Moles and Stoichiometry
|
|
- Andrew Newman
- 7 years ago
- Views:
Transcription
1 Moles and Stoichiometry I. Moles A. Definition. 1. A mole (mol) = x units x = Avogadro's constant, with a dimension of particles mol 1 or mol 1 b. A mass in grams equal to the formula mass of a substance contains 6.02 x formula units. c. The practical unit of formula mass or atomic mass = grams/mol. 2. The following should be familiar to you. Type of Formula Composition of 10.0 g of substance Formula substance mass in moles _ in no. of units Na element 23.0 g/mol 10.0 /23.0= (0.435)(6.02 x10 23 ) = mol 2.62 x atoms H 2 molecular 2.0 g/mol 5.0 moles H x molecules of H moles H 6.02 x atoms of H C 2 H 4 molecular 28.0 g/mol mol C 2 H x molecules C 2 H mol C 4.30 x atoms C mol H 8.60 x atoms H Al 2 O 3 ionic 102 g/mol mol Al 2 O x formula units mol Al x Al 3+ ions mol O x O 2- ions 1
2 3. Additional Example. Consider a g sample of the compound C 4 H 10 O 2 a. How many moles of C 4 H 10 O 2 are present? MM of C 4 H 10 O 2 = 4(12.0) + 10(1.0) + 2(16.0) = 90.0g/mol 48.60g Mol C 4 H 10 O 2 = = mol 90.0g/mol b. How many moles and how many grams of C are present in the sample? The formula shows 4 C atoms for every molecule there are 4 moles of C for every mole of C 4 H 10 O 2. Moles C = mole C 4 H 10 O 2 x 4 mol C 1 mol C 4 H 10 O 2 = 2.16 mol C grams C = (2.16 mol C)(12.0 g/mol C) = g or 48.0 g C grams C =48.60g 90.0 g compound = 25.92g C g moles C = = 2.16 mol C 12.0 g/mol c. How many O atoms are present in the sample? 2 mol O Mol O = mole C 4 H 10 O 2 x = 1.08 mol O 1 mol C 4 H 1 0 O 2 Atoms O = 1.08 mol O x 6.02x10 23 atoms/mol = 6.50x10 23 d. How many atoms of H and how many grams of H are present in the sample? 10 mol H Moles of H = mole C 4 H 10 O 2 x = 5.40 mol H 1 mol C 4 H 10 O 2 grams H = 5.40 molx1.0g/mol = 5.40g Note: mol H could also be obtained by using the moles of C: Mol H = 2.16 mol Cx 10mol H = 5.40 mol 4 mol C 2
3 B. Empirical formulas from analysis data. 1. A compound was analyzed and found to contain 29.1% Na, 40.6% S, and 30.3% O by mass. Calculate the empirical formula. moles of Na = 29.1g / 23.0 g/mol= mol > > 2 moles of S = 40.6g / 32.1g/mol = mol > > 2 moles of O = 30.3g / 16.0g/mol = mol > > 3 formula : Na 2 S 2 O 3 Sodium thiosulfate 2. A g sample of a compound was analyzed and found to contain g of C and g of H. In another experiment the molar mass was estimated to be equal to 30. Calculate the empirical and molecular formulas of the compound. moles C = 0.320g / 12.0g/mol = > > 1 moles H = 0.081g / 1.0 g/mol = > > 3 empirical formula = CH 3 empirical formula mass (EFM) = (1.0) = = 2 = empirical formula units / molecule C 2H 6 = molecular formula. 3. Analysis by combustion. A g sample of a compound containing only C, H, and O was burned to give g of CO 2 and g of H 2 O. Calculate the empirical formula. + O 2 (C x H y O z ) > CO 2 + H 2 O g g g all the C goes to form CO 2 mol C = mol CO 2 = g 44.0 g/mol = mol grams C in the sample = ( mol )(12.0 g/mol) = g all the H ends up in H 2 O 2(1.199 g) mol H = 2xmol H 2 O = = mol 18.0 g/mol grams H in sample = ( mol)(1.0 g/mol) = g total mass of sample = g = g of C + g of H + g of O therefore g of O = g g g = g g mol O = = mol 16.0 g/mol mol C = >
4 mol H = > 8.01 mol O = > 1.00 empirical formula = C 3 H 8 O II. Stoichiometry: Calculations from balanced chemical equations. A. Information in a balanced chemical equation: 4Al(s) + 3O2(g) > 2Al2O3(s) 1. In terms of atoms and molecules. 4 Al atoms + 3 O 2 molecules -----> 2 Al 2 O 3 formula units. 2. In terms of atomic mass units. (4x27) = 108 u of Al + (3x32) = 96 u of O > (2x102) = 204 u of Al 2 O 3 Note the Law of Conservation of mass (mass of products = mass of reactants) 3. In terms of moles. 4 moles of Al + 3 moles of O > 2 moles of Al 2 O 3 4. In terms of grams. 108 g of Al + 96 g of O > 204 g Al 2 O 3 5. In terms of any mass unit. 108 lb. of Al + 96 lb. of O > 204 lb. of Al 2 O slugs of Al + 96 slugs of O > 204 slugs of Al 2 O 3 B. Calculations from balanced equations. 