of C = mass of 1 mole Cu

M = mass/mol = g/mol

One Mole of:

C S

Cu Fe

Hg

3.2 ¹⁹

32.07 g 12.01 g

200.6 g

55.85 g 63.55 g

One mole of these substances contain = 6.022 x 10²³ atoms but is not equal because they have different molar masses

THUS:

one mole of H atoms has 6.022 x 10

²³

atoms

&

One mole of H

₂

molecules has 6.022 x 10

²³

molecules

22

How many amu are there in 8.4 g?

1 g = 6.022 x 10²³ amu 8.4 g = 5.1 x 10²⁴ amu

What is the mass in grams of 13.2 amu ?

13.2 amu = 2.2 x 10^-

23 g

1 amu = 1.66 x 10^-24 g

ورداجوفأ ددع مادختسا نكمي هنأ نلاثملا حضوي ةيرذلا ةلتكلا ليوحتل

حيحص سكعلاو مارجلاب ةلتك ىلإ

1 g 6.022 x 10²³ amu 8.4 g ?

No. of moles تلوملا ددع

Mass of element

رصنععلا ةلتك No. of atoms تارذلا ددع

Molar mass

ةيرلوملا ةلتكلا Avogadro's No.

ورداجوفأ ددع

) (

) (

) /

(

) ) (

( N atoms

atoms N

mol g

M

g mol m

n

A





EXAMPLE 3.2

How many moles of He atoms are in 6.46 g of He ?

How many grams of Zn are in 0.356 mole of Zn?

mol mol g

g M

He m

n 1 . 61

/ 003

. 4

46 .

) 6

(   

g 3 . 23 g/mol

39 .

65 mol

356 .

0 ) (











x m

nM M m

Zn m n

N (atoms) = m (g) × N_A (atoms) / M (g/mol) N = 16.3 × 6.022×10²³ / 32.07

= 3.06×10²³ S atoms

N (atom)=? m =16.3g

) (

) (

) /

(

) ) (

( N atoms

atoms N

mol g

M

g mol m

n

A





m (g) = n (mol) × M (g/mol) m = 0.356 × 65.39

= 23.28 g Zn

n=0.356 mole m (g)=?

) (

) (

) /

(

) ) (

( N atoms

atoms N

mol g

M

g mol m

n

A





3.3 MOLECULAR MASS.

Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule.

SO₂

1S 32.07 amu

2O + 2 x 16.00 amu SO₂ 64.07 amu

For any molecule

molecular mass (amu) = molar mass (grams) 1 molecule SO₂ = 64.07 amu

1 mole SO₂ = 64.07 g SO₂

Molecular mass = (No. of atoms× atomic mass) +( No. of atoms× atomic mass)

element I element II

Molar mass (g/mol)

Atomic mass

molecular mass

Atoms

molecules Al, As, Ni, H

HCl, H2, O2, NaCl

)

لودجلا نم ةيرذ ةلتك يرودلا)

ةلتك ةيئيزج

It must be noted that:

Example 3.5

a- SO₂

= 32.07 amu + 2(16.00 amu

)

= 64.07 amu

b- C₈H₁₀N₄O₂

= 8(12.01 amu) + 10(1.008 amu) + 4(14.01 amu) + 2(16.00 amu)

= 194.20 amu

Do You Understand Molecular Mass?

How many H atoms are in 72.5 g of C₃H₈O ?

N (atoms) = m (g) × N_A (atoms) / M (g/mol)

= 72.5 × 6.022×10²³ × 8 / 60.09

= 5.81 × 10²⁴ H atoms

m= 72.5 g C₃H₈O No. of H atoms ?

M g/mol

calculate 60.09

No. of H atoms in molecule

) (

) (

) /

(

) ) (

( N atoms

atoms N

mol g

M

g mol m

n

A





3.5 PERCENT COMPOSITION

OF COMPOUNDS.

Percent composition of an element in a compound = is the percent by mass of each element in a compound

n x molar mass of element

molar mass of compound x 100%

=

n is the number of moles of the element in 1 mole of the compound

%H = 3 x (1.008 g)

97.99 g x 100% = 3.086%

%P = 30.97 g

97.99 g x 100% = 31.61%

%O = 4 x (16.00 g)

97.99 g x 100% = 65.31%

Molar mass of H₃PO₄ = 3(1.008) + 1(30.97) + 4(16.00) = 97.99 g

37

H.W. Calculate the percent of nitrogen in Ca(NO₃)₂:

a) 12.01%.

b) 17.10%.

c) 18%

d) 16%.

H.W. All of the substances listed below are fertilizers that contribute nitrogen to the soil

.

