# Moles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations

Size: px
Start display at page:

Download "Moles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations"

Transcription

1 Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations

2 Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = amu 16 O = amu 3.1

3 The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly grams of 12 C 1 mol = N A = x Avogadro s number (N A ) 3.2

4 Molar mass is the mass of 1 mole of Na atoms Pb atoms Kr atoms Li atoms in grams 1 mole 12 C atoms = x atoms = g 1 mole 12 C atoms = g 12 C 1 mole lithium atoms = g of Li For any element atomic mass (amu) = molar mass (grams) 3.2

5 Species H Quantity 1 mole Number of H atoms x 10 23

6 Species H 2 Quantity 1 mole Number of H 2 molecules x 10 23

7 Species Na Quantity 1 mole Number of Na atoms x 10 23

8 Species C 6 H 6 Quantity 1 mole Number of C 6 H 6 molecules x 10 23

9 1 mol of atoms = x atoms 1 mol of molecules = x molecules 1 mol of ions = x ions

10 Converting Units Unit Factor Label Method I. Write Unit of Answer II. Write Starting Quantity from Problem III. Add appropriate unit factor(s) to cancel out starting quantity and put in unit of answer.

11 Understanding Molar Mass How many atoms are in g of potassium (K)? 1 mol K = g K 1 mol K = x atoms K g K x 1 mol K g K x x 1023 atoms K 1 mol K = 8.49 x atoms K 3.2

12 How many moles of iron does 25.0 g of iron represent? Atomic mass iron = Conversion sequence: grams Fe moles Fe Set up the calculation using a conversion factor between moles and grams. (grams Fe) 1 mol Fe g Fe (25.0 g Fe) 1 mol Fe = g Fe mol Fe

13 What is the mass of 3.01 x atoms of sodium (Na)? Molar mass Na = g Conversion sequence: atoms Na grams Na Set up the calculation using a conversion factor between grams and atoms g Na (atoms Na) x atoms Na g Na = 11.5 g Na 23 (3.01 x 10 atoms Na) x atoms Na

14

15 Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO 2 1S 2O SO amu + 2 x amu amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = amu 1 mole SO 2 = g SO 2 3.3

16 Calculate the molar mass of C 2 H 6 O. 2 C = 2(12.01 g) = g 6 H = 6(1.01 g) = 6.06 g 1 O = 1(16.00 g) = g g

17 Calculate the molar mass of LiClO 4. 1 Li = 1(6.94 g) = 6.94 g 1 Cl = 1(35.45 g) = g 4 O = 4(16.00 g) = g g

18 Calculate the molar mass of (NH 4 ) 3 PO 4. 3 N = 3(14.01 g) = g 12 H = 12(1.01 g) = g 1 P = 1(30.97 g) = g 4 O = 4(16.00 g) = g g

19 In dealing with diatomic elements (H 2, O 2, N 2, F 2, Cl 2, Br 2, and I 2 ), distinguish between one mole of atoms and one mole of molecules.

20 Calculate the molar mass of 1 mole of H atoms. 1 H = 1(1.01 g) = 1.01 g Calculate the molar mass of 1 mole of H 2 molecules. 2 H = 2(1.01 g) = 2.02 g

21 How many grams of (NH 4 ) 3 PO 4 are contained in 2.52 moles of (NH 4 ) 3 PO 4? The molar mass of (NH 4 ) 3 PO 4 is g. Conversion sequence: moles (NH 4 ) 3 PO 4 grams (NH 4 ) 3 PO 4 Use the conversion factor: grams (NH ) PO 1 mole (NH ) PO g (NH 4) 3PO4 (2.52 mol (NH 4) 3PO 4)) 1 mol (NH 4) 3PO4 = 376g (NH 4) 3PO4

22 56.04 g of N 2 contains how many N 2 molecules? The molar mass of N 2 is g. Conversion sequence: g N 2 moles N 2 molecules N 2 Use the conversion factors 1 mol N g N 1 mol N 2 (56.04 g N 2) g N x 10 molecules N2 2 1 mol N x 10 molecules N2 1 mol N 24 = x 10 molecules N 2 2

23 56.04 g of N 2 contains how many N 2 atoms? The molar mass of N 2 is g. Conversion sequence: g N 2 moles N 2 molecules N 2 atoms N Use the conversion factors 23 1 mol N x 10 molecules N 2 atoms N g N2 1 mol N 1 molecule N2 1 mol N 2 (56.04 g N 2) g N x 10 molecules N2 1 mol N 2 2 atoms N 1 molecule N 2 24 = x 10 molecules N 2

24 If the formula of a compound is known, a two-step process is needed to calculate the percent composition. Step 1 Calculate the molar mass of the formula. Step 2 Divide the total mass of each element in the formula by the molar mass and multiply by 100.

25 total mass of the element molar mass x 100 = percent of the element

26 Calculate the percent composition of hydrosulfuric acid H 2 S. Step 1 Calculate the molar mass of H 2 S. 2 H = 2 x 1.01g = 2.02 g 1 S = 1 x g = g g

27 Calculate the percent composition of hydrosulfuric acid H 2 S. Step 2 Divide the mass of each element by the molar mass and multiply by g H H: (100) = 5.93% g 32.07g S S: (100) = 94.07% 34.09g S 94.07% H 5.93%

28 Percent Composition From Experimental Data

29 Percent composition can be calculated from experimental data without knowing the composition of the compound. Step 1 Calculate the mass of the compound formed. Step 2 Divide the mass of each element by the total mass of the compound and multiply by 100.

30 A compound containing nitrogen and oxygen is found to contain 1.52 g of nitrogen and 3.47 g of oxygen. Determine its percent composition. Step 1 Calculate the total mass of the compound 1.52 g N 3.47 g O 4.99 g = total mass of product

31 Calculate the percent composition of hydrosulfuric acid H 2 S. Step 2 Divide the mass of each element by the total mass of the compound formed g N (100) = 30.5% 4.99 g O N 69.5% 30.5% 3.47g O (100) = 69.5% 4.99g

32 Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound C 2 H 6 O 2 x (12.01 g) %C = g 6 x (1.008 g) %H = g 1 x (16.00 g) %O = g x 100% = 52.14% x 100% = 13.13% x 100% = 34.73% 52.14% % % = 100.0% 3.5

33 Empirical Formula versus Molecular Formula

34 The empirical formula or simplest formula gives the smallest whole-number ratio of the atoms present in a compound. The empirical formula gives the relative number of atoms of each element present in the compound.

35 The molecular formula is the true formula of a compound. The molecular formula represents the total number of atoms of each element present in one molecule of a compound.

