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International University for Science & Technology College of Pharmacy. General Chemistry (Students of Dentistry) Prof. Dr. M. H. Al-Samman. Chapter 3. CHEMISTRY. Covalent Bond and Molecular Compounds الرابطة المشتركة والمركبات الجزيئية. الروابط المشتركة The Covalent Bond :

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International university for science technology college of pharmacy

International University for Science & TechnologyCollege of Pharmacy

General Chemistry

(Students of Dentistry)

Prof. Dr. M. H. Al-Samman


Chapter 3

Chapter 3

CHEMISTRY


International university for science technology college of pharmacy

Covalent Bond and Molecular Compounds

الرابطة المشتركة والمركبات الجزيئية

الروابط المشتركة The Covalent Bond:

الرابطة المشتركة A single covalent bond:

هي الرابطة التي تنشأ بواسطة زوج من الالكترونات بين ذرتين , الكترونا من كل ذرة.

مثال :

H – H , Cl – Cl ,F – F

الرابطة الثنائية:A double covalent bond

هي الرابطة التي تنشأ بواسطة زوج من الاكترونات بين ذرتين , الكترونين من كل ذرة .

مثال:

O=O , H₂C=CH₂

الرابطة الثلاثية A triple covalent bond:

هي الرابطة التي تنشأ بواسطة ثلاثة أزواج من الاكترونات بين ذرتين , ثلاثة إلكترونات من كل ذرة .


International university for science technology college of pharmacy

Covalent Bond and Molecular Compounds

الرابطة المشتركة والمركبات الجزيئية

  • ملاحظات ضرورية :

  • الرابطة الثلاثة هي أقوى وأقصر من الرابطتين الثنائية والأحادية , وطبعا فإن الرابطة الثنائية , هي اقوى وأقصر من الرابطة الأحادية .

  • - نسمي المركبات التي تحوي روابط مشتركة :

  • المركبات الجزيئية(molecular compounds)

  • الرابطة التساندية :

  • هي الرابطة التي تنشأ بين ذرتين , إحداهما تقدم الزوج الاكتروني والأخرى تقدم المدار الفارغ , ونسمي المركبات التي تحوي روابط تساندية بالمركبات التساندية أو المعقدات .


Molecules and formulas

الجزيئات والصيغ Molecules and Formulas

الصيغة الجزيئية molecular formula :

تعطي العدد الصحيح الواقعي , لكل نوع من الذرات في الجزيء.

empirical formulaالصيغة المبسطة:

تعطي النسبة بين الذرات في المركب بأصغر أعداد صحيحة , وهي أبسطصيغة ممكنة .


Molecular compounds

Ball-and-stick model vs. Space-filling model

EOS

المركبات الجزيئة Molecular Compounds


Empirical and molecular formulas

The elemental ratio C:H:O is 1:2:1, so the empirical formula is CH2O

EOS

الصيغ البسيطة والصيغ المجملة Empirical and Molecular Formulas

empirical formulaالصيغة المبسطة:

تعطي النسبة بين الذرات في المركب بأصغر أعداد صحيحة , وهي أبسطصيغة ممكنة .

Example:

Molecular formula of glucose – C6H12O6

EOS


Structural formulas

الصيغ البنيوية Structural Formulas

Shows how atoms are attached to one another.

EOS


Mass percent composition from chemical formulas

النسب المئوية الوزنية للعناصر في الصيغة الجزيئةMass Percent Composition from Chemical Formulas

النسبة المئوية الوزنية للعنصر في مركب ما :

The mass percent composition

تساوي إلى كتلة العنصر بالغرام , مقسومة على الوزن الجزيئيالغرامي ( مول ) مضروبة بمائة .

-Each g amount of the compound contains g amount of the element

-Each 100 g of compound contains x g of the element .


Percentage composition of butane

النسب المئوية لمكونات البوتانPercentage Composition of Butane


Relating molecular formula to empirical formulas

molecular formula mass

= integer (nearly)

empirical formula mass

علاقة الصيغة الجزيئية المجملة بالصيغة المبسطةRelating Molecular Formula to Empirical Formulas

A molecular formula is a simple integer multiple of the empirical formula.