Which of these is the richest source of nitrogen on a mass percentage basis

? ( a

) Urea, (NH₂)₂CO (

b

) Ammonium nitrate, NH₄NO₃ (

c

) Guanidine, HNC(NH₂)₂ (

d

) Ammonia, NH₃

نيجورتينلل ردصم ىنغاوه داوملا هذه نم ايا نم هينزو هبسن ربكا ىلع هئاوتحا ساسا ىلع

؟نيجورتينلا

(a) %N = 46.6%

(b) %N = 58%

(c) %N = 71.1%

(d) %N = 82.2%

Percent Composition and Empirical Formulas

39

Example

Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent.

K Mn O

% 100g 24.75g 34.77g 40.51g

n=m/M 24.75/39.10

=0.633mol 34.77/54.94

=0.6329mol 40.51/16.00

= 2.532mol

 on smallest no. of mole 0.633/0.632

=1 0.6329/0.632

= 1 2.532/0.632

=4

The empirical formula is K₁ Mn₁ O₄

KMnO₄

لحلا تاوطخ .1 لاؤسلا يف ةروكذملا رصانعلا هيف عضن لودج أشنن

.2 اندنع ناك ولف مارجلاب اهنع ربعم ةيوئملا ةبسنلا نأ ربتعن 100

لا هذهف بكرملا نم مارج 100

ىلع ةعزوم مارج

.اهتبسن بسح رصانعلا .3 اتلوملا ددع دجون n

نوناقلا مادختساب رصنع لكل n=m/M

.

.4 .رصانعلا نم لوم رغصأ ىلع اتلوملا ددع مسقن

.5 لثمت اهيلع لصحن يتلا ماقرلا empirical formula

.قباسلا لاثملا يف امك ةحيحص دادعأ نوكت نأ طرشب

.6 نم أدب دادعأب ةغيصلا يف ةدوجوملا لفسلا يف يتلا ماقرلا برضب موقن ةيرشع دادعأ روهظ ةلاح يف 2

، 3 ...

. ةحيحص دادعأ ىلع لصحن ىتح

40

C H O

% 100g 40.92g 4.58g 54.50g

n=m/M 40.92/12.01

= 3.407mol 4.58/1.008

=4.54.mol 54.50/16.00

= 3.406 mol

 on smallest no. of mole

3.407/3.406

= 1 0.4.54/3.406

= 1.33 3.406/3.406

= 1 Convert into integer x

3 3 3.99 = 4 3

The empirical formula is

C₃ H₄ O₃

C₃H₄O₃

Example 3.9 p90:

Ascorbic acid composed of 40.92% C, 4.58% H, and 54.50%

O by mass. Determine its empirical formula.

Determination of Molecular Formulas

If we determine the empirical formula and the molar mass of the compound is known we can determine the molecular Formula by using:

Molar mass of molecule (compound)

Molar mass of empirical formula = n

Molecular formula = (empirical formula)n

• A sample of a compound containing born (B) and hydrogen (H) contains 6.444g of B and 1.803 g of (H). The molar mass of the compound is about 30g. What is its molecular formula?

PRACTIES EXERICISE 3.10

n_B = 6.444 g B x 1 mol B =0.5961 mol B 10.81 g B

n_H = 1.803g H x 1 mol H =1.7888 mol H 1.008 g H

B : 0.5961 =1.0 0.5961

H : 1.7888

0.5961 = 3

BH₃

Molar mass of empirical formula = 10.81 + 3 x 1.008 = 13.834g

The ratio between molar mass and the molar mass of empirical formula

= molar mass / empirical formula = 30 g / 13.834 g ≈ 2

B₂H₆

3.7 CHEMICAL REACTIONS AND

CHEMICAL EQUATION.

3.7

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

A chemical reaction is a process in which one or more substances is changed into one or more new substances

A chemical equation uses chemical symbols to show what happens during a chemical reaction

reactants products

1- تايئئزجلئئا وأتئارذلئئا ددع 2- تلوملئئا ددع ((تايئزجلا وأ تارذلل 3

- ةيرلوملا ةلتكلا

ةيئايميكلا ةلداعملا ةءارق

How to “Read” Chemical Equations

How to “Read” Chemical Equations

2 Mg + O₂ 2 MgO

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

2(24.31) 2(16.0) 2[24.31+16.0]

reactant =80.6 Product = 80.6

√ √

√

X

بتكن  هحيحصلا هغيصلا

( رسيلا فرطلا ىلع لعافتم لكل )

( نميلا فرطلا ىلع جتان لكلو ) 

بناجب يتلا ماقرلا ريغتب نوكي هيئايميكلا هلداعملا نزو هغيصلا اهتحت يتلا تسيلو

ددعلا سفن رصنعلل نوكي ثيحب

. هلداعملا يفرط ىلع

, 

اروهظ رثكلا رصانعلا نزوت مث اروهظ لقلا رصانعلا لوا نزوت نم ددعلا سفن كيدل نا نم دكأتلا يه هريخلا هوطخلا  هلداعملا يفرط ىلع رصنع لكل تارذلا

ةيئاميكلا ةلداعملا نزو ةقيرط

Balancing Chemical Equations

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 ^C₄^H₁₂

بتكن هحيحصلا هغيصلا

( )

جتان لكلو رسيلا فرطلا ىلع لعافتم لكل

) نميلا فرطلا ىلع )

هغيصلا بناجب يتلا ماقرلا ريعغتب نوكي هيئايميكلا هلداعملا نزو اهتحت يتلا تسيلو

هلداعملا يفرط ىلعع ددعلا سفن رصنعلل نوكي ثيحب.

