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

2. Hydrate Lab – for tomorrow  Pre-lab  Title  Purpose  Materials  Procedure  Data table HydrateAnhydrous

3. Hydrates Water molecules are incorporated into the crystalline structure. Methane hydrate  Hydrates, like zinc acetate dihydrate, Zn(C 2 H 3 O 2 ) 2 * 2H 2 O are commonly found in skin care products such as moisturizer, shampoo and lip balm.

4. What is the % H 2 O in nickel chloride dihydrate, NiCl 2 * 2H 2 O? Element#g/mol (molar mass) TOTAL Ni158.69 g/mol58.69g Cl235.45 g/mol70.90g H2OH2O218.02 g/mol36.04g MOLAR MASS= 165.63g/mol NiCl 2 * 2H 2 O %H 2 O = = 21.76% H 2 O

5. % Mg = x 100 24.31 g 95.21 g Percentage Composition Mg magnesium 24.305 12 Cl chlorine 35.453 17 Mg 2+ Cl 1- MgCl 2 1 Mg @ 24.31 g= 24.31 g 2 Cl @ 35.45 g= 70.90 g 95.21 g 25.52% Mg 74.48% Cl (by mass...not atoms) It is not 33% Mg and 66% Cl % = x 100 part whole

6. Empirical and Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. PERCENT BY WEIGHT Therefore, we can express molecular composition as PERCENT BY WEIGHT. Ethanol, C 2 H 6 O 52.13% C 13.15% H 34.72% O

7. Different Types of Formulas 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 C6H6C6H6 CH

8. Calculating Empirical formula  Percent to mass (assume 100g)  Mass to mole (molar mass)  Divide by the smallest (ratio)  Multiply til’ whole*

9. 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. = 4.17 mol C = 5.59 mol H = 2.77 mol O / 2.77 mol = 1.5 C = 2 H = 1 O C3H4O2C3H4O2 50.04% C 5.59% H 44.37% O 50.04g C 5.59g H 44.37g O Step 1) %  gStep 2) g  mol Step 3) mol mol Step 4) multiply til whole *2 XXXXXX

10. 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. = 1.050 mol Cu = 0.3500 mol P = 1.401 mol O / 0.3500 mol =3 Cu = 1 P = 4 O Cu 3 PO 4 66.75% Cu 10.84 % P 22.41 % O 66.75g Cu 10.84g P 22.41g O copper (I) phosphate Step 1) %  gStep 2) g  mol Step 3) mol mol Cu 3 PO 4 XXXXXX

11. 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. = 1.408 mol Na = 0.708 mol S = 2.812 mol O / 0.708 mol = 2 Na = 1 S = 4 O Na 2 SO 4 32.38% Na 22.65% S 44.99% O 32.38 g Na 22.65 g S 44.99 g O sodium sulfate Step 1) %  gStep 2) g  mol Step 3) mol mol Na 2 SO 4 XXXXXX