The Mole. Relative Atomic Mass Ar

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1 STOICHIOMETRY

2 The Mole Relative Atomic Mass Ar Relative Molecular Mass Mr Defined as mass of one atom of the element when compared with 1/12 of an atom of carbon-12 Some Ar values are not whole numbers as most naturally occurring samples of an element contain a mixture of isotopes with different relative abundance No units as it is a ratio Can be found from periodic table Formula: mass of 1 molecule of an element Mass of 1/12 of an atom of carbon-12 To calculate Mr, add up the relative atomic masses of each atom in the chemical formula No units as it is a ratio Eg. Mr of H 2 SO 4 = (2 X 1) (4 X 16) = 98

3 Avogadro s Number The Avogadro Number (or Avogadro Constant) is defined as the number of atoms in 12g of the carbon-12 isotope The value of Avogadro Number is 6.02 x Examples 24g of magnesium would contain 6.02 x magnesium atoms 56g of iron would contain 6.02 x iron atoms 18g of water would contain 6.02 x water molecules

4 The Mole Concept

5 Mole and Mass The mass of one mole of molecules is its relative molecular mass in grams. For example, 18g of water will be one mole and so will contain Avogadro Number of water molecules Examples 1 mole of H 2 molecules has mass of 1 x 2 = 2g 1 mole of O 2 molecules has a mass of 16 x 2 = 32g 1 mole of CuSO 4.5H 2 O has a mass of (16 x 4) + 5[(1x2) + 16] =250g

6 Empirical vs Molecular Formula Empirical formula of a compound Shows the type of elements present in it Shows the simplest ratio of the different types of atoms in it Can be calculated from the masses of the elements combined together and the relative atomic masses of its elements Steps to find empirical formula of a compound Find percentage, mass or relative atomic mass Find number of moles (Mass, % or Ar / Molar mass) Divide by smallest ratio

7 Empirical Formula Example: Given that a molecule contains 88.89% oxygen and 11.11% hydrogen what is its empirical formula? 1. Divide % by Ar H: 11.11/1 = O: 88.89/16 = Simplest ratio H: 11.11/5.55 = 2 O: 5.55/5.55 = 1 3. Empirical formula = H 2 O

8 % Composition of Elements in a compound Steps Write the chemical formula of the substance Find the Mr Divide the atomic mass of the element you want to calculate the % composition of by the Mr and multiply result by 100% Example % composition of Hydrogen in H 2 SO 4 2/98 x 100 = 2.04% % composition of Oxygen in H 2 SO 4 64/98 x 100 = 65.31%

9 Mass of element in a compound Mass of element = % of element x mass of sample = Ar of the atoms of the element in formula Mr of compound X mass of sample Example Calculate the mass of sodium in 10g of sodium carbonate crystals (Na 2 CO 3.10H 2 O) Ar of Sodium atoms in Na 2 CO 3.10H 2 O = 2 X 32 = 46 Mr of compound Na 2 CO 3.10H 2 O = 286 Mass of sodium in 10g of sodium carbonate crystals = (46/286) x 10 = 1.61g

10 Chemical equation Reactants are written on left side of equation while products are written on right side. Steps to write chemical equation using example of formation of water 1. Word equation: Hydrogen + oxygen -- > water 2. Chemical formula: H 2 + O 2 -- > H Balance the equation: 2H > 2H Add state symbols: 2H 2 (g) + O 2 (g) -- > 2H 2 O (l)

11 Ionic equation If the reaction involves reactants in aqueous state, then ionic equation is written to show which particles actually took part in the chemical reaction Steps to write ionic equation 1. Write the overall chemical equation HCl (aq) + NaOH (aq) -- > NaCl (aq) + H 2 O (l) 2. The chemicals in aqueous state before and after the reaction are split into their respective ions [H + Cl - (aq)] + [Na + OH - (aq)] -- > [Na + Cl - (aq)] + H2O (l) 3. The ions which do not take part in the reaction (those ions same before and after the reaction) are removed [H + Cl - (aq)] + [Na + OH - (aq)] -- > [Na + Cl - (aq)] + H2O (l) 4. This leaves the essential ionic equation H + (aq) + OH - (aq)_ -- > H 2 O (l)

