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

What’s the Percent composition of sugar? Sucrose Formula C12H22O11

Percentage Composition (by mass...not atoms) 24.31 g 95.21 g % Mg = x 100 Mg magnesium 24.305 12 Cl chlorine 35.453 17 70.90 g 95.21 g % Cl = x 100 25.53% Mg Mg2+ Cl1- Law of definite proportions states that a chemical compound always contains the same proportion of elements by mass Percent composition — the percentage of each element present in a pure substance—is constant Calculation of mass percentage 1. Use atomic masses to calculate the molar mass of the compound 2. Divide the mass of each element by the molar mass of the compound and then multiply by 100% to obtain percentages 3. To find the mass of an element contained in a given mass of the compound, multiply the mass of the compound by the mass percentage of that element expressed as a decimal 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

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).

What is the % H2O in nickel chloride dihydrate, NiCl2 ∙ 2H2O? Element # g/mol (molar mass) TOTAL Ni 1 58.69 g/mol 58.69g Cl 2 35.45 g/mol 70.90g H2O 18.02 g/mol 36.04g MOLAR MASS= 165.63g/mol NiCl2 * 2H2O %H2O = = 21.76% H2O

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 C6H6 CH

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 Empirical formula gives only the relative numbers of atoms in a substance in the smallest possible ratio Molecular formula gives the actual number of atoms of each kind present per molecule

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

Empirical Formula Find the empirical formula of this compound. Quantitative analysis shows that a compound contains 50.04% carbon, 5.59% hydrogen, and 44.37% oxygen. Find the empirical formula of this compound. 50.04% C 5.59% H 44.37% O 50.04g C 5.59g H 44.37g O = 4.17 mol C / 2.77 mol = 1.5 C X x2 = 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

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

Empirical Formula Find the empirical formula of this compound. 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 32.38 g Na 22.65 g S 44.99 g O X = 1.408 mol Na / 0.708 mol = 2 Na = 0.708 mol S = 1 S Na2SO4 = 2.812 mol O = 4 O Step 1) %  g Step 2) g  mol Step 3) mol mol