chapter-03 - Have to do experiment to get actual yield...

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CHEM 1331 CHAPTER 3 Definition of Mole 6.02 x 10 23 Definition of Molar Mass : 1 mole of a substance weights the atomic/molecular mass in grams Important Equations moles = grams / molar mass Molecules = 6.02 x 10 23 x moles Molarity molarity = moles of solute / liters of solution Stoichiometry BALANCED EQUATIONS Relate relative numbers of moles of substances: If 2 substances A and B have coefficients a and b: Moles of B = (b / a) x moles of A ROADMAP STRESSES RELATIONSHIPS Moles of A n A Moles of B n B Particles N A Particles N B grams of A g Volume of Solution, L grams of B g Volume of Solution, L n B = (b/a)n A M = n / L n = g / MM N = 6.02 x 10 23 n N = 6.02 x 10 23 n Limiting Reagents If given information about more than one reactant: (a) calculate moles of “B” for each reactant individually (b) smaller number of moles of B comes from limiting reactant (c) use this number to finish problem
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Percent Yield %yield = {[actual yield] / [theoretical yield]} x 100 Calculate theoretical yield as above
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Unformatted text preview: Have to do experiment to get actual yield Empirical Formula Ratio of number of each type of atom Related to MOLECULAR FORMULA by whole number multiple Calculate from relative masses of elements For each element in substance: (a) convert to moles (b) convert moles to nice numbers by dividing by smallest number (c) if necessary, multiply ALL by 2, 3 etc to clear fractions If given PERCENT of each element, use percent as grams If not given one element, subtract other masses from total sum For combustion of C, H compounds or C, H, O compounds Sometimes given g of CO 2 and H 2 O, Convert to: g of C by multiplying CO 2 mass by (12 / 44) g of H by multiplying H 2 O mass by (2 / 18) Percent by Mass From formula: %mass of element = mass of element in 1 mole x 100 Molar mass of compound Dilutions Diluting a solution: Old molarity x old volume = new molarity x new volume...
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chapter-03 - Have to do experiment to get actual yield...

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