12 Solubility of Ionic Compounds

13 Calculations from chemical equation A balanced chemical equation shows the mole ratio of the reactants and products involved in the reaction 3 steps to calculating mass of a substance reacted or produced Convert mass of each given substance into number of moles Compare the mole ratio from equation Convert the number of mole to the mass of the substance you want to find

14 Calculating volume of reacting gases Equal volumes of all gases at the same temperature and pressure contain the same number of particles Eg. volume of one mole of O 2 at r.t.p. is 24000cm 3 Eg. volume of one mole of Cl 2 at r.t.p. is 24000cm 3 From the chemical equation Since 1 mole of any gas occupies 24dm 3 at r.t.p, the volume of gas is proportional to the number of moles of the gas 1dm 3 = 1000cm 3

15 Limiting Reactants A balanced chemical equation is used to calculate the exact amounts of reactants used up and products formed using its molar ratio The reactant that is completely used up is known as the limiting reactant. Determines or limits the amount of product formed Once the limiting reactant is used up, the reaction stops Always use the limiting reactant to calculate the product!

16 Calculation on concentration of solutions The concentration of a solution gives the amount of solute in 1dm 3 of solution. Can be expressed in g/cm 3 or mol/dm 3 (molar conc)

17 Acid-Base Titration Calculations

18 Summary

19 Percentage Yield and Percentage Purity

20 Stoichiometry is the calculation of quantities in chemical reactions. Equations tell chemists what amounts of reactants to mix and what amounts of products to expect. You can determine the quantities of reactants and products in a reaction from the balanced equation. When you know the quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created in the reaction. Quantity usually means the amount of a substance expressed in grams or moles. But quantity could just as well be in liters, tons, or molecules. Basic Steps to solving Stoichiometry problems: 1. Write out the balanced equation. This is extremely important! 2. Find the info given to you. 3. Convert that info (mass, concentration and volume, gaseous volume) into moles. 4. Use mole ratio from the equation to find out the number of moles of another substance. 5. Convert this number back into whatever the question wants.

21 Interpreting chemical equations Example 1 Ammonia is widely used as a fertilizer. It is produced industrially by the reaction of nitrogen with hydrogen. The following shows the equation for the reaction N 2 (g) + 3H 2 (g) 2NH 3 (g)

22 Conservation of mass Mass and atoms are conserved in every chemical reaction. The mass of the reactants equals the mass of the products. The number of atoms of each reactant equals the number of atoms for that reactant in the product(s). Unlike mass and atoms, however, molecules, formula units, moles and volumes of gases will not necessarily be conserved - although they may be. Only mass and atoms are conserved in every chemical reaction.

23 Interpreting Chemical Reactions Again, the equation for the formation of ammonia from hydrogen and nitrogen is 3H 2 + N 2 2 NH 3 The most important interpretation of this equation is that 1 mol of nitrogen reacts with 3 mol of hydrogen to form 2 mol of ammonia. With this interpretation, you can relate moles of reactants to moles of product. The coefficients from the balanced equation are used to write conversion factors called mole ratios.

24 Mole-Mole Calculations The mole ratios are used to calculate the number of moles of product from a given number of moles of reactant or to calculate the number of moles of reactant from a given number of moles of product. Three of the mole ratios for this equation are 1 mol N 2 2 mol NH 3 3 mol H 2 3 mol H 2 1 mol N 2 2 mol NH 3

25 Mole-Mole Calculations In the mole ratio below, W is the unknown quantity. The value of a and b are the coefficients from the balanced equation. Thus a general solution for a mole-mole problem is given by From balanced equation b mol W xb x mol G x = mol W a mol G a Given Mole ratio Calculated

26 Example: Using the ammonia reaction, answer the following question. How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? 0.60 mol N 2 x Given 2 mol NH 3 1 mol N 2 Mole Ratio = 1.2 mol NH 3

27 MASS-MOLES CALCULATIONS Balances don t tell you numbers in moles but in grams. As such, there are two related stoichiometry calculations: Moles Mass & Mass - Moles

28 In a mole-mass problem you are asked to calculate the mass (usually in grams) of a substance that will react with or be produced from a given number of moles of a second substance. moles A moles B mass B moles A x mole ratio of B A x molar mass of B If, in an example, you are told something in is excess, just ignore that substance and solve the problem with the needed substances.

