Chemical Reactions

Warm-up 2/25 and 2/29 Since atoms are too small to be seen chemists typically will determine how much of a substance is present by measuring the mass. To convert from mass to the number of atoms or molecules a chemist must be able to answer the following questions: Does every atom have the same mass? Why or why not? Chemical Reactions

Learning Target #1 Define the mole and use in calculations Chemical Reactions

One mole = 6.02 × 1023 This number is called Avogadro’s number This is similar to a “dozen” One dozen = 12 One mole of carbon-12 atoms (6 protons, 6 neutrons, and 6 electrons) has a mass of 12 g One mole of pennies divided evenly between everyone in the world would give everyone 860 billion dollars! Chemical Reactions

The average mass of each atom can be found on the periodic table and is called the atomic mass. The unit is grams per mole. The molar mass isthe sum of all the atomic masses within a compound. Check for understanding – What is the molar mass of water H2O? 1.0 g/mol for each H atom + 16.0 g/mol from the O atom = 18.0 g/mol Chemical Reactions

Atomic mass and molar mass can be used as a conversion factor between mass and moles. Mass (g) ÷ atomic/molar mass (g/mol) = Moles (mol) Moles (mol) x atomic/molar mass (g/mol) = Mass (g) Chemical Reactions

Chemical Reactions Check your understanding – How many moles are in 10 g of carbon? 10 g x 1 mol/12 .0g = 0.83 mol Check your understanding – How many grams are in 2 moles of gold? 2 mol x 197.0 g/1 mol = 394 g

Reflection: Make your own analogy about moles! The average person is about 1.7 meters tall. The sun is 149,600,000,000 meters away from earth. How many times would one mole of people standing feet-to-head reach the sun? 1 mole = 6.02 x 1023 Chemical Reactions

Warm-up 3/1 and 3/2 How many moles of H2O are in 1000 g (about a liter) of H2O? Using your answer from question #1, how many molecules of H2O are in 1000 g (about a liter) of H2O? Chemical Reactions

Avogadro’s number (6.02 x 1023) is used to convert from moles to the number of actual particles (atoms, molecules, ions, etc). Check for understanding – How many eggs are in two dozen eggs? 2 dozen x 12 eggs/dozen = 24 eggs Check for understanding – How many atoms are in two moles of atoms? 2 moles x 6.02 x 1023 atoms/mole = 1.20 x 1024 atoms Chemical Reactions

Note! Be very careful with vocabulary during mole/particle conversions. Atoms refer to individual elements Ions are a charged atom within a compound Molecules are the entire compound Example! The compound CH4 is one molecule and contains five atoms Chemical Reactions

Reflection 3/1 and 3/2 What does the mole concept allow chemists to do? Do heavier elements (by atomic mass) or lighter elements contain more atoms per gram? Why? Chemical Reactions

Warm-up 3/3 and 3/7 Define “atomic mass” How do you find an atomic mass? Define “molar mass” How do you find a molar mass? Define “mole” How can you calculate moles from a mass? From a number of particles? Chemical Reactions

When converting from mass to number of particles two steps are required. Moles is always your intermediate step! Chemical Reactions

Sample problem: How many atoms are in 10 g of H2O? 1. Identify knowns and unknowns Known = 10 g Unknown = ? Atoms 2. Create conversion plan Mass to moles to molecules to atoms 3. Find molar mass/atomic mass/# of atoms Molar mass = 1.0 x 2 + 16.0 = 18.0 g/mol 3 atoms/molecule 4. Solve! 10 g x 1 mol/18.0 g x 6.02 x 1023 molecules/mol x 3 atoms/molecule = 1.00 x 1024 atoms Chemical Reactions

Agenda 3/4 • Finish Mole Conversion Packet! • Have Mr. B check your work • Turn in notebook • Take Mole Conversion Quiz • You will use the class period to work. Period. Chemical Reactions

Agenda 3/11 • Turn in notebooks • Take “Mole Calculation” quiz • Continue lab work OR • Work on missing assignments Chemical Reactions

Warm-up 3/28 and 3/29 What percentage of this class is boys? What percentage of this class is girls? How do you calculate that? Chemical Reactions

Learning Target #2 Calculate the percent composition by mass, the empirical formula, and the molecular formula of varying compounds. Chemical Reactions

Percent Composition (by mass) General equation Mass of the part ÷ Total mass × 100 = Chemistry specific equation Mass of an atom ÷ Mass of the compound × 100 = Note: remember that percent means per one hundred Chemical Reactions