36 Molecular Formula C 2 H 4 Empirical Formula CH 2 Smallest Whole Number Ratio C:H 1:2

37 Molecular Formula C 6 H 6 Empirical Formula CH Smallest Whole Number Ratio C:H 1:1

38 Molecular Formula H 2 O 2 Empirical Formula HO Smallest Whole Number Ratio H:O 1:1

39

40 Two compounds can have identical empirical formulas and different molecular formulas.

41

42 Calculating Empirical Formulas

43 Step 1 Assume a definite starting quantity (usually g) of the compound, if not given, and express the mass of each element in grams. Step 2 Convert the grams of each element into moles of each element using each element s molar mass.

44 Step 3 Divide the moles of atoms of each element by the moles of atoms of the element that had the smallest value If the numbers obtained are whole numbers, use them as subscripts and write the empirical formula. If the numbers obtained are not whole numbers, go on to step 4.

45 Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers Use these whole numbers as the subscripts in the empirical formula. FeO 1.5 Fe 1 x 2 O 1.5 x 2 Fe 2 O 3

46 The results of calculations may differ from a whole number. If they differ ±0.1 round off to the next nearest whole number Deviations greater than 0.1 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number.

47 A gaseous compound containing carbon and hydrogen was analyzed and found to consist of 83.65% carbon by mass. Determine the empirical formula of the compound. 1.Obtain the mass of each element present (in grams). Assume you have 100 g of the compound % C = g C ( ) 16.35% H = g H

48 2. Determine the number of moles of each type of atom present.

49 3. Divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. If all of the numbers so obtained are integers (whole numbers), these are the subscripts in the empirical formula. If one or more of these numbers are not integers, go on to step 4.

50 4. Multiply the numbers you derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula.

51 The balanced chemical equation represents the ratio Of one species to any others in the equation: NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O (l) 1mole of NaOH produces 1mole of NaCl 1 mol NaCl 1mol NaOH 2H 2 O 2 (l) 2 H 2 O (l) + O 2 (g) 2 moles of H 2 O 2 (l) produces 1 mole of O 2 gas 1 mol O 2 2 mol H 2 O 2 (l)

52 Mole composition gives ratio of each element In a compound moles of Al 2 (CO 3 ) 3 has the following moles of Al in the compound: Mol ratio of Al in Al 2 (CO 3 ) 3 : mol Al 2 (CO 3 ) 3 X 2 mol Al = mol Al 1 mol Al 2 (CO 3 ) 3 Moles of compound X mol ratio (element/molecule) = mol of element

53 Mass of Aluminum is obtained from the molar mass: mol Al X g = g Al mol Al Mol X grams/mol = mass The mass of any substance can be determined if The molar mass is known. The molar mass comes From the periodic table.

54 g/mol X mol = g Al 2 (CO 3 ) 3 Al 2 (CO 3 ) 3 continued: C: mol X 3 mol C = 1.04 mol C X g/mol 1 mol Al 2 (CO 3 ) 3 = g C O: mol X 9 mol O = 3.11 mol O X g/mol 1 mol Al 2 (CO 3 ) 3 = g O Formula Weight converted to mass from moles: 2 mol X 26.98g/mol (Al) =53.96g 3mol X g/mol (C) =36.033g 9 mol X g/mol (O) = g total molar mass g in 1 mol Al 2 (CO 3 ) 3

55 The empirical formula for a compound is the Smallest whole number ratio between the Elements that make up the compound. This is not necessarily the actual formula, especially For non-ionic compounds such as organic compounds. To determine the empirical formula, we must know The percent composition of each element in a compound The percent composition is a MASS PERCENT and is Determined for 100 grams of the substance. This has The effect of converting the percent of each element Into a RATIO!

56 Determination of Empirical Formula from Mass % 1. Assume 100 grams of material 2. Determine moles of each element in compound 3. Divide all subscripts in empirical formula by lowest number because the number of atoms must be an integer

57 A compound is discovered to be 63.6% nitrogen And 36.4% oxygen. What is the empirical formula? Convert to mass: 63.6% N = 63.6 g N 100 g Compound 36.4% O= 36.4 g O 100 g compound II. Convert to moles: 63.6 g N X 1 mol N = 4.54 mol N 14.0 g N 36.4 g O X 1 mol O = 2.28 mol O g O The ratio of N to O atoms will be the same as the Ratio of N to O moles: N: 4.54 = 1.99 O: 2.28 = III Divide all mols by the SMALLEST number of mols Empirical Formula= N 2 O

58 For organic compounds and some non-organic Compounds, the molecular formula is the actual Ratio of elements. This is a multiple of the empirical Formula. We can discover the molecular formula from the Empirical one if we determine the molar mass from An experiment. We simply compare the experimental Molar mass with the empirical one.

59 Determination of Molecular Formula from Empirical Formula Only Empirical Formula can be Determined by Mass Percent Multiplier = MoleculeMolarMass EmpiricalMolarMass

60 A phosphorous compound has an empirical formula of P 2 O 5. It has a molar mass of 284 g/mol, determined in the laboratory. The empirical formula weight is: (2 x g + 5 x g)/mol= g/mol Compare the masses in the molar ratio: 284 g/mol = 2; therefore:(p 2 O 5 ) 2 is the molecular g/mol Formula: P 4 O 10 Molar Mass from Experiment = multiplier for E.F. Empirical formula weight

61 The real usefulness of balanced equations can be Observed in the following series of predictions: I Predict moles of any products of a reaction II Predict the amount of any and all reactants needed for a given amount of product III Determine precisely how much of all reactants are used, and which are present in excess. The basis of chemical calculations for reactions Is the balanced reaction equation.

62 So far, we have seen mole ratio s from an equation That tells us how much to expect in the reaction: N 2 (g) + H 2 (g) NH 3 (g) N 2 (g) + 3H 2 (g) 2NH 3 (g) (Balanced equation) Think: mols reactant x moles product = mols product mols reactant What we are asked for goes on top of the ratio. What we Are given goes on bottom. How many moles of NH 3 can be produced from 5.00 mol H mol H 2 (given) x 2 mol NH 3 = 3.33 mol NH 3 formed 3 mol H 2

63 In the laboratory, we must measure the mols by Mass. This requires an initial conversion to mols From the given mass of substance: How many moles of NH 3 are formed from 33.6 g Of N 2 : 33.6 g x 1 mol N 2 x 2 mol NH 3 = 2.40 mol NH g N 2 1 mol N 2 Mass reactant x 1 x mol product = mol product F.W react. mol reactant

