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)

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

More information

Chapter 8: Chemical Equations and Reactions

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

More information

Chapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGraw-Hill 2009 1

Chapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGraw-Hill 2009 1 Chapter 3 Insert picture from First page of chapter Stoichiometry: Ratios of Combination Copyright McGraw-Hill 2009 1 3.1 Molecular and Formula Masses Molecular mass - (molecular weight) The mass in amu

More information

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

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

More information

Chapter 3: Stoichiometry

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

More information

Chemical Reactions. Chemical Arithmetic

Chemical Reactions. Chemical Arithmetic Chemical Reactions Chapter 6 Chemical Arithmetic Balancing Equations The mole Gram - mole conversions Mole - mole relationships in chemical equations Mass relationships in chemical equations Per cent yeilds

More information

Sample Exercise 3.1 Interpreting and Balancing Chemical Equations

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.

More information

CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS

CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS FOLLOW UP PROBLEMS 3.1A Plan: The mass of carbon must be changed from mg to g. The molar mass of carbon can then be used to determine the number of moles.

More information

Chemical Reactions and Equations. Chapter 8

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

More information

Chapter 8 Review and Study Guide

Chapter 8 Review and Study Guide Name: Class: Date: Chapter 8 Review and Study Guide Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Knowledge about what products are produced in a chemical

More information

Chapter 6 Chemical Calculations

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

More information

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

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

More information

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

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

More information

Ch. 9 Chemical Reactions

Ch. 9 Chemical Reactions Ch. 9 Chemical Reactions I. Intro to Reactions (p. 282 285) I II III IV V Chemical Reaction Chemical change Atoms of one or more substances (reactants) are rearranged into new substances (products) Signs

More information

Test 4: Equations and Math of Equations Review

Test 4: Equations and Math of Equations Review Name: Tuesday, November 27, 2007 Test 4: Equations and Math of Equations Review 1. Given the balanced equation: 2KClO 2KCl + 3O 3 2 Which type of reaction is represented by this equation? 1. synthesis

More information

Stoichiometry Chapter 9 Assignment & Problem Set

Stoichiometry Chapter 9 Assignment & Problem Set Stoichiometry Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. Stoichiometry 2 Study Guide: Things You Must Know Vocabulary (know the definition

More information

Subscripts and Coefficients Give Different Information

Subscripts and Coefficients Give Different Information Chapter 3: Stoichiometry Goal is to understand and become proficient at working with: 1. Chemical equations (Balancing REVIEW) 2. Some simple patterns of reactivity 3. Formula weights (REVIEW) 4. Avogadro's

More information

W1 WORKSHOP ON STOICHIOMETRY

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

More information

Do Now CHEMICAL REACTIONS. Chemical Reaction. chemical reaction. Chemical Equations. Evidence

Do Now CHEMICAL REACTIONS. Chemical Reaction. chemical reaction. Chemical Equations. Evidence Do Now CHEMICAL REACTIONS What are some signs that a chemical change may have taken place? Where are the reactants and products in a reaction? What do they represent? Describe the law of conservation of

More information

2. In a double displacement reaction, 3. In the chemical equation, H 2 O 2 H 2 + O 2, the H 2 O 2 is a

2. In a double displacement reaction, 3. In the chemical equation, H 2 O 2 H 2 + O 2, the H 2 O 2 is a What are the missing coefficients for the skeleton equation below? Al 2 (SO 4 ) 3 + KOH Al(OH) 3 + K 2 SO 4 2. In a double displacement reaction, A. 1,6,2,3 B. 2,12,4,6 C. 1,3,2,3 D. 4,6,2,3 E. 2,3,1,1

More information

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

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,

More information

Name Class Date. Chapter: Chemical Equations and Reactions

Name Class Date. Chapter: Chemical Equations and Reactions Assessment Chapter Test A Chapter: Chemical Equations and Reactions In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. 1. You

More information

GHW#9. Louisiana Tech University, Chemistry 100. POGIL Exercise on Chapter 4. Quantities of Reactants and Products: Equations, Patterns and Balancing