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

C₂H₆ + O₂ CO₂ + H₂O

3.7

start with C or H but not O

2 carbon on left

1 carbon

on right multiply CO₂ by 2 C₂H₆ + O₂ 2CO₂ + H₂O

6 hydrogen on left

2 hydrogen

on right multiply H₂O by 3 C₂H₆ + O₂ 2CO₂ + 3H₂O

لقلا رصانعلا لوأ نزوت اروهظ

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

3.7

2 oxygen on left

4 oxygen (2x2)

C₂H₆ + O₂ 2CO₂ + 3H₂O

+ 3 oxygen (3x1)

multiply O₂ by 7 2

= 7 oxygen on right C₂H₆ + O7 ₂ 2CO₂ + 3H₂O

2 remove fraction

multiply both sides by 2 2C₂H₆ + 7O₂ 4CO₂ + 6H₂O

2

اروهظ رثكلا رصانعلا نزوت مث

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

3.7

2C₂H₆ + 7O₂ 4CO₂ + 6H₂O

Reactants Products 4 C

12 H 14 O

4 C 12 H 14 O

4 C (2 x 2) 4 C

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

تارذلا نم ددعلا سفن كيدل نا نم دكأتلا يه هريخلا هوطخلا هلداعملا يفرط ىلع رصنع لكل

Example 3.12

PRACTIES EXERICISE 3.12

Al + O₂ → Al₂O₃

2Al + O₂ → Al₂O₃

2Al + 3/2O₂ → Al₂O₃

2(2Al + 3/2O₂ → Al₂O₃) 4Al + 3O₂ → 2Al₂O₃

 Fe₂O₃ + CO → Fe +CO₂

 Fe₂O₃ + CO → 2Fe +CO₂

 Fe₂O₃ +1/3CO→ 2Fe+1/3CO₂

 3(Fe₂O₃ +1/3CO→ 2Fe+1/3CO₂)

 Fe₂O₃ + 3CO→ 2Fe +3CO₂

H.W. What is the coefficient of H₂O when the equation is balanced:

_ Al₄C₃ + _ H₂O  _ Al(OH)3 + 3CH₄ a. 13

b. 4 c. 6 d. 12

H.W. What are the coefficients of Al4C3 ,H2O and Al(OH)3,respectively, when the equation is balanced:

_ Al₄C₃ + _ H₂O  _ Al(OH)₃ + 3CH₄ a. 4,1,5

b. 1,12,4 c. 1,24, 4 d. 4,12,1

3.8 AMOUNTS OF REACTANTS AND

PRODUCTS .

Use gram ratio of A and B From balanced

equation

دحأ ةيمولعمب ةجتانلا وأ ةلعافتملا داوملا ةفرعمل • :يتلاب موقن ةجتانلا وا تلعافتملا

•

هنوزوم ةلداعملا نوكت نأ دبل هاطعملا ةداملا يددح • given

هداملا مث

ةبولطملا Required

يلهاجت و مهنيب ةقلع يلمعا و

امامت يقابلا •

تلوم ةقلع هنوزوملا هلداعملا يف ةقلعلا و تلوم ىلا اهلوحن تامارجلاب هاطعملا هداملا تناكولف •

ةوطخلا هذه ىلا جاتحن ل تلوملاب تناكاذا دحلا اذهب ىفتكن تلوملاب ةبولطملا هداملا تناك اذا • ىلا تلوملا لوحن تامارجلاب ةبولطملا هداملا تناك اذا • .تامارج

Mole Method

Mass Method

3.9 LIMITING REAGENTS

Limiting Reagent Excess Reagents

Limiting Reagents is the reactant used up first in a reaction

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

3.9

2NO + 2O₂ 2NO₂

NO is the limiting reagent O₂ is the excess reagent

Limiting Reactant

•

هجتانلا هداملا ةيمك ددحي يذلا فشاكلا وه ددحملا فشاكلا

•

هرم لك هسفن وه ددحملا فشاكلا نوكي نأ طرتشي ل •

هدوجوم ىرخلا هداملا و لقأ تلوم ددعب دوجوم ددحملا فشاكلا امئاد هدايزب لك كيطعي هنأ ةقباسلا لئاسملا نع فلتخت ددحملا فشاكلا ةلأسم •

