Chemistry 2202 Review Final Exam Chapter 2: The Mole 1. Isotope – atoms of an element that have the same number of protons but different numbers of neutrons. Ex. The three forms of oxygen are called oxygen-16, oxygen-17, and oxygen-18. They all have 8 electrons and are written as 16/8 O (8 protons + 8 neutrons), 17/8 O (8 protons + 9 neutrons), and 18/8 O (8 protons + 10 neutrons). 2. 3. Mass Number – the sum of the protons and neutrons in the nucleus of one atom of a particular element. Atomic Mass Unit (u) – a unit of mass that is 1/12 of the mass of carbon-12 atom; equal to 1. 66 ? 10-24 g.
Isotopic Abundance – the relative amount of an isotope of an element; expressed as a percent or a decimal fraction. Average Atomic Mass – the average of the masses of all naturally occurring isotopes of an element weighted according to their abundance. Ex. Aam = (% abundance ? mass) + (% abundance ? mass) 100 4. Practice Problems 5. Mole – the amount of substance that contains as many particles (atoms, molecules, or formula units) as exactly 12g of carbon-12. – one mole (1 mol) of a substance contains 6. 02 ? 10exp23 particles of the substance. It is called Avogadro’s constant without the units mol, and avogardro’s number with the mol units. . Converting Moles to Number of Particles n = # of particles/ Avogadro’s Number (6. 02 ? 10exp23) 7. Practice Problems 8. Molar Mass (M) – the mass of one mole of a substance, numerically equal to the element’s average atomic mass; expressed in g/mol. – to find the molar mass of a compound add up all individual compounds. Ex. Na2 Cl2 = Na = 2 ? 22. 99 g/mol = Cl = 2 ? 35. 45 g/mol 116. 88 g/mol 9. Convert from Moles to Mass and Mass to Moles where m = equals mass (g), n = # of moles (mol), M = molar mass (g/mol) m = nM n = m/M 10. Practice Problems 11. Converting Between Moles, Mass.
And Number of Particles n = m/M, n = v/V, n = # particles/6. 02 ? 10exp23 – where n = # moles (mol) , m = mass given (g), M = molar mass (g/mol), v = volume given (L), V = molar volume (L/mol) 12. Practice Problems 13. Pressure – the force that is exerted on an object, per unit of surface area. STP – (standard temperature and pressure) 0 °C and 101. 325 kPa. 14. Molar Volume – the amount of space that is occupied by 1 mol of a substance; equal to 22. 4 L for a gas at STP; V = 22. 4 L/mol. 15. ni = nf Vi Vf ; initial mole is always 1. 0 mol. 16. Practice Problems Chapter 3: Chemical Proportions in Compounds . Percent Composition – the relative mass of each element in a compound (max. 100%). To Calculate Percent Composition of a Compound – % Composition = total mass of element ? 100% mass of compound Ex. Calculate the % composition of water (H20). – H = 2. 02 g/mol, O = 16. 00 g/mol = Total 18. 02 g/mol – % H = 2. 02/18. 02 ? 100% = 11. 2%, %0 = 16. 00/18. 02 ? 100% = 88. 8% 2. Practice Problems 3. Empirical Formula – shows the lowest whole number ratio of atoms of each element in a compound. Molecular Formula – a formula that gives that gives the actual number of atoms of each element in a molecule.
Molecular = actual, Empirical = simplest Ex. H2O2, HO 4. Calculation of Empirical Formulas Ex. What is the empirical formula of a compound that is 25. 9% N and 74. 1% O by mass? Step 1: Assume 100g of compound Step 2: Convert percent to mass N = 25. 9g O = 74. 1g Step 3: Find molar mass (M) of each element and divide by their mass. N = 25. 9 g 14. 01 g/mol = 184 mol O = 74. 1 g 16. 00 g/mol = 4. 63 mol Step 3: Divide by smallest number and if it is not a whole number double all digits. N = 25. 9 g 14. 01 g/mol = 1. 84 mol O = 74. 1 g1. 84 mol = 1 ? 2 = 2 16. 00 g/mol = 4. 63 mol 1. 84 mol = 2. 5 ? 2 = 5 Empirical Formula = N205 . Practice problems 6. Calculation of Molecular Formulas Ex. 7. 36 g of compound is decomposed to give 0. 43g Hand 6. 93 g of O. The molar mass is 34 g/mol. Calculate the molecular mass. Step 1: Get Empirical Formula H = 0. 43g * g/mol = 0. 43 mol O = 6. 93 g0. 43 mol = 1 16. 00 g/mol = 0. 43 mol 0. 43 mol = 1 Empirical formula = HO Step 2: Divide molar mass of the empirical formula (17. 01 g/mol) into the mass of the molecule given (34 g/mol). 34 g/mol 17. 01 g/mol = 2 step 3: Multiply the empirical formula by 2. Molecular Formula = HO ? 2 = H202 7. Practice problems Chapter 4: Quantities in Chemical Reactions . 2. Practice problems 3. Stoichiometry – the study of the relative quantities of reactants and products in chemical reactions. Gas Stoichiometry – stoichiometric analyses involving the volume of gases. Solid (gravimetric) Stoichiometry – stoichiometric analyses involving mass. Solution Stoichiometry – the calculation of quantities of substances in chemical reactions that take place in solutions. 4. Process for Solving Stoichiometric Problems Ex. Passing chlorine gas through molten sulphur produces liquid disulfur dichloride. How many molecules of chlorine react to produce 50. 0 g of disulfur dichloride?