GHW#9. Louisiana Tech University, Chemistry 100. POGIL Exercise on Chapter 4. Quantities of Reactants and Products: Equations, Patterns and Balancing GHW#9. Louisiana Tech University, Chemistry 100. POGIL Exercise on Chapter 4. Quantities of Reactants and Products: Equations, Patterns and Balancing Why? In chemistry, chemical equations represent changes

More information

CHM1 Review for Exam 9

CHM1 Review for Exam 9 Topics 1. Reaction Types a. Combustion b. Synthesis c. Decomposition d. Single replacement i. Metal activity series ii. Nonmetal activity series e. Double replacement i. Precipitates and solubility rules

More information

Chemistry B11 Chapter 4 Chemical reactions

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

More information

Chapter 1 The Atomic Nature of Matter

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.

More information

Types of Chemical Reactions (rxns.)

Types of Chemical Reactions (rxns.) Types of Chemical Reactions (rxns.) Chemical reactions occur when bonds (between the electrons of atoms) are formed or broken Chemical reactions involve l changes in the chemical composition of matter

More information

Chapter 3 Mass Relationships in Chemical Reactions

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

More information

Stoichiometry Dr. M. E. Bridge

Stoichiometry Dr. M. E. Bridge Preliminary Chemistry Course Stoichiometry Dr. M. E. Bridge What is stoichiometry? The meaning of the word: The word stoichiometry comes from two Greek words: (meaning element ) and (meaning measure )

More information

1. Balance the following equation. What is the sum of the coefficients of the reactants and products?

1. Balance the following equation. What is the sum of the coefficients of the reactants and products? 1. Balance the following equation. What is the sum of the coefficients of the reactants and products? 1 Fe 2 O 3 (s) + _3 C(s) 2 Fe(s) + _3 CO(g) a) 5 b) 6 c) 7 d) 8 e) 9 2. Which of the following equations

More information

Chapter 8 - Chemical Equations and Reactions

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

More information

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

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

More information

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

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)

More information

Formulas, Equations and Moles

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

More information

Name Class Date. Chapter: Chemical Equations and Reactions

Name Class Date. Chapter: Chemical Equations and Reactions Assessment Chapter Test B Chapter: Chemical Equations and Reactions PART I In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question

More information

Chapter 7 Chemical Reactions. Chemical & Physical Changes

Chapter 7 Chemical Reactions. Chemical & Physical Changes Chapter 7 Chemical Reactions Chemical & Physical Changes In a physical change, the chemical composition of the substance remains constant. Examples of physical changes are the melting of ice or the boiling

More information

Chemical reactions. Classifications Reactions in solution Ionic equations

Chemical reactions. Classifications Reactions in solution Ionic equations Chemical reactions Classifications Reactions in solution Ionic equations Learning objectives Distinguish between chemical and physical change Write and balance chemical equations Describe concepts of oxidation

More information

Balancing Equations Notes

Balancing Equations Notes . Unit 9 Chemical Equations and Reactions What is a Chemical Equation? A Chemical Equation is a written representation of the process that occurs in a chemical reaction. A chemical equation is written

More information

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 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

More information

Chapter 8 Chemical Equations and Reactions

Chapter 8 Chemical Equations and Reactions Chapter 8 Chemical Equations and Reactions 1 Section 8.1 Objectives: List 3 observations that suggest that a chemical reaction has taken place. List 3 requirements for a correctly written chemical reaction.

More information

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations CHE11 Chapter Chapter Stoichiometry: Calculations with Chemical Formulas and Equations 1. When the following equation is balanced, the coefficients are. NH (g) + O (g) NO (g) + H O (g) (a). 1, 1, 1, 1

More information

Chapter 7: Chemical Equations. Name: Date: Period:

Chapter 7: Chemical Equations. Name: Date: Period: Chapter 7: Chemical Equations Name: Date: Period: 7-1 What is a chemical reaction? Read pages 232-237 a) Explain what a chemical reaction is. b) Distinguish between evidence that suggests a chemical reaction

More information

Chapter 4 Chemical Composition

Chapter 4 Chemical Composition Chapter 4 Chemical Composition 4.1 (a) mole; (b) Avogadro s number; (c) empirical formula; (d) solute; (e) molarity; (f) concentrated solution 4. (a) molar mass; (b) percent composition by mass; (c) solvent;