64 Finally, we must determine the mass of a product In the laboratory, since we cannot evaluate moles: What mass of H 2 is needed to produce 119 g of NH 3 What is asked for: mass of H 2 What is given: mass of NH 3 Mass reactant x 1 x mol product x FW prod =mass F.W react. mol reactant product 119 g NH 3 x1 mol NH 3 x 3 mol H 2 x 2.02 g H 2 = 21.2 g H g NH 3 2 mol NH 3 mol H 2

65 Memorize these steps: Mass react x 1 mol react x mol product x prod. g = mass grams react mol react mol prod.

66 Mass react x 1 mol react x mol product x prod. g = mass grams react mol react mol prod. SiO 2 is etched with HF (aq): SiO 2 (s) + 4 HF (aq) SiF 4 (g) + 2 H 2 O (l) If there is 4.86 moles HF and 60 g SiO 2, will there be Any glass left over? 4.86 mol HF x 1 mol SiF 4 x 104 g SiF 4 = 126 g SiF 4 ; 4 mol HF mol SiF g SiF 4 x 1 mol SiF 4 x SiO 2 x 60g SiO 2 = 72.7 g SiO g SiF 4 SiF 4 mol SiO 2 Since I have only 60g (1 mol) of SiO 2, I will run out of Glass before I use up the acid! SiO 2 is LIMITING

67 4 Ag (s) + 2 H 2 S (g) + O 2 (g) 2 Ag 2 S (s) + 2 H 2 O (l) IF there is mol Ag, mol H 2 S, and excess O 2, how much Ag 2 S is produced and what mass of Reactant is in excess? Ag: mol x 2 mol Ag 2 S = mol Ag 2 S 4 mol Ag H 2 S: mol H 2 S x 2 mol Ag 2 S = mol Ag 2 S 2 mol H 2 S Ag is limiting and H 2 S is present in excess. We must use the limiting reagent quantity to predict actual products: mol Ag 2 S x 248 g Ag 2 S = 18.0 g Ag 2 S formed mol Ag 2 S

68 A reaction in the laboratory results in a product mass That is not quite what the balanced equation calculation Predicts. The difference between the theoretical amount and the Actual one is called Percent Yield: Actual Yield (mass) x 100% = percent yield Theoretical yield (mass)

69 2S (s) + 3O 2 (g) ->2 SO 3 (g) S O Reactant Mixture Product Mixture

70 Mass Changes in Chemical Reactions 1. Write balanced chemical equation 2. Convert quantities of known substances into moles 3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4. Convert moles of sought quantity into desired units

71 Memorize these steps: mol ratio from balanced equation molar mass of product Mass react x 1 mol react x mol product x product g = mass grams react mol react mol prod. molar mass of reactant

72 Stoichiometric Relationships Molarity of reactant a,b,c or d From balanced equation Molarity of product Density (g/ml) (mol/g) Volume (L) Volume -1 (L -1 ) (g/mol) 1/density (ml/g) Volume reactant Grams reactant Moles reactant * moles product moles reactant * Moles product Grams of product Volume of product 1/N a N a Atoms or molecules Of reactant Atoms or molecules Of products a Reactant A + b Reactant B c Product C + d Product D or a Product A + b Product B c Reactant C + d Reactant D 2004 Randal Hallford

73 Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OH moles CH 3 OH moles H 2 O grams H 2 O molar mass CH 3 OH coefficients chemical equation molar mass H 2 O 209 g CH 3 OH 1 mol CH 3 OH x 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH 18.0 g H 2 O x = 1 mol H 2 O 235 g H 2 O

74 2S (s) + 3O 2 (g) ->2 SO 3 (g) Limiting Reagent S O Reactant Mixture Excess Reagent Product Mixture

75 Limiting Reagent Problems 1. Balance chemical equation 2. Determine limiting reagent Do two separate calculations for the amount of product each reactant would produce if they were the limiting reagent The reactant that gives the lower number is the limiting reagent. 3. The amount of product produced is the number calculated by limiting reagent

76 Limiting Reagent Problems 4. Determination of the amount of excess reagent left over Calculate the amount of excess reagent (ER) used in chemical reaction Subtract the ER used from original amount of ER.

77 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant PROBLEM: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine(n 2 H 4 ) and dinitrogen tetraoxide(n 2 O 4 ), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x10 2 g of N 2 H 4 and 2.00x10 2 g of N 2 O 4 are mixed? PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given. In this case one of the reactants is in molar excess and the other will limit the extent of the reaction. mass of N 2 H 4 divide by M mass of N 2 O 4 limiting mol N 2 multiply by M mol of N 2 H 4 molar ratio mol of N 2 O 4 g N 2 mol of N 2 mol of N 2

78 Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant SOLUTION: 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O(l) 1.00x10 2 g N 2 H mol N 2 H 4 4 = 3.12mol N2 H g N 2 H mol N 2 H 4 3 mol N 2 = 4.68mol N 2 2mol N 2 H 4 mol 2.00x10 2 g N 2 O 4 N 2 O 4 = 2.17mol N 2 O g N 2 O 4 N 2 H 4 is the limiting reactant because it produces less product, N 2, than does N 2 O mol N 2 mol N g N 2 = 131g N mol N 2 O 4 3 mol N 2 = 6.51mol N 2 mol N 2 O 4

79 Limiting Reagents 124 g of Al are reacts with 601 g of Fe 2 O 3 : 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. g Al mol Al mol Fe 2 O 3 needed g Fe 2 O 3 needed OR g Fe 2 O 3 mol Fe 2 O 3 mol Al needed g Al needed 124 g Al x 1 mol Al 27.0 g Al x 1 mol Fe 2 O 3 2 mol Al 160. g Fe 2 O x 3 = 367 g Fe 1 mol Fe 2 O 2 O 3 3 Start with 124 g Al need 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g); Al is limiting reagent

80 Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al 2 O 3 g Al 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 124 g Al x 1 mol Al 27.0 g Al x 1 mol Al 2 O 3 2 mol Al 102. g Al 2 O x 3 = 234 g Al 1 mol Al 2 O 2 O 3 3

81 Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100

82 Calculating Percent Yield PROBLEM: PLAN: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0kg of sand are processed, 51.4kg of SiC are recovered. What is the percent yield of SiC in this process? SOLUTION: write balanced equation SiO 2 (s) + 3C(s) SiC(s) + 2CO(g) find mol reactant & product 100.0kg SiO g SiO 2 kg SiO 2 mol SiO 2 = 1664 mol SiO g SiO 2 find g product predicted actual yield/theoretical yield x 100 percent yield mol SiO 2 = mol SiC = g SiC kg 1664mol SiC mol SiC 10 3 g = 66.73kg 51.4kg x100 =77.0% 66.73kg

