Chemistry Notebook: Percent Composition, Hydrates, Empirical and Molecular Formulas, Study notes of Chemistry

Download Chemistry Notebook: Percent Composition, Hydrates, Empirical and Molecular Formulas and more Study notes Chemistry in PDF only on Docsity! Name Date Percent Composition, Hydrates, Empirical and Molecular Formulas! Percent composition: the percentage by mass of each element in a compound. • Percentage = (part / total)x100% Hydrate: a chemical compound that contains chemically bound water molecules. Empirical formula: simplest whole-number ratio of the atoms in the compound. Molecular formula: whole-number multiple of the empirical formula. • Shows the true composition. • Molar mass of a compound = molar mass of the empirical formula x a whole number, n. Part 1 (Review). Finding Percent composition. Divide the mass of the individual element within the compound by the entire mass of the compound. Multiply by 100 to get the percent! (Part/whole *100%) Ex. 1 mole of Fe(NO3)3 Contains 1 mole of iron, 3 moles of Nitrogen, and 9 moles of Oxygen. Total molar mass = 55.845 + 3*14.007 + 9*15.999 = 241.85 g/mol 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 1 𝑚𝑚𝑜𝑜𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚 𝑀𝑀𝑎𝑎𝑜𝑜𝑚𝑚𝑀𝑀 𝑀𝑀𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚(𝑁𝑁𝑂𝑂3)3  55.845𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 241.85𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 ∗ 100% = 23.1% 𝐹𝐹𝐹𝐹 𝑏𝑏𝑏𝑏 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚! 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 3 𝑚𝑚𝑜𝑜𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑁𝑁 𝑀𝑀𝑎𝑎𝑜𝑜𝑚𝑚𝑀𝑀 𝑀𝑀𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚(𝑁𝑁𝑂𝑂3)3  3∗14.007𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 241.85𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 ∗ 100% = 17.4% 𝑁𝑁 𝑏𝑏𝑏𝑏 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚! 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 9 𝑚𝑚𝑜𝑜𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑂𝑂 𝑀𝑀𝑎𝑎𝑜𝑜𝑚𝑚𝑀𝑀 𝑀𝑀𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜 𝐹𝐹𝑚𝑚(𝑁𝑁𝑂𝑂3)3  9∗15.999𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 241.85𝑔𝑔/𝑚𝑚𝑜𝑜𝑚𝑚 ∗ 100% = 59.5% 𝑂𝑂 𝑏𝑏𝑏𝑏 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚! Now you try. Find the percent composition of Aluminum sulfate: Al2(SO4)3 Part 2 % composition  Empirical Formula 1. Convert %  g. 2. Convert g  mol : # g x 1𝑚𝑚𝑜𝑜𝑚𝑚 𝑚𝑚𝑜𝑜𝑚𝑚𝑀𝑀𝑀𝑀 𝑚𝑚𝑀𝑀𝑀𝑀𝑀𝑀 𝑔𝑔 = #𝑚𝑚𝑚𝑚𝑚𝑚 3. Compare these amounts in mol to find the simplest whole-number ratio among the elements. (Divide each by the smallest number of moles. If you need to, double or even triple the answers to make them into whole number ratios.) These ratios will be the subscripts. 4. Ex. Copper bromide is 28.5% Copper and 71.5% Bromide by mass. What is its empirical formula? Cu: 28.5 𝑔𝑔 𝐶𝐶𝐶𝐶 ∗ 1 𝑚𝑚𝑜𝑜𝑚𝑚 𝐶𝐶𝐶𝐶 63.546 𝑔𝑔 𝐶𝐶𝐶𝐶 = 0.448 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶 → 0.448 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶 0.448 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶 = 1 Br: 71.5 𝑔𝑔 𝐵𝐵𝐵𝐵 ∗ 1 𝑚𝑚𝑜𝑜𝑚𝑚 𝐵𝐵𝑀𝑀 79.904 𝑔𝑔 𝐵𝐵𝑀𝑀 = 0.895 𝑚𝑚𝑚𝑚𝑚𝑚 𝐵𝐵𝐵𝐵 → 0.895 𝑚𝑚𝑚𝑚𝑚𝑚 𝐵𝐵𝐵𝐵 0.448 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶 = 1.997≈ 2 Ex. Aluminum oxide is 52.9% Aluminum and 47.1% Oxygen by mass. What is its empirical formula? Al: 52.9 𝑔𝑔 𝐴𝐴𝑚𝑚 ∗ 1 𝑚𝑚𝑜𝑜𝑚𝑚 𝐴𝐴𝑚𝑚 26.982 𝑔𝑔 𝐴𝐴𝑚𝑚 = 1.96 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑚𝑚 → 1.96 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑚𝑚 1.96 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑚𝑚 = 1 ∗ 2 = 2 O: 47.1 𝑔𝑔 𝑂𝑂 ∗ 1 𝑚𝑚𝑜𝑜𝑚𝑚 𝑂𝑂 15.999 𝑔𝑔 𝑂𝑂 = 2.94 𝑚𝑚𝑚𝑚𝑚𝑚 𝑂𝑂 → 2.94 𝑚𝑚𝑚𝑚𝑚𝑚 𝑂𝑂 1.96 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝑚𝑚 = 1.5 ∗ 2 = 3 Now you try. Find the empirical formula of a compound that is 46.6% Nitrogen and 53.4% Oxygen. Fe N O CuBr2 Al2O3