29 Example: Plants use carbon dioxide and water to form glucose (C 6 H 12 O 6 ) and oxygen. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide?

30 Answer: 1. Write the balanced equation 6CO 2 (g) + 6H 2 O(l) -> C 6 H 12 O 6 (s) + 6O 2 (g) 2. Determine what you need to find/know. Unknown: mass of C 6 H 12 O 6 produced Given: amount of H 2 O = 3.00 mol 3. Determine conversion factors moles H 2 O x moles C 6 H 12 O 6 moles H 2 O x grams C 6 H 12 O 6 1 mole C 6 H 12 O 6 = grams C 6 H 12 O 6 4. Solve 1 mol C 3.00 moles H 2 O x 6 H 12 O 6 x 6 moles H 2 O 180 g C 6 H 12 O 6 1 mole C 6 H 12 O 6 = 90.0 g C 6 H 12 O 6

31 In a mass-mole problem you are asked to calculate the moles of a substance that will react with or be produced from a given number of grams of a second substance. mass A moles A moles B mass A x 1 mole A molar mass A x mole ratio B A

32 Mass-Mass Calculations No laboratory balance can measure substance directly in moles Instead, the amount of a substance is usually determined by measuring its mass in grams From the mass of a reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated The mole interpretation of a balanced equation is the basis for this conversion

33 If the given sample is measured in grams, the mass can be converted to moles by using the molar mass Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown If it is the mass of the unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass. As in mole-mole calculations, the unknown can be either a reactant or a product

34 Mass-mass problems can be solved in basically the same way as mole-mole problems. 1. The mass G is changed to moles of G (mass G mol G) by using the molar mass of G. Mass G X 1 mol G molar mass G = mol G 2. The moles of G are changed to moles of W (mol G mol W) by using the mole ratio from the balanced equation. Mol G X b mol W a mol G = mol W 3. The moles of W are changed to grams of W (mol W mass W)

35 The route for solving mass-mass problems is: mass A moles A moles B mass B 1 mole A moles B mass A x x x grams A moles A given molar mass of A mole ratio grams B 1 mole B molar mass of B

36 Example: Calculate the number of grams of NH 3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen.

37 Solution: 1. Write the balanced equation N 2 + 3H 2 2NH 3 2. Write what you know: Unknown: g NH 3 ; g H 2 -> g NH 3 Given: 5.40 g H 2 3. Determine conversion factors 4. Solve g H 2 mol H 2 mol NH 3 g NH 3 1 mol H 5.40 g H 2 x 2 2 mol NH x 3 x 2.0 g H 2 3 mol H 2 given = 30.6 g NH 3 changes given to moles mole ratio 17.0 g NH 3 1 mol NH 3 change moles of wanted to grams

38 The following equation summarizes these steps for a typical stoichiometric problem ag (given quantity) bw (wanted quantity)

39 Using the ammonia reaction equation, determine the number of liters of ammonia that can be produced from 5 grams of nitrogen at STP. N 2 + 3H 2 2NH 3 5g N 2 x 1 mole N 2 28g N 2 x 2 mole NH 3 1 mole N 2 x 22.4 L NH 3 1 mole NH 3 =

40 If the law of conservation of mass is true, how is it possible to make 30.6 g NH 3 from only 5.40 g H 2? Looking back at the equation for the reaction, you will see that hydrogen is not the only reactant. Another reactant, nitrogen, is also involved. If you were to calculate the number of grams of nitrogen needed to produce 30.6 g NH 3 and then compare the total masses of reactants and products, you would have an answer to this question

41 LIMITING REAGENT As you know, a balanced equation is a chemist s recipe - a recipe that can be interpreted on a microscopic scale (interacting particles) or on a macroscopic scale (interacting moles). The coefficients used to write the balanced equation give both the ratio of representative particles and the mole ratio. Recall the equation for the preparation of ammonia. N 2 (g) + 3H 2 (g) 2NH 3 (g)

42 When one molecule of N 2 reacts with three molecules of H 2, two molecules of NH 3 are produced. What would happen if two molecules of N 2 reacted with three molecules of H 2? Would more than two molecules of NH 3 be formed? Before the reaction takes place, nitrogen and hydrogen are present in a 2:3 molecule ratio. According to the balanced equation, one molecule of N 2 reacts with three molecules of H 2 to produce two molecules of NH 3. At this point, all the hydrogen has been used up, and the reaction stops. One molecule of unused nitrogen is left over, in addition to the two molecules of ammonia that have been produced.