83 If a sample of acetylene is burned, and releases 550 KJ of Energy, how many grams of CO 2 can we expect to form? 2 C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O (l) H=-2602 KJ We can relate the moles of CO 2 to the amount of heat, since It is part of the balanced equation: 550 KJ x 4 mol CO 2 x 44.0 g CO 2 = 37.2 g CO KJ mol CO 2

84 Chemical Equations reactants products 3 ways of representing the reaction of H 2 with O 2 to form H 2 O

85 Balancing Chemical Equations 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C 2 H 6 NOT C 4 H 12

86 Balancing Chemical Equations 3. Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 2 carbon on left C 2 H 6 + O 2 1 carbon on right 2CO 2 + H 2 O multiply CO 2 by 2 6 hydrogen on left C 2 H 6 + O 2 2 hydrogen on right 2CO 2 + 3H 2 O multiply H 2 O by 3

87 Balancing Chemical Equations 4. Balance those elements that appear in two or more reactants or products. C 2 H 6 + O 2 2CO 2 + 3H 2 O multiply O 2 by oxygen on left 4 oxygen (2x2) + 3 oxygen (3x1) = 7 oxygen on right C 2 H O 2 2CO 2 + 3H 2 O 2 remove fraction multiply both sides by 2 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O

88 Balancing Chemical Equations 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 4 C (2 x 2) 4 C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) Reactants 4 C 12 H 14 O Products 4 C 12 H 14 O

89 I. Formulas show chemistry at a standstill. Equations show chemistry in action. A. Equations show: 1. the reactants which enter into a reaction. 2. the products which are formed by the reaction. 3. the amounts of each substance used and each substance produced. B. Two important principles to remember: 1. Every chemical compound has a formula which cannot be altered. 2. A chemical reaction must account for every atom that is used. This is an application of the Law of Conservation of Matter which states that in a chemical reaction atoms are neither created nor destroyed. C. Some things to remember about writing equations: 1. The diatomic elements when they stand alone are always written H 2, N 2, O 2, F 2, Cl 2, Br 2, I 2 2. The sign, ----->, means "yields" and shows the direction of the action. 3. A small delta, ( ), above the arrow shows that heat has been added. 4. A double arrow, <----->, shows that the reaction is reversible and can go in both directions. 5. Before beginning to balance an equation, check each formula to see that it is correct. NEVER change a formula during the balancing of an equation. 6. Balancing is done by placing coefficients in front of the formulas to insure the same number of atoms of each element on both sides of the arrow.

90 7. Always consult the Activity Series of metals and non-metals before attempting to write equations for replacement reactions. 8. If a reactant or product is solid, place (s) after the formula 9. If the reactant or product is a liquid, place (l) after the formula 10. If the reactant or product is a gas, place (g) after the formula 11. If the reactant or product is in water, place (aq) after the formula 12. A category of reaction will produce an unstable product which decomposes: H 2 CO 3 (aq) H 2 O (l) + CO 2 (g) carbonic acid H 2 SO 3 (aq) H 2 O (l) + SO 2 (g) sulfurous acid NH 4 OH (aq) NH 3 (g) + H 2 O (l) ammonium hydroxide

91 Rules for Writing Chemical Equations: 1. Write down the formula(s) for the reactants and add a + between them then put a yields arrow ( ) at the end 2. Examine the formula(s) to determine which of four basic types of reactions will occur. On the basis of your decision, write down correct formula(s) for the products on the right side of the yield arrow. 3. Place a coefficient in front of each species in the reactants and the products to ensure that the conservation of matter is observed: Balance the reaction

92 6. A few nonmetals combine with each other. 2P + 3Cl ----> 2PCl II. Four Basic Types of Chemical reactions A. Synthesis two or more elements or compounds combine to give a more complex product Examples of Synthesis reactions: 1. Metal + oxygen -----> metal oxide 2Mg (s) + O 2(g) ----> 2MgO (s) 2. Nonmetal + oxygen -----> nonmetallic oxide C (s) + O 2(g) ----> CO 2(g) 3. Metal oxide + water -----> metallic hydroxide MgO (s) + H 2 O (l) ----> Mg(OH) 2(s) 4. Nonmetallic oxide + water -----> acid CO 2(g) + H 2 O (l) ----> ; H 2 CO 3(aq) 5. Metal + nonmetal -----> salt 2 Na (s) + Cl 2(g) ----> 2NaCl (s)

93 B. Decomposition: A single compound breaks down into simpler compounds. Basic form: AX -----> A + X Examples of decomposition reactions (when heated): 1. Metallic carbonates form metallic oxides and CO 2(g). CaCO 3(s) ----> CaO (s) + CO 2(g) 2. Most metallic hydroxides decompose into metal oxides and water. Ca(OH) 2(s) ----> CaO (s) + H 2 O (g) 3. Metallic chlorates decompose into metallic chlorides and oxygen. 2KClO 3(s) ----> 2KCl (s) + 3O 2(g) 4. Some acids decompose into nonmetallic oxides and water. H 2 SO > H 2 O (l) + SO 3(g) 5. Some oxides decompose. 2HgO (s) ----> 2Hg (l) + O 2(g) 6. Some decomposition reactions are produced by electricity. 2H 2 O (l) ----> 2H 2(g) + O 2(g) 2NaCl (l) ----> 2Na (s) + Cl 2(g)

94 C. Replacement: a more active element takes the place of another element and frees the less active one. Basic form: A + BX -----> AX + B or AX + Y -----> AY + X Examples of replacement reactions: 1. Replacement of a metal in a compound by a more active metal. Fe (s) + CuSO 4(aq) ----> FeSO 4(aq) + Cu (s) 2. Replacement of hydrogen in water by an active metal. 2Na (s) + 2H 2 O (l) ----> 2NaOH (aq) + H 2(g) Mg (s) + H 2 O (g) ----> MgO (s) + H 2(g) 3. Replacement of hydrogen in acids by active metals. Zn (s) + 2HCl (aq) ----> ZnCl 2(aq) + H 2(g) 4. Replacement of nonmetals by more active nonmetals. Cl 2(g) + 2NaBr (aq) ----> 2NaCl (aq) + Br 2(l) NOTE: Refer to the Activity Series for metals and nonmetals to predict products of replacement reactions. If the free element is above the element to be replaced in the compound, then the reaction will occur. If it is below, then no reaction occurs.

95 If the free element is above the element to be replaced in the compound, Then the reaction will occur. If the free element is below it, no reaction occurs.

