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## Ch. 3 Stoichiometry: Calculations with Chemical Formulas

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**Law of Conservation of Mass**• Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged.**Stoichiometry**• The quantitative nature of chemical formulas and chemical reactions**Reactants**• The chemical formulas on the left of the arrow that represent the starting substances 2H2 + O2 2H2O Reactants**Products**• The substances that are produced in the reaction and appear to the right of the arrow 2H2 + O2 2H2O Products**Because atoms are neither created nor destroyed in any**reaction a chemical equation must have the same number of atoms of each element on either side of the arrow**Balancing Chemical Equations**CH4 + O2 CO2 + H2O C=1 C=1 H=4 H=2 O=2 O=3**Balancing Chemical Equations**CH4 + O2 CO2 + 2H2O C=1 C=1 H=4 H=2 X 2 =4 O=2 O=3**Balancing Chemical Equations**CH4 + O2 CO2 + 2H2O C=1 C=1 H=4 H=2 x 2 = 4 O=2 O= 2 + 2x1 = 4**Balancing Chemical Equations**CH4 + 2O2 CO2 + 2H2O C=1 C=1 H=4 H=2 x 2 = 4 O=2 x 2 = 4 O= 2 + 2x1 = 4**Combustion Reactions**• Rapid reactions that produce a flame. • Most combustion reactions involve O2 as a reactant • Form CO2 and H2O as products**Combustion Reactions**C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) C= 3 C=1 X 3 = 3 H=8 H=2 X 4 = 8 O= 2 X 5 = 10 O=(2 X3)+(1X4)=10**Combination Reactions (synthesis)**• 2 or more substances react to form 1 product.**Combination Reactions (synthesis)**2Mg(s) + O2(g) 2MgO(s) Mg=1 x 2=2 Mg= 1 x 2=2 O= 2 O=1 x 2 = 2**Decomposition Reaction**• 1 substance undergoes a reaction to produce 2 or more substances**Decomposition Reaction**CaCO3 (s) CaO (s) + CO2(g) Ca=1 Ca=1 C=1 C=1 O=3 O=1+2=3**3 Methods of Measuring**• Counting • Mass • Volume**Example 1**• If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?**Example 1**• Count: 1 dozen apples = 12 apples • Mass: 1 dozen apples = 2.0 kg apples • Volume: 1 dozen apples = 0.20 bushels apples Conversion Factors: • 1 dozen2.0 k.g1 dozen 12 apples 1 dozen 0.20 bushels**Example 1**• 0.50 bushel x 1 dozen x 2.0 kg = 0.20 bushel 1 dozen = 5.0 kg**Avogadro’s Number**• Named after the Italian scientist Amedo Avogadro di Quaregna • 6.02 x 10 23**Mole (mol)**• 1 mol = 6.02 x 10 23 representative particles • Representative particles: atoms, molecules ions, or formula units (ionic compound)**Mole (mol)**• Moles= representative x 1 mol particles 6.02 x 10 23**Example 2 (atoms mol)**• How many moles is 2.80 x 10 24 atoms of silicon?**Example 2**• 2.80 x 10 24 atoms Si x 1 mol Si 6.02 x 10 23 atoms Si = 4.65 mol Si**Example 3 (mol molecule)**• How many molecules of water is 0.360 moles?**Example 3**• 0.360 mol H2O x 6.02 x 10 23 molecules H2O 1 mol H2O =2.17 molecules H2O**The Mass of a Mole of an Element**• The atomic mass of an element expressed in grams = 1 mol of that element = molar mass Molar mass S Molar mass Hg Molar mass C Molar mass Fe**6.02 x 10 23 atoms S**6.02 x 10 23 atoms Hg 6.02 x 10 23 atoms C 6.02 x 10 23 atoms Fe**Example 4 (mol gram)**• If you have 4.5 mols of sodium, how much does it weigh?**Example 4**• .45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na 1 mol Na**Example 5 (grams atoms)**• If you have 34.3 g of Iron, how many atoms are present?**Example 5**• 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms 55.8 g Fe 1 mol Fe =3.70 x 10 23 atoms Fe**The Mass of a Mole of a Compound**• To find the mass of a mole of a compound you must know the formula of the compound • H2O H= 1 g x 2 O= 16 g 18 g = 1 mole = 6.02 x 10 23 molecules**Example 6 (gram mol)**• What is the mass of 1 mole of sodium hydrogen carbonate?**Example 6**• Sodium Hydrogen Carbonate = NaHCO3 • Na=23 g • H=1 g • C=12 g • O=16 g x3 • 84 g NaHCO3 = 1 mol NaHCO3**Mole-Volume Relationship**• Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical**Avogadro’s Hypothesis**• States that equal volumes of gases at the same temperature and pressure contain the same number of particles • Even though the particles of different gases are not the same size, since the gas particles are spread out so far the size difference is negligible**Standard Temperature and Pressure (STP)**• Volume of a gas changes depending on temperature and pressure • STP= 0oC (273 K) 101.3 kPa (1 atm)**Standard Temperature and Pressure (STP)**• At STP, 1 mol = 6.02 X 1023 particles = 22.4 L of ANY gas= molar volume**Conversion Factors**• AT STP • 1 mol gas22.4 L gas 22.4 L gas 1 mol gas**Example 7**• At STP, what volume does 1.25 mol He occupy?**Example 7**• 1.25 mol He x 22.4 L He = 28.0 L He 1 mol He**Example 8**• If a tank contains 100. L of O2 gas, how many moles are present?**Example 8**• 100. L O2 X 1 mol O2 = 4.46 mol O2 22.4 L O2**Calculating Molar Mass from Density**• The density of a gas at STP is measured in g/L • This value can be sued to determine the molar mass of gas present**Example 9**• A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.**Example 9**• 1 mol gas x 22.4 L gas X 3.58 g gas = 1 mol gas 1 L gas Molar Mass= 80.2 g**Percent Composition**• The relative amounts of the elements in a compound • These percentages must equal 100**Percent Composition**• %element = mass of element x 100 mass of compound