Percent Composition (by mass) Example! – What is the percent composition of hydrogen and oxygen in H2O? Step 1: Calculate molar mass (total mass) 1.0 g/mol x 2 + 16.0 g/mol = 18.0 g/mol Step 2: Divide mass of each atom by molar mass and multiply by 100 1.0 g/mol x 2 ÷ 18.0 g/mol x 100 = 11% 16.0 g/mol ÷ 18.0 g/mol x 100 = 89% Chemical Reactions

Warm-up 3/30 and 3/31 What is the percent by mass composition of glucose (C6H12O6)? What is the empirical formula of glucose (C6H12O6)? Chemical Reactions

Empirical formula from Percent Composition Remember this rhyme . . . Percent to mass Mass to mole Divide by small Multiply until whole Chemical Reactions

Empirical formula from Percent Composition Example: What is the empirical formula of a compound containing 27% carbon and 73% oxygen by mass? Chemical Reactions

Empirical formula from Percent Composition Example: What is the empirical formula of a compound containing 27% carbon and 73% oxygen by mass? Step 1: Percent to mass 27% = 27 g C 73% = 73 g O Chemical Reactions

Empirical formula from Percent Composition Example: What is the empirical formula of a compound containing 27% carbon and 73% oxygen by mass? Step 2: Mass to mole 27 g C ÷ 12 g/mol = 2.25 mol C 73 g O ÷ 16 g/mol = 4.5625 mol O Chemical Reactions

Empirical formula from Percent Composition Example: What is the empirical formula of a compound containing 27% carbon and 73% oxygen by mass? Step 3: Divide by small Divide EVERY mole number by the smallest mole number, this may or may not yield our whole # ratios for the subscripts 2.25 mol C ÷ 2.25 mols = 1 C 4.5625 mol O ÷ 2.25 mols = 2.03 O Chemical Reactions

Empirical formula from Percent Composition Example: What is the empirical formula of a compound containing 27% carbon and 73% oxygen by mass? Step 4 : Multiply until whole After step 3 check to see if your values round easily to whole #s, if so that will be your final step 1 C = 1 C 2.03 O ≈ 2 O Empirical formula is CO2 Chemical Reactions

Empirical formula from Percent Composition Step 4 : Multiply until whole If you end up further than 0.1 from a whole number you need to multiply until whole using the following table . . . 0.5 = multiply by 2 0.33 or 0.66 = multiply by 3 0.25 or 0.75 = multiply by 4 0.2, 0.4, 0.6, 0.8 = multiply by 5 0.167, 0.833 = multiply by 6 0.14, 0.29, 0.43, 0.57, 0.71, 0.86 = multiply by 7 0.125, 0.375, 0.625, 0.875 = multiply by 8 0.11, 0.22, 0.44, 0.56, 0.78, 0.89 = multiply by 9 Chemical Reactions

Molecular formula from Percent Composition To calculate the molecular formula given percent composition by mass . . . 1.) Follow the same steps used to calculate the empirical formula 2.) Divide the GIVEN molar mass by the molar mass of your calculated empirical formula 3.) Multiply the subscripts of the calculated empirical formula by your answer from Step #2 Chemical Reactions

Molecular formula from Percent Composition To calculate the molecular formula given percent composition by mass . . . Example: Given molar mass is 180 g/mol and the calculated empirical formula is CH2O 180 g/mol ÷ (12.0 + 1.0×2 + 16.0) g/mol = 6 1 C × 6 = 6 C, 2 H × 6 = 12 H, 1 O × 6 = 6 O Molecular formula = C6H12O6 Chemical Reactions

Warm-up 4/1 An unknown compound is composed of47.0% potassium, 14.5% carbon, and 38.5% oxygen by mass and has a molar mass of 166.22 g/mol. What is the empirical formula? What is the molecular formula? Chemical Reactions

Warm-up 4/4 and 4/5 What does Dalton’s Law of Conservation of Mass (think baking soda and vinegar experiment) tell us? Does the following chemical equation showing the decomposition of water demonstrate this law? Why or why not? H2O → H2 + O2 Chemical Reactions

Learning Target #3 – Recognize and balance different kinds of chemical reactions. Chemical Reactions

Balancing chemical equations We know from Dalton’s law of conservation of mass that no atoms will be created or destroyed during a chemical reaction. This is demonstrated in how chemists write equations to represent chemical reactions. Combustion of octane 2 C8H18 + 25 O2 → 18 H2O + 16 CO2 Chemical Reactions