More information

Unit 10A Stoichiometry Notes

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

More information

10/21/2013. Chemical REACTIONS you should know. Chemical Reactions. 1 st Write Reaction

10/21/2013. Chemical REACTIONS you should know. Chemical Reactions. 1 st Write Reaction Chapter 3: Chemical Stoichiometry Chemical Equations (Write, Balance, Interpret) Reactions You Should Know Formula Weights (Must know chemical formula Avogadro s number and the Mole Limiting Reactants

More information

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

More information

MOLE CONVERSION PROBLEMS. 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles]

MOLE CONVERSION PROBLEMS. 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles] MOLE CONVERSION PROBLEMS 1. What is the molar mass of MgO? [40.31 g/mol] 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles] 3. How many moles are present in 2.5 x 10 23 molecules of CH

More information

Ch 8 Notes: Chemical Equations and Reactions

Ch 8 Notes: Chemical Equations and Reactions Name: Ch 8 Notes: Chemical Equations and Reactions I. Chemical Equations A properly written chemical equation can summarize any chemical change. The following requirements will help you write and read

More information

Chemical bonding is the true difference between compounds and mixtures. Atomic elements:

Chemical bonding is the true difference between compounds and mixtures. Atomic elements: Chapter 3 - Molecules, compounds, and chemical equations Elements and compounds Chemical bonding is the true difference between compounds and mixtures Atomic elements: Ionic bond: attraction of oppositely

More information

Chemical calculations

Chemical calculations Chemical calculations Stoichiometry refers to the quantities of material which react according to a balanced chemical equation. Compounds are formed when atoms combine in fixed proportions. E.g. 2Mg +

More information

6/27/2014. Periodic Table of the ELEMENTS. Chemical REACTIONS you should know. Brief Review for 1311 Honors Exam 2

6/27/2014. Periodic Table of the ELEMENTS. Chemical REACTIONS you should know. Brief Review for 1311 Honors Exam 2 Brief Review for 3 Honors Exam 2 Chapter 2: Periodic Table I. Metals. Representative Metals Alkali Metals Group Alkaline Earth Metals. Group 2 2. Transition Metals II. Metalloids Chapter 3: All Chapter

More information

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

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

More information

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 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

More information

Stoichiometry Review

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

More information

Mass and Moles of a Substance

Mass and Moles of a Substance Chapter Three Calculations with Chemical Formulas and Equations Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows

More information

Mole - Mass Relationships in Chemical Systems

Mole - Mass Relationships in Chemical Systems Chapter 3: Stoichiometry Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Calculating the Amounts

More information

Chapter 5 Chemical Quantities and Reactions

Chapter 5 Chemical Quantities and Reactions Chapter 5 Chemical Quantities and Reactions 1 Avogadro's Number Small particles such as atoms, molecules, and ions are counted using the mole. 1 mole = 6.02 x 10 23 items Avogadro s number 602 000 000

More information

Stoichiometry. Stoichiometry Which of the following forms a compound having the formula KXO 4? (A) F (B) S (C) Mg (D) Ar (E) Mn

Stoichiometry. Stoichiometry Which of the following forms a compound having the formula KXO 4? (A) F (B) S (C) Mg (D) Ar (E) Mn The Advanced Placement Examination in Chemistry Part I Multiple Choice Questions Part II - Free Response Questions Selected Questions from 1970 to 2010 Stoichiometry Part I 1984 2. Which of the following

More information

IB Chemistry. DP Chemistry Review

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

More information

CHEMICAL REACTIONS. Chemistry 51 Chapter 6

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,

More information

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

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

More information

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) 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)

More information

Calculating Atoms, Ions, or Molecules Using Moles

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

More information

The formation of a gas is evidence of. 1. A chemical change. 2. A physical change. 3. No change in energy 4. Both (1) and (2)

The formation of a gas is evidence of. 1. A chemical change. 2. A physical change. 3. No change in energy 4. Both (1) and (2) Chapter 8 Review The formation of a gas is evidence of 1. A chemical change. 2. A physical change. 3. No change in energy 4. Both (1) and (2) The formation of a gas is evidence of 1. A chemical change.