43 In this reaction, only the hydrogen is completely used up. It is called the limiting reagent (reactant). As the name implies, the limiting reagent (reactant) limits or determines the amount of product that can be formed in a reaction. The reaction occurs only until the limiting reagent (reactant) is used up. By contrast, the reactant that is not completely used up in a reaction is called the excess reagent (reactant). In this example, nitrogen is the excess reagent (reactant) because some nitrogen will remain unreacted.

44 Example: Sodium chloride can be prepared by the reaction of sodium metal with chlorine gas. 2Na(s) + Cl 2 (g) 2NaCl(s) Suppose that 6.70 mol of Na reacts with 3.20 mol Cl 2. a. what is the limiting reagent (reactant)? b. b. how many moles of NaCl are produced?

45 Solution: 1. List the knowns and unknowns for a. known: moles of sodium = 6.70 mol Na moles of chlorine = 3.20 mol Cl 2 2 mol Na = 1 mol Cl 2 (from balanced equation) unknown: limiting reagent (reactant) 2. Solve for the unknown: 1 mol Cl 6.70 mol Na x 2 = 3.35 mol Cl 2 mol Na 2 Given Mole Required amount ratio amount

46 This calculation indicates that 3.35 mol Cl 2 is needed to react with 6.70 mol Na. Because only 3.20 mol Cl 2 is available, however, chlorine must be the limiting reagent (reactant). Sodium, then, must be in excess. Now, then, to find the amount of product (NaCl) made. To find the amount of product, you first have to identify the limiting reactant. Once that is done, you use that reactant to determine the amount of product made, because once the limiting reagent runs out, no more product is made.

47 1. List the knowns and unknowns for b. known: amount of limiting reagent: 3.20 mol Cl 2. 1 mol Cl 2 = 2 mol NaCl (from balanced equation) unknown: amount (moles) of NaCl produced. 2. Solve 3.20 mol Cl 2 x 2 mol NaCl 1 mol Cl 2 = 6.40 mol NaCl

48 PERCENT YIELD When an equation is used to calculate the amount of product that will form during a reaction, a value representing the theoretical yield is obtained. The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants. In contrast, the amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield. The actual yield is often less than the theoretical yield.

49 PERCENT YIELD The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent. The percent yield measures the efficiency of the reaction. actual yield Percent yield = x 100 % theoretical yield A percent yield should not normally be larger than 100%. Many factors can cause percent yields to be less than 100%.

50 Example: Calcium carbonate is decomposed by heating, as shown in the following equation. CaCO 3 (s) CaO(s) + CO 2 (g) a. what is the theoretical yield of CaO if 24.8 g of CaCO 3 is heated? b. What is the percent yield if 13.1 g CaO is produced?

51 Solution: 1. List the knowns and unknowns in a. known: mass of CaCO 3 = 24.8 g 1 mol CaCO 3 = 1 mol CaO (from balanced equation) 1 mol CaCO 3 = g (molar mass) 1 mol CaO = 56.1 g (molar mass) unknown: theoretical yield of CaO =? g CaO

52 2. Solve for the unknown. 1 mol CaCO 24.8 g CaCO 3 x 3 x 1 mol CaO x 56.1 g CaO g CaCO 3 1 mol CaCO 3 1 mol CaO given amount molar mass mole ratio molar mass = 13.9 g CaO Again, this is the theoretical yield, the amount you would make if the reaction were 100% accurate.

53 3. Determine % yield for b. actual yield = 13.1 g CaO theoretical yield = 13.9 g CaO actual yield Percent yield = x 100% theoretical yield Percent yield = x 100% = 94.2% 13.1 g CaO 13.9 g CaO

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