Understanding the Law of Conservation of Mass Who Discovered this Law? • • • • • • 1789, France Antoine Lavoisier Nobleman Statesman Scientist Used one of the first analytical mass balances to prove this law. • Executed on the guillotine during the French Revolution. • He is known as the “Father of Chemistry” because he made it a quantitative science. What does the Law of Conservation of Mass State? • During any chemical reaction, matter is neither created nor destroyed. Mass is conserved from reactants to products. • Therefore, MASS REACTANTS = MASS PRODUCTS What does the law really mean? CH4 (g) + methane gas 2O2 (g) oxygen gas CO2 (g) + carbon dioxide gas 2H2O (l) water Reactants: methane gas (CH4) and oxygen gas (O2) Products: carbon dioxide gas (CO2) and water (H2O) “” means yields. Little numbers (subscripts) – tell how many of a particular type of atom are inside of a molecule. (ex. 4 hydrogen atoms per methane molecule) Big numbers (coefficients) – tell how many of each particle is involved in the reaction. (ex. 2 molecules of oxygen react with 1 molecule of methane) How would you draw this reaction as particles and show conservation of mass? CH4 (g) + methane gas 2O2 (g) oxygen gas CO2 (g) + carbon dioxide gas 2H2O (l) water How does this picture show that particles and therefore mass are conserved from reactant’s side to product’s side? What is all that really happens to the particles in a chemical reaction? Can atoms of one type be changed into (transformed) atoms of another type during a chemical reaction? Note about showing “conservation” particle diagrams in If you have the reaction: A2 + B2 A3B You would show conservation by drawing + 3A2 + 1B2 2A3B Do not simply add stray “atoms” to molecules. It changes them to a different substance. Solving simple math problems involving the Law of Conservation of Mass – copy both solution and answer! MASS REACTANTS = MASS PRODUCTS • If 16 grams of CH4 reacts completely with 64 grams of O2, what mass of products should form? CH4 (g) + methane gas 2O2 (g) oxygen gas 16g + 64g 80 g CO2 (g) + carbon dioxide gas = = Xg X 2H2O (l) water Another Problem… • If 32 grams of CH4 reacts completely with 128 g of O2, and 88 g of CO2 forms, how many grams of H2O form? CH4 (g) + methane gas 2O2 (g) oxygen gas CO2 (g) + carbon dioxide gas 32 g + 128 g = 88 g 160 g = 88 g 72 g = X + + 2H2O (l) water X X Another Problem… • If 8 grams of CH4 react completely with oxygen, and 22 g of CO2 and 9 g of H2O form, how much oxygen (O2) was consumed? CH4 (g) + methane gas 8g 8g 2O2 (g) + + oxygen gas X X X = = = CO2 (g) + carbon dioxide gas 22 g 23 g 2H2O (l) water + 9g 31 g Last Problem • 4 grams of CH4 reacts with 20 g of O2. The CH4 is used up completely but there is some O2 left over. Given that 20 grams of product was formed, how much oxygen was used up? CH4 (g) + methane gas 2O2 (g) oxygen gas CO2 (g) + carbon dioxide gas 2H2O (l) water 4g + X = 20 g X = 16 grams oxygen used up and 20 g oxygen available – 16 grams used = 4 g oxygen left over.