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Topic 7 – Moles and mole calculations Key Concept Definitions and facts 1. Equation shows changes that are taking place in substances 2. Physical equation shows changes that do not change compositions of substances (physical change) 3. Chemical equation Shows changes in compositions of one or more substances (chemical changes) 4. Nuclear equation Shows changes of nucleus contents of one or more atoms (nuclear change or transmutation) 5. Chemical reactions ways by which chemical changes occur 6. Reac tants Starting substances that will be changed in chemical reaction 7. Products Substances remaining after a change had occurred in reaction 8. Coefficients whole-numbers in front of substances in equations. Indicate number of moles (how many) of the substance 9. Arrow Separates Reactants from Products, and can read as “yields” or “produces” Examples H2O(l) ---- > H2O(s) H2(g) + O2(g) ---> H2 O(l) 220 Fr ---- > 4He + 216At 1 Topic 7 – Moles and mole calculations Types of reactions Explanations 10. Synthesis (combination) Two or more reactants combined to make one product 11.Decomposition (Analysis) One reactant breaks to make two or more products Examples + 2H2 + O2 -- > 2H2O + 2H2O ----> 2H2 + O2 12. Single Replacement 13. Double Replacement 14. Combustion (oxidation) 15. Law of Conservation One (a more reactive) reactant replaces a less reactive element of a compound + + Zn + 2HCl --->ZnCl2 +H2 Ions from compound reactants + + switche with each other to form NaCl + AgNO3 ---->NaNO3 + AgCl different compounds a carbon compound burns in the present of oxygen CH4 + O2 --- > CO2 + H2O ) During chemical reactions: neither atoms, mass, charge or energy is created nor destroyed. The total number of atoms, charge, and energy are the same before and after a reaction (conserved) 16. A balanced chemical equation An equation that shows conservation N2 + H2 ---- > NH3 allows 2N 2N this equation to be 6H 6H balanced. atoms of atoms of = reactants 2 products Topic 7 – Moles and mole calculations Key concept 17. 36.Law of conservation of mass Example During a chemical reaction, mass of substances before and after the reaction is the same. 3Fe 20.9 g + 2O2 ------- > Fe3O4 : 8.0 g = 28.9 g NOTE: The total mass of Fe and O2 (reactants) is the same as the total mass of Fe3O4 (product) 18. Balancing equation 37. An equation Balancing is balanced equationwhen usingit acontains the correct combination of smallest RAPwhole-number table. coefficients. The coefficients allow number of each atomThe on both use ofsides a Table of the is optional equation to be the same. To balance an equation: One or more coefficients in front of the substances must be changed in the equation. The right combination of coefficients will make the number of atoms on both sides equal. Suggestion to balancing equations . Make a table (optional) to keep track of number of atoms as coefficients are changed . Try balancing one atom at a time . Every time a coefficient is changed, RECOUNT the number of each atom affected by the change. NOTE changes on the table. (be sure to count atoms correctly) . Always change coefficients of free elements (Na, Cl2) last . Always put parenthesis around polyatomic ions and count it as one unit . Be sure that coefficients are in smallest whole-number ratio 19. Sums of all coefficient: After you had balance an equation, you may be asked for the sum of all coefficients. (which may not be written in front of the substances) Examples of balanced equations and sums of coefficients are given on the next page. . 