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

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

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.

What is the % H 2 O in nickel chloride dihydrate, NiCl 2 * 2H 2 O? Element#g/mol (molar mass) TOTAL Ni g/mol58.69g Cl g/mol70.90g H2OH2O g/mol36.04g MOLAR MASS= g/mol NiCl 2 * 2H 2 O %H 2 O = = 21.76% H 2 O

% Mg = x g g Percentage Composition Mg magnesium Cl chlorine Mg 2+ Cl 1- MgCl g= g g= g g 25.52% Mg 74.48% Cl (by mass...not atoms) It is not 33% Mg and 66% Cl % = x 100 part whole

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

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

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

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 C3H4O2C3H4O % 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

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. = mol Cu = mol P = mol O / mol =3 Cu = 1 P = 4 O Cu 3 PO % Cu % P % 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

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