Balancing chemical equations 2 C8H18 + 25 O2 → 18 H2O + 16 CO2 The numbers in front of each compound are called coefficients and provide the mole ratios of each compound and/or atom used and produced in the reaction. Chemical Reactions

Balancing chemical equations 2 C8H18 + 25 O2→ 18 H2O + 16 CO2 The reactants of a chemical reaction are the compounds and/or atoms on the left side of the reaction arrow. This is the starting material for the reaction. The products of a chemical reaction are the compounds and/or atoms of the right side of the reaction arrow. This is what the reaction produces. Chemical Reactions

Balancing chemical equations 2 C8H18 + 25 O2 → 18 H2O + 16 CO2 To count atoms in a chemical equation you simply multiply each atom subscript by the corresponding coefficient Example: On the reactant side there are 8 x 2 = 16 carbon atoms and on the product side there are 1 x 16 = 16 carbon atoms. Chemical Reactions

Balancing chemical equations 2 C8H18 + 25 O2 → 18 H2O + 16 CO2 To count atoms in a chemical equation you simply multiply each atom subscript by the corresponding coefficient Check your understanding: How many H and O atoms are there on each side of the reaction arrow? 2 x 18 = 36 H on left 18 x 2 = 36 on right 25 x 2 = 50 O on left 18 x 1 + 16 x 2 = 50 O on right Chemical Reactions

Balancing chemical equations 2 C8H18 + 25 O2→ 18 H2O + 16 CO2 Important Note #1: The number of each atom remains unchanged when you move from reactant side to product side in a balanced equation. Reactants: Products: 16 Carbon atoms 16 Carbon 36 Hydrogen atoms 36 Hydrogen atoms 50 Oxygen atoms 50 Oxygen atoms Chemical Reactions

Balancing chemical equations 2 C8H18 + 25 O2 → 18 H2O + 16 CO2 Important Note #2: A chemical reaction can be scaled to any size as long as the mole ratios remain constant. Chemical Reactions

Balancing chemical equations If the # of each atom does not remain constant from the reactant to the product side the equation must be balanced. Decomposition of water H2O → H2 + O2 Reactants Products 2 H atoms 2 H atoms 1 O atom! 2 O atoms! Chemical Reactions

Balancing chemical equations H2O → H2 + O2 Balancing chemical equations can be frustrating; it is a trial and error process that requires practice and persistence. Chemical Reactions

Balancing chemical equations 2 H2O → 2 H2 + O2 Tips! You may ONLY change coefficients No coefficient = coefficient of one Balance each atom type one at a time Balance polyatomic ions that appear on both sides of the equation as single units Balance H atoms and O atoms last Visual aids/tallying helps! Chemical Reactions

Balancing chemical equations Practice Simulation URL: https://phet.colorado.edu/sims/html/balancing-chemical-equations/latest/balancing-chemical-equations_en.html Chemical Reactions

Chemical Reactions

Warm-up 4/13 and 4/14 Predict the product! (remember no atoms will be created or destroyed) 1.) 4 Fe + 3 O2→ ??? 2.) 2 H2O → ??? 3.) CuSO4 + Fe → ??? 4.) NaHCO3 + HC2H3O2→ ??? Chemical Reactions

Warm-up 4/13 and 4/14 Record your observations of each reaction! (Work in groups of 2-3) 1.) 4 Fe + 3 O2→ ??? 2.) 2 H2O → ??? 3.) CuSO4 + Fe → ??? 4.) NaHCO3 + HC2H3O2→ ??? Chemical Reactions

Warm-up 4/13 and 4/14 Revise your prediction after observation? 1.) 4 Fe + 3 O2→ ??? 2.) 2 H2O → ??? 3.) CuSO4 + Fe → ??? 4.) NaHCO3 + HC2H3O2→ ??? Chemical Reactions

Warm-up 4/13 and 4/14 1.) 4 Fe + 3 O2 → 2 Fe2O3 Synthesis 2.) 2 H2O → 2 H2 + O2 Decomposition 3.) CuSO4 + Fe → FeSO4+ Cu Single replacement 4.) NaHCO3 + HC2H3O2 → NaC2H3O2 + H2CO3 H2CO3 → H2O + CO2 Double replace/Decomposition Chemical Reactions

Chemical Reactions