الصيغة الجزيئية أو المجملة هي مضاعف بسيط للصيغة المبسطة

That is, an empirical formula of CH2 means that the molecular formula is CH2, or C2H4, or C3H6, or C4H8, etc.

So: we find the molecular formula by:

We then multiply each subscript in the empirical formula by the integer.


Mass percent composition and chemical formulas

النسبة المئوية الوزنية للمكونات والصيغ الكيميائيةMass Percent Compositionand Chemical Formulas

Example :

Phenol, a general disinfectant, has the composition 76.57% C, 6.43% H, and 17.00% O by mass. Determine its empirical formula.

Answer:

Moles of C percent :76.57/12.011 = 6.374 %

Moles of H percent : 6.43/1.0079 = 6.379 %

Moles of O percent :17.00/15.998=1.062 %

- We divide each figure by the smallest figure to obtain the ratio between the elements in the formula:

The number of O atoms = 1.062/1.062 = 1 atom

The number of C atoms = 6.347/ 1.062 = 6 atom

The number of H atoms = 6.379/1.062 = 6 atom

The empirical formula of phenol is : C₆H₆O


International university for science technology college of pharmacy

النسبة المئوية الوزنية للمكونات والصيغ الكيميائيةMass Percent Compositionand Chemical Formulas

Example:

Diethylene glycol, used in antifreeze, as a softening agent for textile fibers and some leathers, and as a moistening agent for glues and paper, has the composition 38.70% C, 9.67% H, and 51.61% O by mass.

1-Determine its empirical formula.

2-Determine its molecular formula if the molecular mass of the compound is 62 u

Answer:

Moles of C % = 38.70/12.011 = 3.225 mol

Moles of H % = 9.67/1.0079 = 9.594 mol

Moles of O % = 51.61/15.998 = 3.226 mol


International university for science technology college of pharmacy

النسبة المئوية الوزنية للمكونات والصيغ الكيميائيةMass Percent Compositionand Chemical Formulas

Example: cont.

-Number of O atoms = 1 atom

-Number of C atoms = 1 atom

-Number of H atoms = 2.97=3 atom

-The empirical formula is : CH₃O

  • The empirical formula mass: 12 + ( 3x1) + 16 = 31 u

  • The integer number : 62/31 =2

  • The molecular formula : C₂H₆O₂


International university for science technology college of pharmacy

النسبة المئوية الوزنية للمكونات والصيغ الكيميائيةMass Percent Compositionand Chemical Formulas

Example:

The empirical formula of hydroquinone, a chemical used in photography, is C3H3O, and its molecular mass is 110 u. What is its molecular formula?

Answer:

The empirical formula mass : (12x3)+ (3x1) + 16 = 55 u

The integer number : 110/55 = 2

The molecular formula : C₆H₆O₂


International university for science technology college of pharmacy

النسبة المئوية الوزنية للمكونات والصيغ الكيميائيةMass Percent Compositionand Chemical Formulas

Example :

Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment shows that the molecular mass of the compound is 90 u. Determine (a) the mass percent composition, (b) the empirical formula, and (c) the molecular formula of the compound.

Mass of carbon=0.1935x12/44=0.052779

Mass of hydrogen=0.1000x2/18=0.11111

Percent c=(0.052779/0.1000)x100=52.77%

Percent H=(0.0555/0.1000)x100=11.11%

Percent of O=100 – 63.88 = 36.12%

C mol%=52.77/12=4.3975 mol

H mol %=5.55/1=5.55 mol

O mole %=36.12/16=2.25mol


Mass percent composition and chemical formulas1

النسبة المئوية الوزنية للمكونات والصيغ الكيميائيةMass Percent Compositionand Chemical Formulas

Then by dividing each molar percent by the smallest figure which is 2.25 we get:

-Number of O atoms=1 atom

-Number of C atoms =2 atom

-Number of H atom=5 atom

-The empirical formula : C₂H₅O

-Empirical formula mass : 24 + 5+ 16 = 45 u

  • Integer number: 90/45 = 2

  • The molecular formula : C₄H₁₀O₂


Balancing chemical equations

موازنة المعادلات الكيميائيةBalancing Chemical Equations

Example :