More information

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

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

More information

Chapter 3 Calculation with Chemical Formulas and Equations

Chapter 3 Calculation with Chemical Formulas and Equations Chapter 3 Calculation with Chemical Formulas and Equations Practical Applications of Chemistry Determining chemical formula of a substance Predicting the amount of substances consumed during a reaction

More information

Chapter 3 Stoichiometry of Formulas and Equations

Chapter 3 Stoichiometry of Formulas and Equations 3-1 Chapter 3 Stoichiometry of Formulas and Equations 3-2 Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical

More information

Calculation of Molar Masses. Molar Mass. Solutions. Solutions

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

More information

Types of Reactions. CK12 Editor. Say Thanks to the Authors Click (No sign in required)

Types of Reactions. CK12 Editor. Say Thanks to the Authors Click  (No sign in required) Types of Reactions CK12 Editor Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) To access a customizable version of this book, as well as other interactive content, visit

More information

Balancing Equations Notes

Balancing Equations Notes Unit 7 Chemical Equations and Reactions What is a Chemical Equation? A Chemical Equation is a written representation of the process that occurs in a chemical reaction. A chemical equation is written with

More information

Chapter 3. Stoichiometry of Formulas and Equations

Chapter 3. Stoichiometry of Formulas and Equations Chapter 3 Stoichiometry of Formulas and Equations Chapter 3 Outline: Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing

More information

Name Exam No. C. CHEM 60 Fall 2015 Exam 2 Ch 5-8, 100 points total, however, there are a maximum of 110 points. sodium bicarbonate

Name Exam No. C. CHEM 60 Fall 2015 Exam 2 Ch 5-8, 100 points total, however, there are a maximum of 110 points. sodium bicarbonate Name Exam No. C CHEM 60 Fall 2015 Exam 2 Ch 5-8, 100 points total, however, there are a maximum of 110 points. In the space provided below each problem, show a neat, complete, and logical method for solving

More information

Stoichiometry. Unit Outline

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

More information

Chapter 3 Stoichiometry Mole - Mass Relationships in Chemical Systems

Chapter 3 Stoichiometry Mole - Mass Relationships in Chemical Systems Chapter 3 Stoichiometry Mole - Mass Relationships in Chemical Systems 3.1 Atomic Masses 3.2 The Mole 3.3 Molar Mass 3.4 Percent Composition of Compounds 3.5 Determining the Formula of a Compound 3.6 Chemical

More information

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

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

More information

Chemical Calculations: Formula Masses, Moles, and Chemical Equations

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

More information

Formulae, stoichiometry and the mole concept

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

More information

Chapter 3: Stoichiometry

Chapter 3: Stoichiometry Chapter 3: Stoichiometry Goal is to understand and become proficient at working with: 1. Avogadro's Number, molar mass and converting between mass and moles (REVIEW). 2. empirical formulas from analysis.

More information

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

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

More information

Chemical Reactions. Chemical Equations. Mole as Conversion Factor: To convert between number of particles and an equivalent number of moles:

Chemical Reactions. Chemical Equations. Mole as Conversion Factor: To convert between number of particles and an equivalent number of moles: Quantities of Reactants and Products CHAPTER 3 Chemical Reactions Stoichiometry Application of The Law of Conservation of Matter Chemical book-keeping Chemical Equations Chemical equations: Describe proportions

More information

Classifying Chemical Reactions Chapter 7

Classifying Chemical Reactions Chapter 7 Classifying Chemical Reactions Chapter 7 Classifying Chemical Reactions Chemical reactions can be divided into five categories: I. Combination Reactions II. Decomposition Reactions III. Single-Replacement

More information

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 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

More information

What is a chemical reaction?

What is a chemical reaction? Chapter 5 Chemical Reactions and Equations What is a chemical reaction? How do we know a chemical reaction occurs? Writing chemical equations Predicting chemical reactions Representing reactions in aqueous

More information

CP Chapter 11 Notes Reactions and Equations

CP Chapter 11 Notes Reactions and Equations CP Chapter 11 Notes Reactions and Equations Evidence of Chemical Reactions How can you tell a reaction has taken place? Temperature change Color change Gas/bubbles Appearance of a (precipitate) Chemical

More information

Chapter 3 Chemical Reactions and Reaction Stoichiometry. 許富銀 ( Hsu Fu-Yin)

Chapter 3 Chemical Reactions and Reaction Stoichiometry. 許富銀 ( Hsu Fu-Yin) Chapter 3 Chemical Reactions and Reaction Stoichiometry 許富銀 ( Hsu Fu-Yin) 1 Stoichiometry The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry.