1. General. 108 tons of Al + 96 tons of O > 204 tons of Al 2 O 3 a. A balanced chemical equation gives stiochiometeric information directly in terms of moles of reactants and products. Example, in the above equation, each mole of Al that reacts requires 3 4 mole of O 2 and produces 1 2 mole of Al 2O 3. b. In the laboratory, we do not obtain information directly in terms of moles. The normal laboratory measurements involve obtaining 1) the masses of pure solids and some pure liquids. 2) the volumes of solutions of known concentration. 3) the volumes of gases at known temperatures and pressures. 4) the volume of liquids of known densities. c. In stoichiometric calculations information is given, and asked for, in terms of these laboratory measurements. Therefore, the sequence to be followed is: 4
5 1) CONVERT LABORATORY INFORMATION TO MOLES 2) USE THE BALANCED CHEMICAL EQUATION TO GET INFORMATION IN TERMS OF MOLES. 3) RECONVERT MOLES BACK INTO LABORATORY UNITS. 2. Conversions to moles-these relationships should be familiar to you. a. Direct mass measurements of pure solids and liquids. Moles = b. Pure liquids of known densities. Mass in grams formula mass Mass = (density)x(volume) The densities are usually in g/ml and the volumes are in ml Mass in grams Moles = formula mass c. Volumes of solutions of known Molarity: Molarity (M) = Moles of Solute Liters of Solution = mmol Solute ml of Solution moles of solute = (M)(no. of liters of solution) mmoles of solute = (M)(no. of ml of solution) d. Volumes of gases: Use the Ideal gas Law (see Chapter ) 1) for a pure gases PV = nrt = g MM RT. Where V = volume of the gas in L, P = pressure of the gas, MM= molar mass n = number of moles of the gas, T = temperature in K (= C ), R = gas constant. The value of R depends on the dimensions of pressure. R = 62.4 Torr L mol K = atm L mol K = Pa m3 mol K 2) For mixtures of gases use partial pressures and Dalton s Law. P x V = n x RT for a component x of a gaseous mixture (P x = Partial pressure of x) P tot = P x 3) This is used when collecting a gas over water. A standard laboratory technique for collecting a gaseous product from a reaction is to allow the gas to displace water in a 5
6 container. The gas collected is saturated with water vapor. Therefore the total pressure in the container is due to a mixture of the gas (X) and H 2 O, so that P Tot = P X + P H 2 O P H2 O is the vapor pressure of water and is a function of temperature only, this can be looked up in standard tables and subtracted from P tot (usually atmospheric pressure) to obtain the partial pressure of the gas. C. Stoichiometric calculations. 1. Suppose that 80.0 g of Al was reacted according to the equation 4Al(s) + 3O 2 (g) ----> 2Al 2 O 3 (s) a. How many moles of O 2 was consumed? 80.0 g moles Al = 27.0 g/mol = 2.96 mol of Al moles O 2 = 2.96 mol Al x 3 mol O 2 4 mol Al = 2.22 mol O 2 b. What volume of O 2 measured at 27 C and 750 Torr pressure would be used? V = n Torr L O 2 RT (2.22 mole)(62.4 P = mol K )(300 K) 750 Torr = 55.3 L c. How many grams of Al 2 O 3 would be formed? moles Al 2 O 3 = 2.96 mol Al x ( 2 mol Al 2O 3 4 mol Al ) = 1.48 mol Al 2 O 3 g Al 2 O 3 = (1.48 mol)(102 g/mol) = 151 g Al 2 O 3 6
7 2. Consider the reaction: 4KO 2 (s) + 2H 2 O(l) ----> 4KOH(aq) + 3O 2 (g). Suppose that 7.50 g of KO 2 was placed in 50.0 ml of water. a. What volume of O 2 would be collected H 2 O at 27 C and 757 Torr atmospheric pressure? 7.50 g moles KO 2 = 71.1 g/mol = mol KO 2 moles of O 2 produced = (0.105 mol KO 2 ) x ( 3 mol O 2 4 mol KO 2 ) = 7.875x10-2 = n O2 Since the gas is collected over water, the vapor pressure of water must be subtracted from the atmospheres pressure to get the partial pressure of O 2, therefore P O2 = Patm -P H2 O, where P H2 O = vapor pressure of water. At 27 C, P H2 O = 27 Torr. P O2 = 757 Torr - 27 Torr = 730 Torr V = n o2rt P O2 = (7.875x10-2 )(62.4)(300) 730 = 2.02 L b. How many grams of KOH are formed? moles of KOH formed = mol KO 2 x ( g KOH formed = mol (56.1 g/mol) = 5.89 g c. How many ml of water remains? 4 mol KOH 4 mol KO 2 ) = mol KOH moles H 2 O required = mol KO 2 x ( 2 mol H 2O 4 mol KO 2 ) = mol H 2 O g H 2 O =( mol) x (18.0 g/mol) = g H 2 O. This is equivalent to 1 ml of H 2 O (the density of H 2 O = 1.0 g/ml) ml of H 2 O will remain. d. What is the final concentration of KOH? M KOH = mol 49.0x10-3 L = 2.14M 7