96 D. Ionic or Double Displacement: occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following: 1. a precipitate 2. a gas 3. water or some other non-ionized substance. Basic form: AX + BY -----> AY + BX Examples of ionic reactions: 1. Formation of precipitate. NaCl (aq) + AgNO 3(aq) ----> NaNO 3(aq) + AgCl (s) BaCl 2(aq) + Na 2 SO 4(aq) ----> 2NaCl (aq) + BaSO 4(s) 2. Formation of a gas. HCl (aq) + FeS (s) ----> FeCl 2(aq) + H 2 S (g) 3. Formation of water. (If the reaction is between an acid and a base it is called a neutralization reaction.) HCl (aq) + NaOH (aq) ----> NaCl (aq) + H 2 O (l) 4. Formation of a product which decomposes. CaCO 3(s) + HCl (aq) ----> CaCl 2(aq) + CO 2(g) + H 2 O (l)

97 Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction. Equations of this type are called Net Ionic equations.

98 Combustion of Hydrocarbons: Another important type of reaction, in addition to the four types above, is that of the combustion of a hydrocarbon. When a hydrocarbon is burned with sufficient oxygen supply, the products are always carbon dioxide and water vapor. If the supply of oxygen is low or restricted, then carbon monoxide will be produced. This is why it is so dangerous to have an automobile engine running inside a closed garage or to use a charcoal grill indoors. Hydrocarbon (C x H y ) + O 2(g) -----> CO 2(g) + H 2 O (g) CH 4(g) + 2O 2(g) ----> CO 2(g) + 2H 2 O (g) 2C 4 H 10(g) + 13O 2(g) ----> 8CO 2(g) + 10H 2 O (g) NOTE: Complete combustion means the higher oxidation number is attained. Incomplete combustion means the lower oxidation number is attained. The phrase "To burn" means to add oxygen unless told otherwise.

99 Conservation of matter requires identical numbers of atoms on both sides Begin by inspecting the number of each type of atom on each side. Nitrogen must separate before the reaction Hydrogen must separate before the reaction. Re-combine in a different chemical species Same atoms, different ratio!

100

101 D. Ionic or Double Displacement: occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following: 1. a precipitate 2. a gas 3. water or some other non-ionized substance. Basic form: AX + BY -----> AY + BX Examples of ionic reactions: 1. Formation of precipitate. NaCl (aq) + AgNO 3(aq) ----> NaNO 3(aq) + AgCl (s) BaCl 2(aq) + Na 2 SO 4(aq) ----> 2NaCl (aq) + BaSO 4(s) 2. Formation of a gas. HCl (aq) + FeS (s) ----> FeCl 2(aq) + H 2 S (g) 3. Formation of water. (If the reaction is between an acid and a base it is called a neutralization reaction.) HCl (aq) + NaOH (aq) ----> NaCl (aq) + H 2 O (l) 4. Formation of a product which decomposes. CaCO 3(s) + HCl (aq) ----> CaCl 2(aq) + CO 2(g) + H 2 O (l)

102 Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction. Equations of this type are called Net Ionic equations.

103

### Atomic mass is the mass of an atom in atomic mass units (amu)

Micro World atoms & molecules Laboratory scale measurements Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00

### Chapter 8: Chemical Equations and Reactions

Chapter 8: Chemical Equations and Reactions I. Describing Chemical Reactions A. A chemical reaction is the process by which one or more substances are changed into one or more different substances. A chemical

### Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT

Lecture Presentation Chapter 3 Chemical Reactions and Reaction James F. Kirby Quinnipiac University Hamden, CT The study of the mass relationships in chemistry Based on the Law of Conservation of Mass

### Sample Exercise 3.1 Interpreting and Balancing Chemical Equations

Sample Exercise 3.1 Interpreting and Balancing Chemical Equations The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms.

### Chapter 3: Stoichiometry

Chapter 3: Stoichiometry Key Skills: Balance chemical equations Predict the products of simple combination, decomposition, and combustion reactions. Calculate formula weights Convert grams to moles and

### Chapter 6 Chemical Calculations

Chapter 6 Chemical Calculations 1 Submicroscopic Macroscopic 2 Chapter Outline 1. Formula Masses (Ch 6.1) 2. Percent Composition (supplemental material) 3. The Mole & Avogadro s Number (Ch 6.2) 4. Molar

### Chapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry

Chapter 3! : Calculations with Chemical Formulas and Equations Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2

### Calculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu

Calculations and Chemical Equations Atomic mass: Mass of an atom of an element, expressed in atomic mass units Atomic mass unit (amu): 1.661 x 10-24 g Atomic weight: Average mass of all isotopes of a given

### Chapter 5. Chemical Reactions and Equations. Introduction. Chapter 5 Topics. 5.1 What is a Chemical Reaction

Introduction Chapter 5 Chemical Reactions and Equations Chemical reactions occur all around us. How do we make sense of these changes? What patterns can we find? 1 2 Copyright The McGraw-Hill Companies,

### Chapter 8 - Chemical Equations and Reactions

Chapter 8 - Chemical Equations and Reactions 8-1 Describing Chemical Reactions I. Introduction A. Reactants 1. Original substances entering into a chemical rxn B. Products 1. The resulting substances from

### Formulas, Equations and Moles

Chapter 3 Formulas, Equations and Moles Interpreting Chemical Equations You can interpret a balanced chemical equation in many ways. On a microscopic level, two molecules of H 2 react with one molecule

### Chapter 3 Mass Relationships in Chemical Reactions

Chapter 3 Mass Relationships in Chemical Reactions Student: 1. An atom of bromine has a mass about four times greater than that of an atom of neon. Which choice makes the correct comparison of the relative

### Moles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:

Like a recipe: Balancing Eqns Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) coefficients subscripts Balancing Eqns Balancing Symbols (s) (l) (aq) (g) or Yields or Produces solid liquid (pure liquid)

### Concept 1. The meaning and usefulness of the mole. The mole (or mol) represents a certain number of objects.

Chapter 3. Stoichiometry: Mole-Mass Relationships in Chemical Reactions Concept 1. The meaning and usefulness of the mole The mole (or mol) represents a certain number of objects. SI def.: the amount of

### Chapter 1 The Atomic Nature of Matter

Chapter 1 The Atomic Nature of Matter 6. Substances that cannot be decomposed into two or more simpler substances by chemical means are called a. pure substances. b. compounds. c. molecules. d. elements.

