Write the formula for aluminum sulfate Al2 SO4 3 What is its formula weight 2 x Al 53 96 g mol of Al 3 x S 96 21 g mol of S 12 x O 192 00 g mol of O 342 17 g mol of Al2 SO4 3 How many g are in 1 2 moles How many atoms of Al are there O atoms 1 2 mol 342 17 g mol of Al2 SO4 3 4 1 x 102 g 1 2 moles Al2 SO4 3 6 022 x 10 23 formula unit mole 2 Al 1 formula unit 1 4 x 1024 atoms of Al 1 2 moles Al2 SO4 3 6 022 x 10 23 formula unit mole 12 O 1 formula unit 8 7 x 1024 atoms of o Q1 Write the formula for Fe III phosphate Fe3 PO43What is its formula weight to one decimal place Enter that into clicker Fe x 1 55 85 g mol P x 1 30 97 g mol O x 4 64 00 g mol 150 8 g mol How many moles are in 50 0 g of sodium hydroxide NaOH Na 22 99 O 16 00 H 1 00 39 99 g mol 50 0 g 39 99 g mol 1 25 mol NaOH Percent composition and formulas of compounds Percent composition of water H20 Formula weight 18 00 g mol H mass of H mass of H2O x 100 O mass of O mass of H2O x 100 H 2 00 g H 18 00 g x 100 11 1 O 16 00 g O 18 00 g x 100 88 9 always check that it adds up to 100 approximately What is the composition of C in butane Molecular formula C4H10 Formula weight 58 00 g mol of C 48 00g 58 00g x 100 82 8 C Q2 Calculate the composition of carbon monoxide Enter only the O into the clicker to one decimal place CO C x 12 00 O x 16 00 28 00 g mol 16 00 g 28 00 g x 100 57 1 O 12 00 g 28 00 g x 100 42 9 C Derivation of formulas from elemental composition Empirical simplest formula is the smallest whole number ratio of atoms present Example benzene molecular formula C6H6 so empirical formula is CH H2O2 HO Molecular formula indicates the actual numbers of atoms present in a molecule of the compound Calculate the composition of potassium sulfite K2SO3 2 x K 78 2 g S x 1 32 07 g O x 3 48 00 g 158 27 78 2 158 27 x 100 49 4 K 32 07 158 27 x 100 20 3 S 48 158 27 x 100 30 3 O The analysis of a salt shows that it contains 56 58 potassium 8 68 carbon and 34 78 oxygen Calculate the empirical formula for this substance 56 58 g 39 10 g mol 1 447 mol K 1 447 7233 2 00 8 68 g 12 00 g mol 0 7233 mol C 7233 7233 1 00 34 78 g 16 g mol 2 173 mol O 2 173 7233 3 00 K2CO3 63 6 N 36 4 O calculate the empirical formula 63 6 g 14 00 g mol 4 543 mol N 4 543 2 275 2 00 N 36 4 g 16 00 g mol 2 275 mol O 2 275 2 275 1 00 O Empirical formula N2O Determination of Molecular formulas A compound is found to contain 85 63 C and 14 37 H by mass In another experiment its molar mass is found to be 56 1 g mol What is its molecular formula CH2 14 g mol 56 1 14 4 4 x CH2 C4H8 Short cut to previous problem 1 mole 56 1 g 85 63 C 08563 x 56 1 48 038 g C 12 4 14 37 H 1437 x 56 1 8 06 g H 1 8 Molecular Formula C4H8 Law of Multiple Proportions It is possible for two elements A and B to combine to form more than one compound The ratios of the masses of element B that combine with a given mass of element A in each compound can be expressed by small whole numbers 13 Some other interpretations of chemical formulas What mass of phosphorus is contained in 45 3 grams of NH4 3PO4 148 97 g mol 45 3 g 148 97 g mol 304 mol NH4 3PO4 304 mol NH4 3PO4 1 P 1 NH4 3PO4 304 mol of P 304 mol x 30 97 g mol 9 42 g of P What mass of ammonium phosphate NH4 3PO4 FW 148 97 g mol would contain 15 0 g of N Purity of Samples The percent purity of a sample of a substance is always represented as mass of pure substance purity 100 mass of sample mass of sample includes impurities 16 A bottle of sodium phosphate Na3PO4 is 98 3 pure Na3PO4 What is the mass of Na3PO4 in 250 0 g of this sample of Na3PO4 Chapter 3 Chemical Equations and Reaction Stoichiometry Chemical equations are used to describe chemical reactions and they show 1 the substances that reacts called reactants 2 the substances formed called products and 3 the relative amounts of the substances involved The in front of cpds are called coefficients and represent the of molecules of each reactant or product needed to balance the equation Methane reacts with oxygen to produce carbon dioxide and water Write the chemical equation then balance it A balanced chemical equation must always include the same number of each kind of atom on both sides of the equation Law of Conservation of Matter provides the basis for balancing chemical equations because matter is neither created nor destroyed during a chemical reaction Law of Conservation of Matter There is no detectable change in quantity of matter in an ordinary chemical reaction Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation This law was determined by Antoine Lavoisier 21 Iron III oxide reacts with carbon monoxide to yield pure elemental iron and carbon dioxide NH3 burns in oxygen to form NO and water