1. Chemical Quantities – What's the matter? – Matter is made up of either elements or compounds. – Elements are pure substances made of one kind of atom. They cannot be broken down into simpler chemical substances. – Compounds are two or more elements chemically combined into an unchangeable proportion. They can be broken down into simpler substances. Demonstration: Sugar(C 12 H 22 O 11 ) & Sulfuric Acid( H 2 SO 4 ) (draw in your notes) – Sugar is made from carbon, hydrogen and oxygen. The reaction causes bonds to break(absorb energy) and form new ones(release energy) into carbon and water(H 2 O) Everything!

2. Elements & Compounds Review: Chemical symbols indicate the element. Chemical formulas indicate how many atoms of each element. Formulas are ratios Example: A molecule of sucrose, chemical formula - C 12 H 22 O 11 12 atoms of carbon 22 atoms of hydrogen 11 atoms of oxygen How much of the sugar molecule is made from carbon? Or hydrogen? Or oxygen? Some elements are smaller in mass than others Example: hydrogen has 22 atoms in sucrose nearly twice as that of oxygen or carbon Only makes up 6.49% of the sucrose molecule

3. Elements & Compounds Atomic Mass: The weighted average of the masses of all the isotopes of an element. Found on the Periodic Table Measured in amu (atomic mass units) Example: Atomic mass ofcarbon hydrogen oxygen Formula Mass: the sum of the atomic masses of all the atoms in a chemical formula. Example:C 12 H 22 O 11 carbon:12.01amu x12 atoms= hydrogen: 1.01amu x22 atoms= oxygen:16.00amu x11 atoms= = 16.00 amu 144.12amu 22.22amu + 176.00amu 342.34amu Find the formula masses for: Na 2 SO 4 CH 2 O Fe 2 O 3 C 6 H 12 O 6 = 12.01 amu = 1.01 amu

4. Percent Composition  Percent composition: percent by mass of a element that is contained in a compound  total mass of the element divided by the total mass of the compound, multiplied by 100.  Example: What percent does hydrogen make up in Sucrose, C 12 H 22 O 11 ? Carbon:12 atoms x 12.01amu = Hydrogen:22 atoms x 1.01amu = Oxygen: 11 atoms x 16.00amu = +144.12amu + 22.22amu +176.00amu 342.34amu % comp = 22.22amu 342.34amu x100 % comp = 6.49%

5. Percent Composition  % composition by mass  Formula works with just mass in grams Example: 200.grams of sugar was decomposed into 84.5grams of carbon. What percent is carbon in the sucrose? 84.5g carbon 200.g sucrose % mass of carbon = x 100 % mass of carbon =42.25%

6. Percent Composition Hydrate – a substance that has a specific number of water molecules bonded to each of its formula units. Often a salt Example: sodium carbonate decahydrate: Na 2 CO 3 ∙10H 2 O The dot represents a loose bond between Na 2 CO 3 and H 2 O Coefficient shows how many water molecules are in the formula How much water is in the hydrate Na 2 CO 3 ∙10H 2 O ? Use % composition formula

7. Density ●Review: ●Mass – the amount of matter in a given object ●Measured in grams & kilograms ●Volume – the amount of space an object takes up ●Measured in liters or cm 3 (cubic centimeters) ● Mass and volume are extensive properties- properties that depend on how much of it one has ●Density- the amount of mass per unit volume of a substance. ●how much stuff is in a particular substance ●intensive property- physical properties that do not rely on the amount ●for pure substances, density does not change whether one has 1 gram of a substance for 1,000,000 grams. ●Densities of several substances are found on the R.T. ●Formula: Unit is either g/mL or g/cm 3 Density = mass volume

8. Density ● What is the density of 100.0 grams of water that is contained in a 100.0mL graduated cylinder? What is the volume in milliliters of 75 grams of corn oil that has a density of 1.15 g/mL? What would the volume be for 200. grams of aluminum(Aluminum density is listed on Pg.1 of NC R.T.) ?

9. Empirical Formula ● Molecular formula tells us the precise number of atoms in a compound ● Empirical formula – the simplest ratio of atoms in a compound ● Ex. glucose: C 6 H 12 O 6 ● ratio is 6:12:6, which can be reduced to 1:2:1 ● therefore, the empirical formula for glucose is CH 2 O ● Ex:N 2 H 4 → C3H6→C3H6→ Na 2 CO 3 → NH 2 CH 2 Na 2 CO 3

10. Empirical Formula Molecular Formula from Empirical Formula · Determine the molecular formula for hydrazine if the empirical formula is NH 2 and the formula mass is 32.06amu. Step 1: Find the empirical formula mass. nitrogen:1 atomx14.01amu= hydrogen:2 atomsx 1.01amu= Step 2: Divide the formula mass given by the empirical formula mass. Step 3: Take the result and multiply it the number of atoms in the empirical formula 14.01amu +2.02amu 16.03amu formula mass:32.06amu empirical formula mass:16.03amu = 2 N (1x2) H (2x2) =N 2 H 4