8 3. Limiting reagents. a. It is not necessary, and many times not desirable, to mix stoichiometric amounts of reactants. Under these conditions, the reagent used up first (the limiting reagent) determines the amount of product; the other reactant is the excess reactant. In such cases the first step is to decide which reactant is limiting. b. Consider the reaction 3Mg + N > Mg 3 N 2 Suppose that 2.50 g Mg and 1.00 g N 2 are mixed and the reaction takes place. 1) What reactant will be in excess and how many grams of this reactant remains unreacted? 2.50 g moles Mg = 24.3 g/mol = mol moles N 1.00 g 2 = 28.0 g/mol = mol The stoichiometric Mg molar ratio = 3 N 1, the experimental molar ratio is = 2.88 which is less than 3 N 2 is in excess and Mg is the limiting reagent and will determine the amount of product. moles of N 2 required = mol Mg x ( 1 mol N 2 3 mol Mg moles N 2 left = mol mol = mol g in excess = ( mol) x (28.0 g/mol) = 3.92x10-2 g. b. How many grams of Mg 3 N 2 will be formed. 4. Yields. ) = mol moles Mg 3 N 2 = mol Mg x ( 1 mol Mg 3N 2 3 mol Mg ) = 3.43x10-2 mol grams of Mg 3 N 2 = (3.43x10-2 mol) x (100.9 g/mol) = 3.46 g a. The above calculations give the maximum amount of product that could be obtained in a reaction. In practice, this is never obtained, we do not get 100% of what we expect. b. The percent yield, or yield, is: experimental amount of product Yield = x100 theoretical amount of product The theoretical amount of product is that calculated in the above examples. c. There are a number of reasons for low yields. 1) Poor technique. 2) An unfavorable equilibrium. In such cases the yield can be improved by having one of the reactants in excess. 3) Competing reaction or consecutive reactions. 8
9 d. When new reactions are reported in the literature, they are usually described under conditions which maximize the yield. 9
10 Problems 1. On analysis a 2.75 g sample of a compound was found to contain 1.55 g of phosphorus and 1.20 g of sulfur. Calculate the empirical formula of this compound. (P 4 S 3 ) 2. A sample of a compound contains 3.96 g of carbon, 0.66 g of hydrogen, and 3.52 g of oxygen. In another experiment it was found that a 2.05 g sample of the gaseous compound occupied a volume of 584 ml at 125 C and 400 Torr pressure. a. What is the empirical formula of this compound? b. What is the molar mass of this compound? c. What is the molecular formula of this compound? (C 3 H 6 O 2 ; 218; C 9 H 18 O 6 ) 3. A sample of a compound containing only nitrogen and sulfur was burned in oxygen completely converting the nitrogen to N 2 O 3 and the sulfur to SO 2. The oxides formed were trapped and weighed giving g of N 2 O 3 and g of SO 2. What is the empirical formula of the compound? (N 2 O 3 ) 4. Calculate the empirical formulas from the following percent compositions. a. 34.3% Na; 17.9% C; 47.8% O b. 39.3% C; 8.2% H; 52.5% O c. 72.3% Fe; 27.7% O d. 69.9% Fe; 30.1% O e. 77.7% Fe; 22.3% O f. 9.7% Al; 38.4% Cl; 51.9% O (Answer: a.naco 2 b. C 2 H 5 O 2 c. Fe 3 O 4 d. Fe 2 O 3 e. FeO f. AlCl 3 O 9 ) 5. A 5.00 g sample of a compound containing C, H, and N was burned in oxygen to give g of CO2 and g of H2O. Calculate the empirical formula of the compound. (C 2 H 7 N) 6. A g sample of a mixture of NaCl and KCl was dissolved in water and the chloride precipitated as AgCl. If the mass of the AgCl precipitate is g, calculate the mass percent NaCl in the sample. (42.9% NaCl) 7. Calcium hydroxide reacts with H 3 PO 4 to give Ca 3 (PO 4 ) 2. Suppose the g of Calcium hydroxide was placed in 200 ml of a 0.045M H 3 PO 4 solution. What reactant would be in excess and how many grams of Ca 3 (PO 4 ) 2 would be formed. (H 3 PO 4, 1.19 g Ca 3 (PO 4 ) 2 ) 8. What volume of H 2 could be collected over water at 27 C and 774 Torr atmospheric pressure by the reaction of g of Na with excess water to give H 2 and NaOH? (139 ml) 10