### Chemistry B11 Chapter 4 Chemical reactions

Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl

### W1 WORKSHOP ON STOICHIOMETRY

INTRODUCTION W1 WORKSHOP ON STOICHIOMETRY These notes and exercises are designed to introduce you to the basic concepts required to understand a chemical formula or equation. Relative atomic masses of

### SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001

SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001 1. A small pin contains 0.0178 mol of iron. How many atoms of iron are in the pin? 2. A sample

### Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses

Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses B. Calculations of moles C. Calculations of number of atoms from moles/molar masses 1. Avagadro

### Unit 10A Stoichiometry Notes

Unit 10A Stoichiometry Notes Stoichiometry is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations

### CHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS

1 CHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS The Chemical Equation A chemical equation concisely shows the initial (reactants) and final (products) results of

### Part One: Mass and Moles of Substance. Molecular Mass = sum of the Atomic Masses in a molecule

CHAPTER THREE: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Part One: Mass and Moles of Substance A. Molecular Mass and Formula Mass. (Section 3.1) 1. Just as we can talk about mass of one atom of

### Chemical Calculations: Formula Masses, Moles, and Chemical Equations

Chemical Calculations: Formula Masses, Moles, and Chemical Equations Atomic Mass & Formula Mass Recall from Chapter Three that the average mass of an atom of a given element can be found on the periodic

### Stoichiometry Review

Stoichiometry Review There are 20 problems in this review set. Answers, including problem set-up, can be found in the second half of this document. 1. N 2 (g) + 3H 2 (g) --------> 2NH 3 (g) a. nitrogen

### PART I: MULTIPLE CHOICE (30 multiple choice questions. Each multiple choice question is worth 2 points)

CHEMISTRY 123-07 Midterm #1 Answer key October 14, 2010 Statistics: Average: 74 p (74%); Highest: 97 p (95%); Lowest: 33 p (33%) Number of students performing at or above average: 67 (57%) Number of students

### Calculation of Molar Masses. Molar Mass. Solutions. Solutions

Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements

### 1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) 3 (s) + H 2 (g)

1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) A) 1 B) 2 C) 4 D) 5 E) Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) Al (s) + H 2 O (l)? Al(OH)

### IB Chemistry. DP Chemistry Review

DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount

### CHEMICAL REACTIONS. Chemistry 51 Chapter 6

CHEMICAL REACTIONS A chemical reaction is a rearrangement of atoms in which some of the original bonds are broken and new bonds are formed to give different chemical structures. In a chemical reaction,

### 1. What is the molecular formula of a compound with the empirical formula PO and a gram-molecular mass of 284 grams?

Name: Tuesday, May 20, 2008 1. What is the molecular formula of a compound with the empirical formula PO and a gram-molecular mass of 284 grams? 2 5 1. P2O 5 3. P10O4 2. P5O 2 4. P4O10 2. Which substance

### Calculating Atoms, Ions, or Molecules Using Moles

TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary

### Stoichiometry. What is the atomic mass for carbon? For zinc?

Stoichiometry Atomic Mass (atomic weight) Atoms are so small, it is difficult to discuss how much they weigh in grams We use atomic mass units an atomic mass unit (AMU) is one twelfth the mass of the catbon-12

### Chemical Equations. Chemical Equations. Chemical reactions describe processes involving chemical change

Chemical Reactions Chemical Equations Chemical reactions describe processes involving chemical change The chemical change involves rearranging matter Converting one or more pure substances into new pure

### Writing, Balancing and Predicting Products of Chemical Reactions.

Writing, Balancing and Predicting Products of Chemical Reactions. A chemical equation is a concise shorthand expression which represents the relative amount of reactants and products involved in a chemical

### Chapter 3 Stoichiometry

Chapter 3 Stoichiometry 3-1 Chapter 3 Stoichiometry In This Chapter As you have learned in previous chapters, much of chemistry involves using macroscopic measurements to deduce what happens between atoms

### Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O

Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Ans: 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO 2 + 11 H 2 O 3.2 Chemical Symbols at Different levels Chemical symbols represent

### Chem 31 Fall 2002. Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations. Writing and Balancing Chemical Equations

Chem 31 Fall 2002 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations Writing and Balancing Chemical Equations 1. Write Equation in Words -you cannot write an equation unless you

### Stoichiometry. Unit Outline

3 Stoichiometry Unit Outline 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis

### Chapter 5 Chemical Quantities and Reactions. Collection Terms. 5.1 The Mole. A Mole of a Compound. A Mole of Atoms.

Chapter 5 Chemical Quantities and Reactions 5.1 The Mole Collection Terms A collection term states a specific number of items. 1 dozen donuts = 12 donuts 1 ream of paper = 500 sheets 1 case = 24 cans 1

### Writing and Balancing Chemical Equations

Name Writing and Balancing Chemical Equations Period When a substance undergoes a chemical reaction, chemical bonds are broken and new bonds are formed. This results in one or more new substances, often

### Chemistry Final Study Guide

Name: Class: Date: Chemistry Final Study Guide Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The electrons involved in the formation of a covalent bond

### Chemical Equations and Chemical Reactions. Chapter 8.1

Chemical Equations and Chemical Reactions Chapter 8.1 Objectives List observations that suggest that a chemical reaction has taken place List the requirements for a correctly written chemical equation.

### Formulae, stoichiometry and the mole concept

3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be

### IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily.

The Mole Atomic mass units and atoms are not convenient units to work with. The concept of the mole was invented. This was the number of atoms of carbon-12 that were needed to make 12 g of carbon. 1 mole

### Other Stoich Calculations A. mole mass (mass mole) calculations. GIVEN mol A x CE mol B. PT g A CE mol A MOLE MASS :

Chem. I Notes Ch. 12, part 2 Using Moles NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics. 1 MOLE = 6.02 x 10 23 representative particles (representative particles

### THE MOLE / COUNTING IN CHEMISTRY

1 THE MOLE / COUNTING IN CHEMISTRY ***A mole is 6.0 x 10 items.*** 1 mole = 6.0 x 10 items 1 mole = 60, 00, 000, 000, 000, 000, 000, 000 items Analogy #1 1 dozen = 1 items 18 eggs = 1.5 dz. - to convert

### UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS

UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS 4.1 Formula Masses Recall that the decimal number written under the symbol of the element in the periodic table is the atomic mass of the element. 1 7 8 12

### Study Guide For Chapter 7

Name: Class: Date: ID: A Study Guide For Chapter 7 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The number of atoms in a mole of any pure substance

### Steps for balancing a chemical equation

The Chemical Equation: A Chemical Recipe Dr. Gergens - SD Mesa College A. Learn the meaning of these arrows. B. The chemical equation is the shorthand notation for a chemical reaction. A chemical equation

### Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry

Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Why? Chemists are concerned with mass relationships in chemical reactions, usually run on a macroscopic scale (grams, kilograms, etc.). To deal with

### Balancing Chemical Equations Worksheet

Balancing Chemical Equations Worksheet Student Instructions 1. Identify the reactants and products and write a word equation. 2. Write the correct chemical formula for each of the reactants and the products.