3 Topic 7 – Moles and mole calculations 20. Balanced equations and sum of coefficients Li3N Unbalanced ------> Li + N2 Balanced 2Li3N -------> 6 Li + N2 Sum of all coefficients 2 + 1 3Ca(OH)2 + 2H3PO4 ---- > Ca3(PO4)2 + 6H2O Sum of coefficients 3 Unbalanced + 2 C3H4 + + O2 Balanced C3H4 + 4O2 Sum of coefficients 1 4 = 9 Ca(OH)2 + H3PO4 ----- > Ca3(PO4)2 + H2O Unbalanced Balanced + 6 + 4 1 ----- > CO2 + 6= + ------ > 3CO2 + + 3 + 12 H2O 2H2O 2 = 9 Topic 7 – Moles and mole calculations Key concept 21. Mole Definition and fact A unit of quantity equals to 6.02 x 1023 (Avogadro’s number) How many atoms 22. Moles interpretation in formula Examples 1 dozen = 12 1 mole = 6.02 x 1023 (602000000000000000000000) In 1 mole of H2O there are: 2 moles of H atoms 1 mole of O atom In 2 moles of H2O there are: 4 moles of H atoms 2 moles of O atoms 23. Molar mass The mass of 1 mole of Molar mass of H2O is 18 g (18g/mole) a substance Molar mass of NaCl is 58 g ( 58g/mole) 24. Gram-atomic mass The mass of 1 mole of Gram-atomic mass of sodium (Na) is 22.989 g/mole. an element. Gram atomic mass of Gram-atomic mass of chlorine elements can be found (Cl) is 35.5 g /mole on the Periodic Table 25. Gram- formula The mass of one mole (molecular ) of a compound mass The sum of all the atomic masses in a formula The gram-formula mass of NaCl is 58.4 g/mole. It is the sum of the mass of 1 Na (22.9 g ) + the mass of 1 Cl (35.5 g ) The gram-formula mass of H2O is 18 g It is the sum of the mass of 2 H (2 g ) + the mass of 1 O (18 g) 26. Percent composition Portion of a mass of a % by mass of Na in NaCl = 39.3% substance that is due to the mass of individual % by mass of Cl in NaCl = 60.7 % element in its formula 5 Topic 7 – Moles and mole calculations Concept 27. Hydrates Facts and definitions Examples Ionic substances CuSO4 .5H2O (copper containing water in their sulfate pentahydrate) crystalline structure CuSO4 H 2O A diagram of the hydrate CuSO4.5H2O showing 5 water molecules attached to the crystalline structure of CuSO4 28. Anhydrous a hydrate with the water removed In the hydrate (copper sulfate pentahydrate) . The hydrate is CuSO4 5H2O The anhydrous is CuSO4 29. Formula mass The sum of the mass of a hydrate of the anhydrous + the mass of water 30. Percent composition of a hydrate Portion of a hydrate’s mass that is due to the mass of the water. The formula mass of CuSO4 .5H2O is 250 g. This is the sum of the mass of 1CuSO4 ( 160 g) + the mass of 5H2O (90 g) % of water in the hydrate CuSO4.5H2O is 36 % This is the portion of the mass of CuSO4.5H2O (250 g) that is due to the mass of 5H2O (90 g) In the next few pages, you will learn step-by-step how to calculate number of moles of atom, mass, mole, and percent composition of a given formula. 6 Topic 7 – Moles and mole calculations Calculations with formula 31. Determining moles of atoms Given moles x how many atom Examples How many moles of oxygen are in mole of Ca(ClO ) x 6 = 6 moles of oxygen How many moles of oxygen are in mole of Ca(ClO3)2? x 6 = 3 moles of oxygen What is the total number of moles of atoms Total moles of atoms = sum of all atoms in a formula in mole of Ca(ClO3)2? Ca = 1 mole Total moles of atoms Cl = 2 moles 1 + 2 + 6 = 9 moles O = 6 moles 32. Determining Gram-Atomic mass What is the Gram-Atomic mass of silver (Ag)? 107.868 g 107.868 Ag 47 LOOK on Periodic Table - 33. Calculating 7 Gram-Formula mass. What is the Gram-Formula mass of . count each atom in formula Al2(SO4)3 . multiply how many by Atoms Atomic How Total atomic mass of the element mass x many = mass . add up all products (total 27 g 2 54 g mass of each element) to get Al formula mass S 32 g 3 96 g Gram-formula mass = sum of all masses O 16 g 12 Gram- Formula mass = 192 g 342 g of Al2(SO4)3 . 7 Topic 7 – Moles and mole calculations Calculation with formulas 34. Calculating Gram-Formula mass of hydrates NOTE: when an atom appears on both sides of the (.) in a hydrate, you must add them up. Examples What is the Gram-Formula mass of the hydrate CaCO3 4H2O ? Atoms Atomic How Total mass x many =mass Ca 40 1 40 g In example to the right, there are: C 12 1 12 g 3 O in CaCO3 (left of .) and O 16 7 112 g 4 O in 4H2O (right of .) H 1 8 8g Total # of O atoms = 3 + 4 = 7 Gram-Formula mass = 172 g of 35. Calculating mass from moles What is the mass of Al2(SO4)3 ? Mass = moles x formula mass Mass = 2.5 x 342 = 855 g What is the mass of 0.3 mole of water, H2O ? Formula mass of H2O: 2 H 2 (1) = 2 g 18 g 1 O 1 (16) = 16 g Mass = mole x formula mass Mass = 0.3 x 18 = 5.4 g 36. Calculating moles from mass Given mass Mole = Formula mass Use this Table T equation to calculate mass and mole 8 How many moles are represents in 200 grams of Al2(SO4)3 ? 