More information

Chapter 3 Stoichiometry

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

More information

Writing, Balancing and Predicting Products of Chemical Reactions.

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

More information

Unit 2. Molar Mass Worksheet

Unit 2. Molar Mass Worksheet Unit 2 Molar Mass Worksheet Calculate the molar masses of the following chemicals: 1) Cl 2 8) UF 6 2) KOH 9) SO 2 3) BeCl 2 10) H 3 PO 4 4) FeCl 3 11) (NH 4 ) 2 SO 4 5) BF 3 12) CH 3 COOH 6) CCl 2 F 2

More information

Types of Chemical Reactions

Types of Chemical Reactions Why? Types of Chemical Reactions Do atoms rearrange in predictable patterns during chemical reactions? Recognizing patterns allows us to predict future behavior. Weather experts use patterns to predict

More information

Chemistry Final Study Guide

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

More information

Chemical Reactions Chapter 8 Assignment & Problem Set

Chemical Reactions Chapter 8 Assignment & Problem Set Chemical Reactions Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. Chemical Reactions 2 Study Guide: Things You Must Know Vocabulary (know the

More information

STOICHOMETRY UNIT Determine the mass of carbon present in a sample of glucose weighing 5.4 g.

STOICHOMETRY UNIT Determine the mass of carbon present in a sample of glucose weighing 5.4 g. PERCENTAGE COMPOSITION STOICHOMETRY UNIT 3 1. A sample of magnesium weighing 2.246 g burns in oxygen to form 3.724 g of magnesium oxide. What are the percentages of magnesium and oxygen in magnesium oxide?

More information

AP Chemistry Prep - Summer Assignment 2013

AP Chemistry Prep - Summer Assignment 2013 AP Chemistry Prep - Summer Assignment 2013 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which value has only 4 significant digits? a. 6.930 c. 8450

More information

MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY

MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY [MH5; Ch. 3] Each element has its own unique mass. The mass of each element is found on the Periodic Table under the chemical symbol for the element (it is usually

More information

Skeleton Equations equations describing a reaction using formulas. HCl(aq) + Zn(s) ZnCl 2 (aq) + H 2 (g)

Skeleton Equations equations describing a reaction using formulas. HCl(aq) + Zn(s) ZnCl 2 (aq) + H 2 (g) 10.1 Reactions and Equations Evidence of Chemical Reactions Chemical Reaction a process in which one or more substances are converted into new substances with different physical and chemical properties.

More information

CHEMISTRY. (i) It failed to explain how atoms of different elements differ from each other.

CHEMISTRY. (i) It failed to explain how atoms of different elements differ from each other. CHEMISTRY MOLE CONCEPT DALTON S ATOMIC THEORY By observing the laws of chemical combination, John Dalton proposed an atomic theory of matter. The main points of Dalton s atomic theory are as follows: (i)

More information

1. P 2 O 5 2. P 5 O 2 3. P 10 O 4 4. P 4 O 10

1. P 2 O 5 2. P 5 O 2 3. P 10 O 4 4. P 4 O 10 Teacher: Mr. gerraputa Print Close Name: 1. A chemical formula is an expression used to represent 1. mixtures, only 3. compounds, only 2. elements, only 4. compounds and elements 2. What is the total number

More information

Unit 7 A1 Worksheet: Writing and Balancing Chemical Equations

Unit 7 A1 Worksheet: Writing and Balancing Chemical Equations Name Period Unit 7 A1 Worksheet: Writing and Balancing Chemical Equations 1. Describe the following word equation with a statement or sentence: Iron + Oxygen iron (III) oxide 2. In a chemical equation,

More information

Chemical Equations and Chemical Reactions. Chapter 8.1

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.

More information

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 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

More information