### stoichiometry = the numerical relationships between chemical amounts in a reaction.

1 REACTIONS AND YIELD ANSWERS stoichiometry = the numerical relationships between chemical amounts in a reaction. 2C 8 H 18 (l) + 25O 2 16CO 2 (g) + 18H 2 O(g) From the equation, 16 moles of CO 2 (a greenhouse

### 2. DECOMPOSITION REACTION ( A couple have a heated argument and break up )

TYPES OF CHEMICAL REACTIONS Most reactions can be classified into one of five categories by examining the types of reactants and products involved in the reaction. Knowing the types of reactions can help

### Stoichiometry. Lecture Examples Answer Key

Stoichiometry Lecture Examples Answer Key Ex. 1 Balance the following chemical equations: 3 NaBr + 1 H 3 PO 4 3 HBr + 1 Na 3 PO 4 2 C 3 H 5 N 3 O 9 6 CO 2 + 3 N 2 + 5 H 2 O + 9 O 2 2 Ca(OH) 2 + 2 SO 2

### The Mole Concept. The Mole. Masses of molecules

The Mole Concept Ron Robertson r2 c:\files\courses\1110-20\2010 final slides for web\mole concept.docx The Mole The mole is a unit of measurement equal to 6.022 x 10 23 things (to 4 sf) just like there

### Chemical Equations & Stoichiometry

Chemical Equations & Stoichiometry Chapter Goals Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of the term

### 6 Reactions in Aqueous Solutions

6 Reactions in Aqueous Solutions Water is by far the most common medium in which chemical reactions occur naturally. It is not hard to see this: 70% of our body mass is water and about 70% of the surface

### Unit 4 Conservation of Mass and Stoichiometry

9.1 Naming Ions I. Monatomic Ions A. Monatomic ions 1. Ions formed from a single atom Unit 4 Conservation of Mass and Stoichiometry B. Naming Monatomic Ions 1. Monatomic cations are a. Identified by the

### STOICHIOMETRY UNIT 1 LEARNING OUTCOMES. At the end of this unit students will be expected to:

STOICHIOMETRY LEARNING OUTCOMES At the end of this unit students will be expected to: UNIT 1 THE MOLE AND MOLAR MASS define molar mass and perform mole-mass inter-conversions for pure substances explain

### Tutorial 4 SOLUTION STOICHIOMETRY. Solution stoichiometry calculations involve chemical reactions taking place in solution.

T-27 Tutorial 4 SOLUTION STOICHIOMETRY Solution stoichiometry calculations involve chemical reactions taking place in solution. Of the various methods of expressing solution concentration the most convenient

### Problem Solving. Stoichiometry of Gases

Skills Worksheet Problem Solving Stoichiometry of Gases Now that you have worked with relationships among moles, mass, and volumes of gases, you can easily put these to work in stoichiometry calculations.

### 1. How many hydrogen atoms are in 1.00 g of hydrogen?

MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x 10-24 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?

### Chapter Three: STOICHIOMETRY

p70 Chapter Three: STOICHIOMETRY Contents p76 Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions. p70 3-1 Counting by Weighing 3-2 Atomic Masses p78 Mass Mass

### Chapter 7: Chemical Reactions

Chapter 7 Page 1 Chapter 7: Chemical Reactions A chemical reaction: a process in which at least one new substance is formed as the result of a chemical change. A + B C + D Reactants Products Evidence that

### Name Date Class CHEMICAL QUANTITIES. SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296)

10 CHEMICAL QUANTITIES SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296) This section defines the mole and explains how the mole is used to measure matter. It also teaches you how to calculate

### MASS RELATIONSHIPS IN CHEMICAL REACTIONS

MASS RELATIONSHIPS IN CHEMICAL REACTIONS 1. The mole, Avogadro s number and molar mass of an element. Molecular mass (molecular weight) 3. Percent composition of compounds 4. Empirical and Molecular formulas

### Percent Composition and Molecular Formula Worksheet

Percent Composition and Molecular Formula Worksheet 1. What s the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% 2. If the molar mass of the compound in problem 1 is

### Chapter 6 Notes Science 10 Name:

6.1 Types of Chemical Reactions a) Synthesis (A + B AB) Synthesis reactions are also known as reactions. When this occurs two or more reactants (usually elements) join to form a. A + B AB, where A and

### Chapter 1: Moles and equations. Learning outcomes. you should be able to:

Chapter 1: Moles and equations 1 Learning outcomes you should be able to: define and use the terms: relative atomic mass, isotopic mass and formula mass based on the 12 C scale perform calculations, including

### Topic 4 National Chemistry Summary Notes. Formulae, Equations, Balancing Equations and The Mole

Topic 4 National Chemistry Summary Notes Formulae, Equations, Balancing Equations and The Mole LI 1 The chemical formula of a covalent molecular compound tells us the number of atoms of each element present

### Name Class Date. Section: Calculating Quantities in Reactions. Complete each statement below by writing the correct term or phrase.

Skills Worksheet Concept Review Section: Calculating Quantities in Reactions Complete each statement below by writing the correct term or phrase. 1. All stoichiometric calculations involving equations

### Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C-4.4)

Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1 1. 2. 3. 4. 5. 6. Question What is a symbolic representation of a chemical reaction? What 3 things (values) is a mole of a chemical

### Chapter 5, Calculations and the Chemical Equation

1. How many iron atoms are present in one mole of iron? Ans. 6.02 1023 atoms 2. How many grams of sulfur are found in 0.150 mol of sulfur? [Use atomic weight: S, 32.06 amu] Ans. 4.81 g 3. How many moles

### Mole Notes.notebook. October 29, 2014

1 2 How do chemists count atoms/formula units/molecules? How do we go from the atomic scale to the scale of everyday measurements (macroscopic scale)? The gateway is the mole! But before we get to the

### CHAPTER 3 Calculations with Chemical Formulas and Equations. atoms in a FORMULA UNIT

CHAPTER 3 Calculations with Chemical Formulas and Equations MOLECULAR WEIGHT (M. W.) Sum of the Atomic Weights of all atoms in a MOLECULE of a substance. FORMULA WEIGHT (F. W.) Sum of the atomic Weights

### Solution. Practice Exercise. Concept Exercise

Example Exercise 9.1 Atomic Mass and Avogadro s Number Refer to the atomic masses in the periodic table inside the front cover of this textbook. State the mass of Avogadro s number of atoms for each of

### neutrons are present?

AP Chem Summer Assignment Worksheet #1 Atomic Structure 1. a) For the ion 39 K +, state how many electrons, how many protons, and how many 19 neutrons are present? b) Which of these particles has the smallest

### Lecture Topics Atomic weight, Mole, Molecular Mass, Derivation of Formulas, Percent Composition

Mole Calculations Chemical Equations and Stoichiometry Lecture Topics Atomic weight, Mole, Molecular Mass, Derivation of Formulas, Percent Composition Chemical Equations and Problems Based on Miscellaneous

### Chemical Reactions in Water Ron Robertson

Chemical Reactions in Water Ron Robertson r2 f:\files\courses\1110-20\2010 possible slides for web\waterchemtrans.doc Properties of Compounds in Water Electrolytes and nonelectrolytes Water soluble compounds

### Stoichiometry. 1. The total number of moles represented by 20 grams of calcium carbonate is (1) 1; (2) 2; (3) 0.1; (4) 0.2.