Balance the equation

Fe + O2 Fe2O3(not balanced)

2Fe + 3/2 O₂ Fe2O3

4Fe + 3 O₂ 2Fe2O3

Example :

Balance the equation

C2H6 + O2  CO2+ H2O

C2H6 + 7/2 O2  2CO2+ 3 H2O

2C2H6 + 7 O2  4 CO2+ 6 H2O

Example :

Balance the equation

H3PO4 + NaCN  HCN + Na3PO4

H3PO4 + 3 NaCN  3 HCN + Na3PO4


International university for science technology college of pharmacy

موازنة المعادلات الكيميائيةBalancing Chemical Equations

Example:

When 0.105 mol propane is burned in an excess of oxygen, how many moles of oxygen are consumed? The reaction is

C3H8+ 5 O2 3 CO2+ 4 H2O

1mole ----5mole

0.105------?

Number of O moles = 0.105 x 5 /1 = 0.525 mol


International university for science technology college of pharmacy

موازنة المعادلات الكيميائيةBalancing Chemical Equations

Q-

When 11.6 g of butane is burned in an excess of oxygen.

  • how many grams of oxygen are consumed?

  • The reaction is

    C₄H₁₀+ 13/2 O2 4 CO2+ 5 H2O

    58 (13/2)x32

    11.6g x

    The mass of consumed oxygen: 41.6 g


Limiting reactants

المواد المحددة للتفاعل Limiting Reactants

إذا لم تكن المواد المتفاعلة , متساوية المكافئات , ففي نهاية التفاعل , يمكن أن تبقى بعض المواد الداخلة في التفاعل , بغير تفاعل ( زائدة ).

:Limiting Reactant المادة المتفاعلة المحددة للتفاعل

المادة المحددة للتفاعل , هي المادة التي تستهلك كليا في التفاعل , وبانتهائها ينتهي التفاعل .


International university for science technology college of pharmacy

المواد المحددة للتفاعل Limiting Reactants


International university for science technology college of pharmacy

المواد المحددة للتفاعل Limiting Reactants

The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield.

:Theoretical yield المنتوج النظري

هو المردود الناتج الذي نحصل عليه من خلال الحسابات النظرية البحتة .

:The actual yield المردود الناتج الفعلي

هو المردود الناتج الفعلي , الذي نحصل عليه واقعيا .

:The percent yield مردود التفاعل

مردود التفاعل , هو حاصل قسمة الناتج الحقيقي , على الناتج المتوقع نظريا , مضروبا بمائة .


International university for science technology college of pharmacy

المواد المحددة للتفاعل Limiting Reactants

Example:

During a process we burned 11.60 g of butane C₄H₁₀ with 80 g of oxygen , we obtained 30 g of carbon dioxide .

1- write the equation of the reaction.

2- balance the equation.

3- assign the limiting reactant in the reaction.

4- calculate the theoretical yield of carbon dioxide.

5- calculate the percentage yield of the reaction.


International university for science technology college of pharmacy

المواد المحددة للتفاعل Limiting Reactants

Answer:

1- C₄H₁₀ + O₂ → CO₂ + H₂O

2- C₄H₁₀ + O₂ → 4 CO₂ + 5 H₂O

C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O

3- to assign the limiting reagent, we write:

C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O

58 g 208 g

11.6 g x g

The needed amount of oxygen= (11.6x208)/58 = 41.6g

So, Butane is the limiting reactant .


