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UT Arlington CHEM 1441 - Exam 2 Study Guide

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Chemistry 1441 1st EditionExam # 2 Study Guide Chapters: 3 & 4Chapter 3: (Stoichiometry of Formulas and Equations)Mole and Molar Mass1. Mole is a number word:- 6.022x1023 is called Avagadros’ Number, Na- “mole” is often abbreviated as “mol”- These and equalities (conversion factors for dimensional analysis- Ex: How many moles of atoms are present in 9.5x1015 atoms?#mol atoms = 9.5x1015 atoms x 1 mol atoms6.022 ×1023atoms = 1.6x10 -8mol atoms2. Mole relates to mass:- Molar mass: uses to convert moles to mass and mass back to moles- Ex: 1 mol C = 12.01g C- Ex: 1 mol H2O = (2 x 1.008) x (1 x 16.00) = 18.02g H2O- Molar mass formula: M = mass/nM = Molar massN = number of moles- Ex: How many iron atoms are in 675g of iron?¿ Fe atoms=675 g Fe ×1 molFe55.85 gFe×6.022× 1023Fe atoms1 molFe=7.28× 1024Fe atoms- Hint: A simple way to work these problems out is to write down what you wantto find and equal it to what you have and then go from there. 3.Mol-to-Mol ratios can be found- From the subscripts of chemical formulas: magnesium nitrate Mg(NO3)2- Coefficient of Balanced Formulas- THAT’S WHY KNOWING HOW TO BALANCE EQUATIONS IS VITAL BECAUSE YOU NEED TO KNOW THE CORRECT COEFFICIENT TO SOLVE PROBLEMS such as when we get to COMBUSTION REACTIONSMole-to-Mole Ratios from Chemical Formulas- Ex: How many ions are in 15.0 g of aluminum carbonate?NH4¿2CO3¿NH4¿2CO3¿NH4¿2CO3¿1 mol ¿96.09 g ¿1 mol ¿NH4¿2CO3׿¿ ions=15.0 g ¿Balancing Chemical Equations- Ex: 1 Fe3O4 + 4 H23 Fe + 4 H2O- Combustion Reactions:- When a hydrocarbon (compound containing only C and H), or a compouns containing only C, H, and O goes through a complete combustion in oxygen, the products are CO2 and H2O.- Ex: C,Hor C,H,O + O2CO2 + H2OReaction Stoichiometry- For stoichiometry problems we must balance the equation correctly and use the coefficients to determine the quantities in chemical reactions- Ex: (complete combustion reaction)- What mass of oxygen gas is required for the complete combustion of 10.0 g of cyclopentanol, C5H9OH?1 C5H9OH + 7 O25 CO2 + 5 H2O¿ g O2=10.0 g C5H9OH ×1 mol C5H9OH86.13 g C5H9OH×7 mol O21 mol C5H9OH×32.00 g O21mol O2=26.0 g O2Percent Yield- Percent Yield = actual yieldtheoretical yield× 100 %Limiting Reactant- The limiting reactant is the reactant you run out of first- Ex: If you have 20 men and 30 women and you are pairing up couples with the 1:1 ration then you would run out of men and have 10 women left without partners. The men would then be the limiting reagent and the women would be the excess reagent because you have an excess of 10 women.Percent Composition of Compounds- % = (part/whole) x 100%- Ex: What is the percent of hydrogen in water?% H = [mass H/mass H2O] x 100%% H = [(2 x 1.008)/18.02] x 100%% H = 11.2%Empirical and Molecular Formulas- Molecular formula: the # of atoms in each element in a molecule- Empirical formula: the smallest whole-number mol-to-mol ratio of elements- Ex: The molecular formula of benzene is C6H6.The Empirical formula would be CH.6:6 is equivalent to 1:1 ratioChapter 4 (Three Major Classes of Chemical Reactions)Molarity and Solution Stoichiometry- Molarity (M) = # mol solute/Vsolution(L) OR M = n/V- Solution by dilution: M1V1 = M2V2OR MconcVconcentration = MdilVdilutionElectrolytes1) Review:- Names, formulas & charges of all ions- Aqueous solutions, solvents, solutes, solutions2) Go back and review aqueous solutions and the conduction of electricity.3) When ionic compounds dissolve in water they dissociate into ions.4) Review what ionization means.5) Know what electrolyte means.6) Strong Electrolyte7) Weak electrolytes8) Nonelectrolytes9) Strong Acids10) Strong Bases11) Define saltsPrecipitation Reactions- Occurs when 2 soluble ionic compounds form an insoluble product- Known as double replacement reqaction- MUST KNOW SOLUBILITY RULES: (review notes)- Soluble Ionic Compounds- Insoluble Ionic CompoundsEX: Cu(OH)2Insoluble- Strong electrolyte = strong baseAcid-Base Reactions- Double displacement OR Metathesis Reactions- Two types: Precipitation and Acid Base ReactionsSolution Stoichiometry1) Write out the balanced equation2) Write what your given of each compound3) Write what you are looking for and equal it to the volume you are given4) When conduct a dimensional analysis until the unit you don’t need are canceled and youhave to unit you needREFER toNOTES for MEMORIZATION RULES and Chapter 1 & 2


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UT Arlington CHEM 1441 - Exam 2 Study Guide

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