Stoichiometry 1 The total number of moles represented by 20 grams of calcium carbonate is (1) 1; (2) 2; (3) 01; (4) 02 2 A 44 gram sample of a hydrate was heated until the water of hydration was driven

### Unit 9 Stoichiometry Notes (The Mole Continues)

Unit 9 Stoichiometry Notes (The Mole Continues) is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations

### Chapter 4. Chemical Composition. Chapter 4 Topics H 2 S. 4.1 Mole Quantities. The Mole Scale. Molar Mass The Mass of 1 Mole

Chapter 4 Chemical Composition Chapter 4 Topics 1. Mole Quantities 2. Moles, Masses, and Particles 3. Determining Empirical Formulas 4. Chemical Composition of Solutions Copyright The McGraw-Hill Companies,

### Name Date Class STOICHIOMETRY. SECTION 12.1 THE ARITHMETIC OF EQUATIONS (pages 353 358)

Name Date Class 1 STOICHIOMETRY SECTION 1.1 THE ARITHMETIC OF EQUATIONS (pages 353 358) This section explains how to calculate the amount of reactants required or product formed in a nonchemical process.

### Unit 3 Notepack Chapter 7 Chemical Quantities Qualifier for Test

Unit 3 Notepack Chapter 7 Chemical Quantities Qualifier for Test NAME Section 7.1 The Mole: A Measurement of Matter A. What is a mole? 1. Chemistry is a quantitative science. What does this term mean?

### Stoichiometry and Aqueous Reactions (Chapter 4)

Stoichiometry and Aqueous Reactions (Chapter 4) Chemical Equations 1. Balancing Chemical Equations (from Chapter 3) Adjust coefficients to get equal numbers of each kind of element on both sides of arrow.

### CHEM 120 Online: Chapter 6 Sample problems Date: 2. Which of the following compounds has the largest formula mass? A) H2O B) NH3 C) CO D) BeH2

CHEM 120 Online: Chapter 6 Sample problems Date: 1. To determine the formula mass of a compound you should A) add up the atomic masses of all the atoms present. B) add up the atomic masses of all the atoms

### Atomic Masses. Chapter 3. Stoichiometry. Chemical Stoichiometry. Mass and Moles of a Substance. Average Atomic Mass

Atomic Masses Chapter 3 Stoichiometry 1 atomic mass unit (amu) = 1/12 of the mass of a 12 C atom so one 12 C atom has a mass of 12 amu (exact number). From mass spectrometry: 13 C/ 12 C = 1.0836129 amu

### Appendix D. Reaction Stoichiometry D.1 INTRODUCTION

Appendix D Reaction Stoichiometry D.1 INTRODUCTION In Appendix A, the stoichiometry of elements and compounds was presented. There, the relationships among grams, moles and number of atoms and molecules

### Chemistry: Chemical Equations

Chemistry: Chemical Equations Write a balanced chemical equation for each word equation. Include the phase of each substance in the equation. Classify the reaction as synthesis, decomposition, single replacement,

### Chemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.

1 Introduction to Chemistry Atomic Weights (Definitions) Chemical Calculations: The Mole Concept and Chemical Formulas AW Atomic weight (mass of the atom of an element) was determined by relative weights.

### EXPERIMENT 12: Empirical Formula of a Compound

EXPERIMENT 12: Empirical Formula of a Compound INTRODUCTION Chemical formulas indicate the composition of compounds. A formula that gives only the simplest ratio of the relative number of atoms in a compound

### Answers and Solutions to Text Problems

Chapter 7 Answers and Solutions 7 Answers and Solutions to Text Problems 7.1 A mole is the amount of a substance that contains 6.02 x 10 23 items. For example, one mole of water contains 6.02 10 23 molecules

### We know from the information given that we have an equal mass of each compound, but no real numbers to plug in and find moles. So what can we do?

How do we figure this out? We know that: 1) the number of oxygen atoms can be found by using Avogadro s number, if we know the moles of oxygen atoms; 2) the number of moles of oxygen atoms can be found

### Ch. 10 The Mole I. Molar Conversions

Ch. 10 The Mole I. Molar Conversions I II III IV A. What is the Mole? A counting number (like a dozen) Avogadro s number (N A ) 1 mole = 6.022 10 23 representative particles B. Mole/Particle Conversions

### The Mole. 6.022 x 10 23

The Mole 6.022 x 10 23 Background: atomic masses Look at the atomic masses on the periodic table. What do these represent? E.g. the atomic mass of Carbon is 12.01 (atomic # is 6) We know there are 6 protons

### Name: Class: Date: 2 4 (aq)

Name: Class: Date: Unit 4 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The balanced molecular equation for complete neutralization of

### CONSERVATION OF MASS During a chemical reaction, matter is neither created nor destroyed. - i. e. the number of atoms of each element remains constant

1 CHEMICAL REACTINS Example: Hydrogen + xygen Water H + H + + - Note there is not enough hydrogen to react with oxygen - It is necessary to balance equation. reactants products + H + H (balanced equation)

### CHEMICAL FORMULA COEFFICIENTS AND SUBSCRIPTS. Chapter 3: Molecular analysis 3O 2 2O 3

Chapter 3: Molecular analysis Read: BLB 3.3 3.5 H W : BLB 3:21a, c, e, f, 25, 29, 37,49, 51, 53 Supplemental 3:1 8 CHEMICAL FORMULA Formula that gives the TOTAL number of elements in a molecule or formula

### Lecture 5, The Mole. What is a mole?

Lecture 5, The Mole What is a mole? Moles Atomic mass unit and the mole amu definition: 12 C = 12 amu. The atomic mass unit is defined this way. 1 amu = 1.6605 x 10-24 g How many 12 C atoms weigh 12 g?

### = 11.0 g (assuming 100 washers is exact).

CHAPTER 8 1. 100 washers 0.110 g 1 washer 100. g 1 washer 0.110 g = 11.0 g (assuming 100 washers is exact). = 909 washers 2. The empirical formula is CFH from the structure given. The empirical formula