CH 221 Chapter Two Part 2 Concept Guide
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1 CH 221 Chapter Two Part 2 Concept Guide 1. Ion Charges What charge are the following ions expected to have? (a) ionic barium (b) ionic oxygen (c) ionic potassium (a) Barium is expected to form cations. Elements in periodic Group 2A form ions of +2 charge, therefore barium is expected to form Ba 2+. (b) Oxygen is expected to form anions. It is in periodic Group 6A, and forms O 2-. (c) Potassium is expected to form cations. Elements in periodic Group 1A form ions of +1 charge, therefore potassium will form K Ion Charge and Empirical Formula Aluminum acts as a metal and oxygen acts as a nonmetal when they react. Predict the empirical formula for aluminum oxide. Aluminum loses 3 valence electrons to form Al 3+, whereas oxygen gains 2 valence electrons to form O 2-. : When ionic compounds form from elements, the total charge on the cations must balance out the total negative charge on the anions. We will need two Al ions, which give a +6 charge, for every 3 oxygen ions, which give a -6 charge. The empirical formula is: Al 2 O Electrostatic Forces Which compound's ions are held together by stronger forces: LiBr or MgS? The most significant difference between the two compounds is the charges on the individual ions. Li and Br have +1 and -1 charges, respectively. Mg and S have +2 and -2 charges, respectively. The higher charges on Mg and S (+2 and -2, in relation to +1 and -1 in LiBr) lead to stronger electrostatic forces. 4. Ionic Compounds Give the number and identify the constituent ions in the following ionic compounds: (a) NaF (b) CaCl 2 (c) Cu(NO 3 ) 2 (d) NaCH 3 CO 2. Page V-2b-1 / Chapter Two Part Two Concept Guide
2 (a) 1 Na + and 1 F - ion (c) 1 Cu 2+ ion and 2 NO 3 ions (b) 1 Ca 2+ ion and 2 Cl - ions (d) 1 Na + and 1 CH 3 CO 2 ion Nomenclature Give the formula for each of the following ionic compounds: (a) ammonium nitrate (b) cobalt(ii) sulfate (c) nickel(ii) cyanide. (a) NH 4 NO 3 (b) CoSO 4 (c) Ni(CN) 2 6. Nomenclature Name the following ionic compounds: (a) Li 2 CO 3 (b) KHSO 3 (c) CuCl and CuCl 2. (a) Lithium carbonate (b) Potassium hydrogen sulfite (c) Copper(I) chloride and copper(ii) chloride 7. Nomenclature What are the names of each of these molecules? (a) CO 2 (b) S 2 F 10 (c) BF 3 The symbol of the cation is always given first, followed by the anion symbol. The correct names for the above molecules are: (a) carbon dioxide (b) disulfur decafluoride (c) boron trifluoride 8. Nomenclature Give the name for each of the following compounds: (a) PI 3 (b) SCl 2 (c) XeO 3 (a) Phosphorus triiodide (b) Sulfur dichloride (c) Xenon trioxide Page V-2b-2 / Chapter Two Part Two Concept Guide
3 9. Naming Hydrogen-containing Compounds What is the name of HBr? Hydrogen monobromide or hydrobromic acid 10. Nomenclature Give the formula for each of the following compounds: (a) zinc(ii) carbonate (b) sodium phosphate (c) aluminum chloride. (a) ZnCO 3 (b) Na 3 PO 4 (c) AlCl Molar Mass What is the molar mass of NaOH? To find the molar mass of a compound, we must list the elements in the compound's formula and determine the number of atoms of each element in the formula. Then, for each element, we need to look up the molar mass. Once we have found the mass contributed by each element for one mole of the compound, the molar mass is calculated by adding these individual masses. 1 mol Na in NaOH = (1) g/mol = g 1 mol O in NaOH = (1) g/mol = g 1 mol H in NaOH = (1) g/mol = g Molar mass of 1 mol NaOH = g g g = g/mol NaOH 12. Molar Mass What is the molar mass of Cu(NO 3 ) 2? To find the molar mass of a compound, we must list the elements in the compound's formula and determine the number of atoms of each element in the formula. Then, for each element, we need to look up the molar mass. Once we have found the mass contributed by each element for one mole of the compound, the molar mass is calculated by adding these individual masses. Page V-2b-3 / Chapter Two Part Two Concept Guide