جتانلا بلطي و نيلعافتملا ةيلاتلا تاوطخلاب موقن ددحملا فشاكلا ددحن يكل • 1 •

-

تلوم ىلا تلعافتملا داوملا تامارج لوحن

2 •

-

هنوزوملا ةلداعملا يف هداملا لماعم ىلع ةجتانلا تامارجلا مسقن

3 • - ددحملا فشاكلا يه تلوم ددع لقأ هداملا

4 •

-

قباسلا ءزجلا يف انملعت ام بسح هجتانلا هداملا دجون

When 22.0 g NaCl and 21.0 g H₂SO₄ are mixed and react according to the equation below, which is the limiting reagent?

2NaCl + 1H₂SO₄  Na₂SO₄ + 2HCl

 n NaCl = 22/58.5 = 0.376/2=0.188 mol

 n H₂SO₄ = 21/98 = 0.214/1 = 0.241 mol

 n NaCl (0.188 mol)<n H₂SO₄ (0.241 mol)

 So NaCl is the limiting reagent

Example:

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.

3.9

n (mol) = m (g) / M (g/mol)

بكرم لك ةبسن Al = 4.596/2 = 2.298 Fe₂O₃ = 3.756/1 = 3.756

ددحملا فشاكلا وه ايددع رغصلا limiting reagent

نإف جئاتنلا ىلع ءانبو Al

ددحملا فشاكلا وه

ةيمك باسحل Fe₂O₃

ددحملا فشاكلا مدختسن

53.96 g of Al 101.96 g of Al₂O₃ 124 g x g

x = 124×101.96/53.96

= 234.3 g Al₂O₃

• Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide 2NH3 (g) + CO2 (g) → (NH2)2CO (aq) + H2O (ι)

• In on process 637.2 g of NH₃ are treated with 1142 g of CO₂ a) which of the two limiting reagents? b) calculate the mass of (NH2)2CO formed ? C) how much excess reagent ( in gram) is left at the end of the reaction

a)

• n ^NH3 = 637.2 /17.03=37.416 /2mol= 18.71 mol

• n ^CO2 = 1142/ 44.01= 25.95 /1 = 25.95 mol

• n ^NH3 (18.74 mol)<n CO2 (25.95 mol)

• So NH3 is the limiting reagent

• C) How much excess reagent ( in gram) is left at the end of the reaction

Science

NH3 is the LR so CO2 is the excess the amount remain

From chemical eq. 2 mole NH3 = 1 mol CO2 37.482 mol = x

number of moles of CO2 react = 0.493 x 1/2 = 18.74 mol mass of CO2 = 18.74 x 44 = 824.56 ^g

The excess mass of CO2 = total mass CO2^{– reacted} CO2

= 1142 -823.4 = 318.6 g

Dr. Laila Al-Harbi

3.10 REACTION YIELD

Quantities of product calculated represent the maximum amount obtainable (100 % yield)

Most chemical reactions do not give 100 % yield of product because of:

1-Side reactions (unwanted reactions)

2-Reversible reactions ( reactants products) 3-Losses in handling and transferring

Reaction Yield

Theoretical Yield is the amount of product that would result if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtained from a reaction.

% Yield = Actual Yield

Theoretical Yield x 100

3.10

Reaction Yield

The percent Yield is the proportion of the actually yield to the theoretical yield which can be obtained from the following relation:

When 22.0 g NaCl and 21.0 g H₂SO₄ are mixed and react according to the equation below 8.95 g HCl is formed .what is the %yield of HCl?

2NaCl + 1H₂SO₄  Na₂SO₄ + 2HCl

 n NaCl = 22/58.5 = 0.376/2=0.188 mol

 n H2SO4 = 21/98 = 0.214/1 = 0.241 mol

 n NaCl (0.188 mol)<n H₂SO₄ (0.241 mol)

 So NaCl is the limiting reagent

From chemical eq. 2 mole NaCl = 2 mol HCl 0.376 mol = x

number of moles of HCl produced = 0.376 x 2/2 =0.376 mol

mass of HCl produced = n x MM(HCl) = 0.376 x 36.5 =13.724 g

 %yield of HCl = practical/theoriotical x 100

 %yield of HCl = 8.95 /13.724 x 100 = 65.21%

عل ةيرظنلا ةليصحلا باسحل فشاكلا مدختسن Ti

ددحملا

189.7 g of TiCl4 47.88 g of Ti 3.54×10⁷ g x g

x = 47.88 × 3.54×10⁷ / 189.7 x (theoretical yield) = 8.95×10⁶ g Ti

%yield = actual yield/theoretical yield × 100

= (7.91×10⁶ g / 8.95×10⁶ g) × 100

= 88.4%