200 moles = = 0.58 mol 342 How many moles of H2O are in 54 grams of the substance? 54 moles = = 3.0 mol 18 Topic 7 – Moles and mole calculations Calculation with formulas Examples What is the percent composition of Al, S, and O in the formula Al2(SO3)4 37. Percent composition by mass calculation Refer to Set 13 example for total mass of each atom in Al2(SO4)3 See Table T for percent composition equation % of Al = Total mass of an atom %= x 100 Formula mass % of S = % of O = 54 x 100 = 15.8 % 342 96 342 192 342 x 100 = 28.1 % x 100 = 56.1 % What is the percent of water in the hydrate CaCO3.4H2O ? 38. Percent composition of hydrates Find mass of 4 H2O: 8 H = 8 (1) = 8 72 g 4 O = 4 (16) = 64 Total mass of H2O % H2O = x 100 Formula mass of hydrate 72 % H2O = = 42 % 172 Refer to Set 14 example to see how the formula mass of CaCO3.4H2O (172 g) was determined. 39. Percent by volume % = part volume whole volume x 100 A 350 ml alcohol solution contains 190 ml of isopropyl alcohol. What is percent composition of the alcohol in this solution? % = 190 x 100 = 45 % 350 . 9 Topic 7 – Moles and mole calculations Calculation with formulas examples A 3.5 grams sample of a hydrated was heated to remove its water. The mass of the anhydrous remaining was determined to 2.3 grams. What is percent of water in the hydrate 40. Percent of water from lab data Mass of water = mass of hydrate – mass of anhydrous % water = mass of water mass of hydrate x 100 mass of water = 3.5 - 2.3 = 1.2 g 1.2 % water = x 100 = 34.3 % 3.5 A student collected the following data during a hydrate experiment: Mass of crucible = 15.3 g Mass of hydrate + crucible = 19.5 g Mass of anhydrous + crucible = 17.5 g What is the percent of water in the hydrate compound that was used in the experiment? Mass of hydrate = 19.5 – 15.3 = 4.2 g Mass of anhydrous = 17.5 – 15.3 = 2.2 g Mass of water = 4.2 – 2.2 = 2.0 g 2.0 % water = x 100 = 47.6 % 4.2 41. Molecular formula from mass and empirical formula . Determine mass of empirical formula . Determine unit (how many) of empirical formula What is the molecular formula of a compound with a molecular mass of 56 g/mol and an empirical formula of CH2 empirical formula’s mass: 1 C = 12 2 H = 2 14 g molecular mass = 56 = molecular mass 4 empirical mass 14 empirical formula . Determine molecular formula Molecular formula = 4 (CH ) = C H 2 4 8 multiply each subscript of empirical by unit (step 2) 10 Topic 7 – Moles and mole calculations Calculation with equations 42. A balanced chemical equation Remember: Coefficients represent number of moles. Examples Shows mole ratio (proportion) of substances taking part in a chemical reaction. mole ratio is the ratio of coefficients (whole number in front of the substances) In the balanced equation 4 NH3 + 5 O2 ---- > 4NO + 6 H2O Mole ratio of NH3 to O2 is 4 : 5 Mole ratio of NH3 to NO is 1 : 1 (reduced from 4 : 4) Mole ratio of NO to H2O is 2 : 3 (reduced from 4 : 6) 43. mole-mole problems Given the balanced equation below: 4 NH3 + 5 O2 ---- > 4NO + 6 H2O when of 4 NH3 + 5 O2 ---- > 4NO + 6 H2O . Re-write equation . Write and from question under the correct substances. 2 Setup mole proportion 4 . of water, , is produced is reacted? X = 6 2 . Cross multiply and Solve for 4X X X = 12 = 3 mol . 11 Topic 7 – Moles and mole calculations Calculations with equation 44. Volume – Volume problem Examples Given the balanced equation below: 2C4H10 + 13O2 ------>8CO2 + 10H2O is the total combusted to produce . Re-write equation and . Write from question under the correct substances. . Setup volume proportion 2C4H10 2 8 = = = 3Cu + 2H3PO4 ----- > Cu3(PO4)2 + 3H2 is the total by reacting . Re-write equation that must be ? + 13O2 ------>8CO2 + 10H2O . Cross multiply and Solve for 45. Mass – Mass problem of that is produced ? 3Cu + 2H3PO4 ----- > Cu3(PO4)2 + 3H2 . Write and under correct formula . Setup mass proportion Mass of 2H3PO4 6 H = 6(1) = 6 g 2 P = 2(31) = 62 g 8 O = 8(16) = 128 g 198 g Mass of 3H2 = 6 (1) = 6g 12 mass of 2H3PO4 == mass of 3H2 198 == 6 198 == 294