Step 1: Write a balanced equation. Cl2 (g) + 2S (l) – S2Cl2 (l) Step 2: If you if you are given the mass, volume or number of particles of a substance, convert it to the number of moles. Cl2 (g) + 2S (l) – S2Cl2 (l) 50. 0g n = m/M 50. 0 g / 135 g/mol = 0. 370 mol S2Cl2 Step 3: Calculate the number of moles of the required substance based on the number of moles of the given substance, using the appropriate mole ratio. Cl2 (g) + 2S (l) – S2Cl2 (l) 0. 370 mol 1:1 0. 370 mol Step 4: If required, convert the number of moles of the required substance to mass, volume, or number of particles. # particles = n(avogadro’s #) . 370 mol (6. 02 ? 10exp23) = 2. 22 ? 10exp23 molecules of Cl2 5. 6. 7. Practice Problems 8. Limiting Reactant – the reactant that is completely consumed during a chemical reaction, limiting is the amount of product produced. Excess Reactant – a reactant that remains after a chemical reaction is over. 9. Determining Limiting Reactant When you are given amounts of two or more reactants to solve a stoichiometric problem, you first need to identify the limiting reactant. One way to do this is to find out how much product would be produced by each reactant if the other reactant were present in excess.
The reactant that produces the least amount of product is the limiting reactant. Ex. Lithium nitride reacts with water to form ammonia and lithium lydroxide, according to the following balanced chemical equation: Li3N(s) + 3H20(l) – NH3(g) + 3Li0H(aq) If 4. 87 g of lithium nitride reacts with 5. 80g of water, find the limiting reactant. – You need to determine whether lithium nitride or water is the limiting reactant. – Reactant: lithium nitride, Li3N – 4. 87 g – Reactant: water, H20 – 5. 80 g – Convert the given masses into moles, use mole ratios of reactant and products to determine how much ammonia is produced by each amount of reactant.
The limiting reactant is the reactant that produces the smaller amount of product. Li3N = 0. 140 mol H20 = 0. 322 mol Ratio of Li3N (1:1) = 0. 140 mol of NH3 Ratio of H20 (3:1) = 0. 322/3 = 0. 107 mol of NH3 – Therefore water is the limiting reactant and lithium nitride is the excess reactant. 10. Practice Problems 11. 12. Calculation % Yield = Actual/Theoretical Yield ? 100% Ex. 13. Practice problems 14. Practice problems 15. Actual Yield – the measured quantity of product obtained in a chemical reaction. Theoretical Yield – the amountof product that is produced by a chemical reaction as predicted by stoichiometry.
Percent Yeild – the actual yield of a reaction, expressed as a percent of the theorectical yield. 16. Calculation % Yield = Actual/Theoretical Yield ? 100% Ex. Ammonia can be prepared by reacting nitrogen gas, taken from the atmosphere, with hydrogen gas. N2(g) + 3H2(g) – 2NH3(g) When 7. 5 ? 10exp1 g of nitrogen reacts with sufficient hydrogen, the theoretical yield of ammonia is 9. 10g. If 1. 72g of ammonia is obtained by experiment, what is the percentahe yield of the reaction? Actual yield = 1. 72g Theoretical yield = 9. 10 g % yield = 1. 72 g/ 9. 10 g ? 100% = 18. 9% 17. Practice problems
Chapiter 7: Solutions and Solubility 1. Solution – a homogeneous mixture of a solvent and one or more solutes. Solvent – a substance that has other substances dissolved in it. Solute – a substance that is dissolved in a solution. Concentrated solution – has a higher proportion of solute to solvent than a dilute solution. Dilute solution – has a lower proportion of solute to solventthan a concentrated solution. 2. Aqueous solution – a solution in which water is the solvent. Miscible – a term used to describe substances that are able to combine with each other in any proportion.
Immiscible – a term used to describe substances that are not able to combine with each other in a solution. Alloy – a solid solution of two or more metals. Saturated solution – a solution in which no more of a particular solute can be dissolved at a specific temperature. Unsaturated solution – a solution in which more of a particular solute can be dissolved at a specific temperature. Solubility – the mass of solute that dissolves in a given quantity of solvent at a specific temperature. 3. Temperature – the rate of dissolving is greater at higher temperatures.
Agitation – agitating a mixture by stirring or by shaking the container increases the rate of dissolving. Surface area of solute particles – decreasing the size of the particles increases the rate of dissolving. When you break up a large mass into many smaller masses, you increase the surface area that is in contact with the solvent. This allows the solid to dissolve faster. 4. equilibrium constant – the ratio of the concentrations of the products over the concentrations of the reactant, with all concentrations terms raised to the power of the coefficients in the chemical equation.