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Unit #3 CDA Review Material

Unit #3 CDA Review Material. Molar Mass. MgCl 2 Mg – 1 (24.31) = 24.31 Cl – 2 (35.45) = 70.90 or 95.21 g/ mol. 1 mole MgCl 2 =. 95.21 g MgCl 2. Percent Composition.

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Unit #3 CDA Review Material

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  1. Unit #3 CDA Review Material

  2. Molar Mass MgCl2 Mg – 1 (24.31) = 24.31 Cl – 2 (35.45) = 70.90 or 95.21 g/mol 1 mole MgCl2 = 95.21 g MgCl2

  3. Percent Composition The percent BY MASS of each element in a compound divide the element’s total mass (part) by the molar mass (whole) then multiple by 100 to get the percent. Example – Find the % composition of MgCl2 (Using the info from molar mass on the previous slide Molar mass = 95.21 g/mol) Mg - 24.31 / 95.21 x 100 = 25.53% Mg Cl - 70.90 / 95.21 x 100 = 74.47% Cl (PART) (WHOLE)

  4. Percent of an element in a compound Taking the % OF an amount • Change the percent to decimal (move 2x to left) • Multiply that decimal by the given amount What is the mass of chlorine in 27.8g of MgCl2 (Use the % of Cl you just found on the previous slide) 74.47% Cl .7447 x 27.8g = 20.7g Cl

  5. Identifying Empirical & Molecular Formulas ~ Empirical (CANNOT be reduced) ~ Molecular (CAN be reduced) • C2H4 • NO3 • S9Cl12 • C3Cl9 • N4S9 CH2- empirical molecular empirical S3Cl4 - empirical molecular CCl3 - empirical molecular empirical

  6. Calculating Empirical Formula(% comp  empirical formula) • Change % g for each element • Divide by molar mass of that element • Identify the smallest number of moles and divide all by that number • Round each to the nearest whole # (sometimes you have to multiply to get a whole number - special) • The resulting whole #are the subscripts for that element in the empirical formula

  7. Calculating Empirical Formula 63.5% Silver 8.2% Nitrogen 28.3% Oxygen 63.5 g Ag 8.2 g N 28.3 g O 107.87 14.01 16.00 .589 mole Ag .59 mole N 1.77 mole O .589 .589 .589 1 1 3 AgNO3

  8. Calculating Molecular Formula • Find the empirical formula • Calculate the molar mass of your empirical formula • Identify the molar mass of your molecular (GIVEN in the problem every time!) • Divide the molecular mass / empirical mass • Round to the nearest whole # • Multiply the whole # by the subscripts in the Empirical formula

  9. Practice If a compound has an empirical formula of NO3 and a molecular mass of 186g – what is the molecular formula? formula: NO3 molar mass: 62.01g Molecular mass (given) 186g = 3 empirical mass 62.01 3 x NO3 = N3O9

  10. Mole Conversions

  11. Mole Conversions 1 mole = molar mass (grams) 1 mole = 6.02 x 1023 particles (atoms, molecules, fu) 1 mole = 22.4 Liters

  12. Using molar mass • How many grams are in 15.7 mole MgCl2 • How many moles are in 0.75 grams of silver? • What is the mass of 30.7 mole water?

  13. Using Avagadro’s Number • Remember there are 6.02 x 1023 particles (atoms, molecule, f.u) in 1 mole • How many atoms are in 1 mole? • How many atoms are in 2.10 moles of Copper? • How many moles are in 4.21 x 1026 atoms of aluminum?

  14. Using molar volume • How many liters are in 1 mole of any gas? • How many liters are in 12.95 L of oxygen gas? • How many moles are in 0.758 moles of nitrogen gas?

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