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Take out #46 & stampsheet . Homework … Complete prelab

Take out #46 & stampsheet . Homework … Complete prelab. Do Now : If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?. 24.305. 35.453. Mg. Cl. 12. 17. magnesium. chlorine. Percentage Composition. (by mass...not atoms). 24.31 g 95.21 g.

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Take out #46 & stampsheet . Homework … Complete prelab

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  1. Take out #46 & stampsheet. Homework … Complete prelab

  2. Do Now: If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?

  3. 24.305 35.453 Mg Cl 12 17 magnesium chlorine Percentage Composition (by mass...not atoms) 24.31 g 95.21 g % Mg = x 100 70.90 g 95.21 g % Cl = x 100 25.53% Mg Mg2+ Cl1- 74.47% Cl MgCl2 It is not 33% Mg and 66% Cl 1 Mg @ 24.31 g= 24.31 g 2 Cl @ 35.45 g= 70.90 g 95.21 g

  4. Hydrate Lab CuSO4 ∙ ?H2O CuSO4 Hydrate Anhydrous • Water molecules are incorporated into the crystalline structure. • Hydrates, like zinc acetate dihydrate, Zn(C2H3O2)2 ∙ 2H2O are commonly found in skin care products (ex: moisturizer, shampoo & lip balm).

  5. What is the % H2O in nickel chloride dihydrate, NiCl2 * 2H2O? %H2O = = 21.76% H2O

  6. Empirical and Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY WEIGHT. Ethanol, C2H6O 52.13% C 13.15% H 34.72% O

  7. Different Types of Formulas C6H6 • Molecular Formula – shows the real # of atoms in one molecule or formula unit • Empirical Formula – shows smallest whole number mole ratio **Sometimes the empirical & molecular formula can be the same • Structural Formula- molecular formula info PLUS bonding electron and atomic arrangement CH

  8. The Empirical Formula March • Percent to mass • Mass to mole • Divide by smallest • Return to whole

  9. 50.04g C 5.59g H 44.37g O / 2.77 mol / 2.77 mol / 2.77 mol Empirical Formula Quantitative analysis shows that a compound contains 50.04% carbon, 5.59% hydrogen, and 44.37% oxygen. Find the empirical formula of this compound. = 1.5 C = 4.17 mol C 50.04% C 5.59% H 44.37% O • x2 • x2 • x2 X X X = 5.59 mol H = 2 H C3H4O2 = 2.77 mol O = 1 O Step 1) %  g Step 2) g  mol Step 3) mol mol Step 4) return to whole

  10. 66.75g Cu 10.84g P 22.41g O / 0.3500 mol / 0.3500 mol / 0.3500 mol Empirical Formula Quantitative analysis shows that a compound contains 66.75% copper, 10.84% phosphorus and 22.41% oxygen. Find the empirical formula of this compound. =3 Cu 66.75% Cu 10.84 % P 22.41 % O = 1.050 mol Cu X X X Cu3PO4 = 1 P = 0.3500 mol P = 4 O = 1.401 mol O Step 1) %  g Step 2) g  mol Step 3) mol mol

  11. 32.38 g Na 22.65 g S 44.99 g O / 0.708 mol / 0.708 mol / 0.708 mol Empirical Formula Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound. 32.38% Na 22.65% S 44.99% O X X X = 2 Na = 1.408 mol Na Na2SO4 = 0.708 mol S = 1 S = 2.812 mol O = 4 O Step 1) %  g Step 2) g  mol Step 3) mol mol

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