4 1 mol Cu in Cu(NO 3 ) 2 = (1) g/mol = g 2 mol N in Cu(NO 3 ) 2 = (2) g/mol = g 6 mol O in Cu(NO 3 ) 2 = (6) g/mol = g Molar mass of 1 mol Cu(NO 3 ) 2 = g g g = g/mol of Cu(NO 3 ) Converting Mass to Moles What quantity, in moles, does 107 g HBr represent? The molar mass of HBr is g/mol. The number of moles of HBr is: 107 g HBr * 1 mol HBr/ g HBr = 1.32 mol HBr 14. Molar Mass and Moles Which represents a greater number of moles: 4.5 g of carbon dioxide or 4.5 g of sodium chloride? The molar masses are g for CO 2 and g for NaCl. The numbers of moles of each compound is calculated using the molar mass: 4.5 g CO 2 * 1mol CO 2 /44.01 g CO 2 = 0.10 mol CO g NaCl * 1 mol NaCl/58.44 g NaCl = mol NaCl There are more moles of CO 2 than of NaCl. 15. Atomic Mass What is the average atomic mass of chlorine? Mass of 35 Cl amu Mass of 37 Cl amu Isotopic abundance of 35 Cl % Isotopic abundance of 37 Cl % Consider the mass and abundance of the isotopes of chlorine. : Step 1. There are 2 naturally occurring isotopes of chlorine: 35 Cl and 37 Cl. A sample of chlorine shows that the 2 isotopes are not present in equal amounts. The percent abundance is calculated by dividing the number of atoms of a given isotope by the total number of atoms of all isotopes of that element, times one hundred. Page V-2b-4 / Chapter Two Part Two Concept Guide
5 Step 2. The average atomic mass is For chlorine, the atomic mass is Atomic mass = amu * amu * = amu 16. Converting Mass to Atoms The 1989 nutritional recommended dietary allowance (RDA) of iron for a female age is 15 mg. How many iron atoms is this? Mass must be converted to moles, then moles must be converted to atoms. Avogadro s number (6.022 x ) will be needed, as will the molar mass of iron. Note: in molar calculations, all masses must be converted to grams. : Follow the series of multiplication steps below to convert mass to atoms. In 15 mg of iron there are 1.6 x Fe atoms. 17. Converting Mass to Molecules and Atoms How many ozone molecules and how many oxygen atoms are contained in g ozone, O 3? First, write the equation for the synthesis of ozone. Then, to calculate the number of ozone molecules in g ozone, convert mass to molecules using the molecular weight of ozone and Avogadro s number (6.022 x ). Next, to calculate the number of oxygen atoms, start with the mass of ozone, convert this to grams, use a moleto-mole ratio to convert ozone to O 2, and finally, use Avogadro s number to obtain the number of oxygen atoms. : Step 1. Write the equation for the synthesis of ozone, and calculate its mass. 3 O 2 (g) 2 O 3 (g) O 3 = 3( g/mol) = g/mol Step 2. To calculate the number of ozone molecules from its mass, multiply grams of ozone, molecular weight of ozone, and Avogadro s number. Page V-2b-5 / Chapter Two Part Two Concept Guide
6 The number of molecules of ozone in grams is x molecules. Step 3. To calculate the number of atoms of oxygen, multiply three atoms of oxygen per molecule of ozone and the number of molecules of ozone. The number of oxygen atoms in g of ozone is x atoms. 18. Empirical Formula A 1.27 g sample of an oxide contains 0.55 g phosphorus and 0.72 g oxygen. What this oxide's empirical formula? First, it is necessary to determine from the experimental data the number of moles of atoms of each element present. The simplest ratio is then found by dividing the numbers of moles of each element by the number of moles of the element present in the smallest amount. Step 1. Calculate the number of moles of phosphorus and oxygen. (0.55 g P)(1 mol P/30.97 g P) = mol P (0.72 g O)(1 mol O/16.00 g O) = mol O Step 2. Divide the numbers of moles by the number of moles of the element present in the smallest amount: P mol P/0.018 mol P = mol O/0.018 mol P = 2.5 Thus, the empirical formula is P 1.0 O 2.5. Step 3. Double all numbers in the formula to convert the fraction to a whole number. 2(P 1.0 O 2.5 ) = P 2 O 5 The empirical formula is P 2 O Percent Composition A 2.91 g sample of potassium metal when burned in oxygen formed a compound weighing 6.11 g and containing only potassium and oxygen. What is the percent composition of each element in this compound? Page V-2b-6 / Chapter Two Part Two Concept Guide
7 The percent composition is the percent by mass of each element in the compound, which is given by the mass of that element divided by the total mass of the compound, times 100. The percent potassium in this compound is: %K = (2.91 g / 6.11 g compound)(100%) = 47.6 % The percent oxygen in this compound is: 6.11 g compound g K = 3.20 g O %O = (3.20 g / 6.11 g compound)(100%) = 52.4 % The percent composition of the compound is 47.6% potassium and 52.4% oxygen. Page V-2b-7 / Chapter Two Part Two Concept Guide
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