International university for science technology college of pharmacy

المواد المحددة للتفاعل Limiting Reactants

Answer:

4- To calculate the theoretical yield, we write:

C₄H₁₀ + 13/2 O₂ → 4 CO₂ + 5 H₂O

58 g 4x 44 g

11.6 g y g

The theoretical yield= (11.6 x 176)/ 58= 35.2 g

5- to calculate the percentage yield, we write:

The percentage yield=(actual yield/ theoretical yield)x100

The percentage yield = (30/35.2)x100 = 85.22%


International university for science technology college of pharmacy

المواد المحددة للتفاعل Limiting Reactants

  • Example:

  • Part of the SO2 that is introduced into the atmosphere ends up being converted to sulfuric acid, H2SO4. The net reaction is:

    • 2SO2(g) + O2(g) + 2H2O(l)  2H2SO4(aq)

  • Answer:

  • How much H2SO4 can be formed from 5.0 mol of SO2, 1.0 mol O2, and an unlimited quantity of H2O?

  • Assign the limiting reagent of the reaction


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    المواد المحددة للتفاعل Limiting Reactants

    • 2SO2(g) + O2(g) + 2H2O(l)  2H2SO4(aq)

  • 2mol 1 mol 2 mol

  • 1 mol 2mol

  • - The amount of H2SO4 in grams= 2x98 = 196 g

  • - Oxygen is the limiting reagent in the reaction.


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    المواد المحددة للتفاعل Limiting Reactants

    • Consider the following reaction:

      • 2Na3PO4(aq) + 3Ba(NO3)2(aq)  Ba3(PO4)2(s) + 6NaNO3(aq)

    • Suppose that a solution containing 3.50 g of Na3PO4 is mixed with a solution containing 6.40 g of Ba(NO3)2. How many grams of Ba3(PO4)2 can be formed? What is the % yield, if experimentally, only 4.70 g were obtained from the reaction?

    • Solution

      • At first, it must determine which of reactants is completely

      • consumed and is therefore the limiting reactant. The

      • quantity of this reactant, in turn, will determine the quantity of

      • Barium Phosphate Ba3(PO4)2


    International university for science technology college of pharmacy

    • It is needed a grams-to-moles conversion factor to convert from the given reactant masses and moles-to-grams factor to convert to the desired product mass. The quantity of excess reactant can be calculated as the difference between the given mass of this reactant and the mass consumed (reacted) in the reaction

    • The balanced equation is given as following:

      2 Na3PO4(aq)+ 3Ba(NO3)2(aq) Ba3(PO4)2(s)+ 6NaNO3(aq)

    • It can identify the limiting reactant by finding the number of moles Barium Phosphate Ba3(PO4)2produced by assuming first one reactant, and then the other is as the limiting reactant.

      No. moles Na3PO4 = 0.021 moles (MW of Na3PO4 = 164)

      No. moles Ba(NO3)2 = 0.025 moles (MW of Ba(NO3)2 = 261)

      • 0.021 moles Na3PO4 produce 0.011 moles of Ba3(PO4)2

      • 0.025 moles Ba(NO3)2 produce 0.008 moles of Ba3(PO4)2

        Because the amount of product in the second calculation [0.008 mol

        Ba3(PO4)2] is smaller than (0.021 mol Ba3(PO4)2), thus the Barium

        Nitrate is the limiting reactant. So thus when 0.008 mol of Ba3(PO4)2

        has been formed, a quantity 0.025 mol of Ba(NO3)2is completely

        Consumed and the reaction stops, producing a specific mass of

        Ba3(PO4)2


    International university for science technology college of pharmacy

    المواد المحددة للتفاعل Limiting Reactants

    • Having found that the amount of product is 0.0083 mol Ba3(PO4)2, thus the mass of Ba3(PO4)2 is 3.7 grams.

      Number of moles of reacted Na3PO4 is 0.017 moles. The mass of Na3PO4 is 2.8 grams the unreacted of Na3PO4 is 0.71 gr.

    • The theoretical yield = 4.99 gr. Ba3(PO4)2

    • [MW ofBa3(PO4)2= 601]

    • The experimental yield = 4.70 gr.Ba3(PO4)2

    • The percentage yield = (4.70 / 4.99)x100 = 94.2 %


    International university for science technology college of pharmacy

    The End of Chapter 3The test will cover Chapters 1-3, Scheduled Homework: 3.9, 3.11, 3.15, 3.17, 3.19, 3.21, 3.25, 3.27, 3.31, 3.33, 3.43, 3.47


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