The Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.02 X 10 23 cookies 1 dozen cars = 12 cars 1 mole of cars = 6.02 X 10 23 cars 1 dozen Al atoms.

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The Mole 1 dozen cookies = 12 cookies 1 mole of cookies = 6.02 X cookies 1 dozen cars = 12 cars 1 mole of cars = 6.02 X cars 1 dozen Al atoms = 12 Al atoms 1 mole of Al atoms = 6.02 X atoms Note that the NUMBER is always the same, but the MASS is very different!

mole المول : هو كمية المادة التي تحتوي على عدد أفوغادرو من جسيماتها. يرمز للمول بالرمز (n). عدد افجادرو : ثابت عددي يرمز له بالرمز N A وتساوي هذه القيمة 1 mole C= 6.02 x C atoms 1 mole H2O= 6.02 x H 2 O molecules

اختارى الاجابة الصحيحة 1. Number of atoms in mole of Al a) 500 Al atoms b) 6.02 x Al atoms c) 3.01 x Al atoms 2.Number of moles of S in 1.8 x S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1 x mole S atoms

Molar Mass – الكتلة المولية The Mass of 1 mole (in grams) Equal to the numerical value of the average atomic mass (get from periodic table) 1 mole of C atoms= 12.0 g 1 mole of Mg atoms =24.3 g 1 mole of Cu atoms =63.5 g

Molar Mass of Molecules and Compounds Mass in grams of 1 mole equal numerically to the sum of the atomic masses g/mol = 1 mole of CaCl 2 1 mole Ca x 40.1 g/mol + 2 moles Cl x 35.5 g/mol = g/mol CaCl 2 1 mole of N 2 O 4 = 92.0 g/mol

أكملى العبارات التالية A.Molar Mass of K 2 O =.....Grams/mole B.Molar Mass of antacid Al(OH) 3 =......Grams/mole

أجيبى عن السؤال التالى How many grams of Al are in 3.00 moles of Al? 1 mole Al = 27.0 g Al X 3.00 moles Al 3mole of Al X 27 / 1 Answer = 81.0 g Al

Atoms/Molecules and Grams 1- How many atoms of Cu are present in 35.4 g of Cu? 2- How many atoms of K are present in 78.4 g of K? 3- What is the mass (in grams) of 1.20 X molecules of glucose (C 6 H 12 O 6 )?

النسبة المئوية Percent Composition النسبة المئوية Percent Composition What is the percent carbon (C) in C 5 H 8 NO 4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 %C b) 24.3 %C c) 41.1 %C

Chemical Formulas of Compounds الصيغ الكيميائية Empirical FormulaEmpirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic formula are always empirical formula Molecular FormulaMolecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound

A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. require mole ratios so convert grams to moles moles of N = 2.34g of N = moles of N g/mole g/mole moles of O = 5.34 g = moles of O g/mole g/mole Formula=

Calculation of the Molecular Formula مركب صيغته الأولية NO2 ، الوزن الجزيئ لهذا المركب 92 g/mol أوجدى الصيغة الجزيئية 1- calculate of molecular weight of NO2 = 14+2x16 =46 2-repating factor = molecular weight/ molecular weight of NO2 = 92/46 = 2 3 – multply NO2 x 2 = N2O4

المعادلة الكيميائية chemical equation يمكن الاستفادة من المعادلات الكيميائية في اجراءالحسابات المتعلقة بالكميات المستخدمة من التفاعلات وبكميات المواد الناتجة وذلك لما تمتاز به المعادلة الكيميائية في أنها تعبير واضح للدلالة على عدد الذرات أو الجزيئات المتفاعلة أو الناتجة في تفاعل كيميائي معين.

حسابات المعادلة الكيميائية فعلى سبيل المثال لدينا التفاعل التالي : كم كتلة حمض الكبريت التي تتفاعل مع (80g) من النحاس حسب المعادلة السابقة؟ الحل :

وبما أن المعادلة السابقة تقتضي أن يتفاعل مع المول الواحد من النحاس مولان من حمض الكبريت فإن :

When N2O5 is heated, it decomposes 2N2O5(g)  4NO2(g) + O2(g) How many moles of N2O5 were used if 210g of NO2 were produced تحول الجرامات الى مولات n= m/M = 210 /108 = 4.56 mol 2 mole (N2O5) 4 NO X X= mole of N2O5 = 4.56 x2 /4 = 2.24 تحول مولات N2O5 الى جرامات n=m/M 2.24 = m/108 = g

Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Two starting amounts? Where do we start? Hide one

3H 2 (g) + N 2 (g)  2NH 3 (g) 4 g H2 react with 40 g N2 and produced 5 g of NH3 (actual yield) Find the limiting reactant How many gram of NH3 will produced (therotical yield) نحول الجرامات الى مولات n (H2) = 4/2 = 0.5 mol n (N2) = = 1.43 molن نقسم على معاملات كل متفاعل n (H2) 0.5/3=0.166 n (N2) 1,43/1 =1 الأصغر فى المولات هوH2 ومنه يتضح أن H2 هو المتفاعل المحدد الذي يستهلك بأكمله، وأن N2 هو المتفاعل الفائض الذي يستهلك بعضه ويبقى جزء منه فائضاً.

Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Hide Based on: KO 2 = mol O mol KO

تابع حل السؤال نحسب مولات NH3 3 mole (H2) 2 mol NH3 0.5 mole (H2) X X = 0.5 x2 /3 = mole بعد حساب مولات NH3 تحول الى جرامات 5.6 = 0.33 x 17 المحصول المئوي = ( المحصول الحقيقي / المحصول النظري ) × 100 = 5/5.6 x100