11 9. A hydrocarbon was analyzed and found to contain 84.1% C and 15.9% H by mass. In an experiment it was found that a g sample of the gaseous compound occupied a volume of 215 ml at 50 C and 400 Torr. Calculate the molecular formula of the compound. (C 8 H 18 ) 10. Aluminum sulfide reacts with oxygen to give aluminum sulfate. How much oxygen would be consumed when 6.60 g of aluminum sulfide reacts? (8.45 g) 11. Propene, C 3 H 6, burns in oxygen to form carbon dioxide and water. a) Write the balanced equation for this reaction. b) Explain what the equation states in a quantitative way. c) Per mole of propene, how many moles of oxygen would be required? d) Suppose 0.84 g of propene is burned. How many moles of Propene is present? How many moles of oxygen would be required for the complete combustion? How many moles of water and carbon dioxide would be formed? Calculate the grams of oxygen and the grams of carbon dioxide and water formed. e) Show that the masses in (d) are in accordance with the law of conservation of mass. (d mol C 3 H 6 ; O 2, 0.09 mol, 2.88g; CO 2, 0.06 mol, 2.64 g; H 2 O, 0.06 mol, 1.08g) 12. Calcium carbonate decomposes on heating to give calcium oxide and CO 2 (g). What volume of CO 2, measured at 100 C and 757 torr, would be generated by the decomposition of 5.0 g of calcium carbonate? (1.54 L) 13. Naphthalene, C 10 H 8, reacts with O 2 to give CO and H 2 O. What volume of CO could be collected over water at 24 C and 750 torr total pressure by the complete reaction of 1.50 g of naphthalene? (Vapor pressure of water at 24 C = 22 torr.) (2.98 L) 14. a. A compound containing only carbon and hydrogen when reacted with O 2 produced 1.62 g of H 2 O and 2.40 liter of CO 2 gas when measured at 27 C and 700 torr pressure. Assuming that all of the carbon in the compound was converted to CO 2 and all the hydrogen was converted to H 2 O, calculate the empiricalformula of the compound. (CH 2 ) b. In another experiment, a 0.30 g sample of this gaseous compound was found to occupy a volume of 137 ml at 27 C and 730 torr pressure. Calculate the molar mass of the compound. (56.1 g/mol) c) what is the molecular formula of the compound? (C 4 H 8 ) 11
12 d) Write the balanced equation for the reaction of this compound with O 2. (C 4 H O > 4 CO H 2 O) 15. Aluminum reacts with HCl to give AlCl 3 and H 2 (g). What volume of H 2 would be collected over water at 28 C and 748 torr pressure when 5.0 g of Al is placed in 250 ml of a 2.0 M HCl solution? (Vapor pressure of water at 28 C = 28 m torr.) (0.500 mol HCl (LIMITING REAGENT); mol Al; 13.0 L of H 2 ) 16. Zinc sulfide reacts with O 2 (g) to give zinc (II) oxide and SO 2 (g). What volume of O 2 (g), measured at 25 C and 740 torr pressure, would be required to react with 0.25 g of zinc sulfide? What volume of SO 2 (g), measured under the same conditions, would be produced in this reaction? (volume of O 2 = 96.7 ml volume of SO 2 = 64.5 ml) 17. Phosphorus burns in O 2 to give P 2 O 5. Suppose that 10.0 g of P is ignited in a 30.0 L container of O 2 at a temperature of 100 C and a pressure of 400 torr. a. What reactant is in excess, and how many moles of that reactant will be left unreacted? (O 2 in excess, mol left) b. How many grams of P 2 O 5 will be formed? (22.9 g) 18. CO 2 can be removed from a gaseous mixture by reacting it with Na 2 O (s) to give Na 2 CO 3 (s). A mixture of CO 2 (g) and an inert gas in a 5.0 liter container originally exerted a pressure of 500 torr at 25 C. After the gas mixture was exposed to Na 2 O, the pressure in the container decreased to 200 torr. a) What was the partial pressure of CO 2 in the gas mixture? (300 Torr) b) How many grams of CO 2 was in the gas mixture? (3.55 g) c) How many grams of Na 2 CO 3 was formed? (8.55 g) 19. Consider the following reaction: Al + H 2 SO > Al 2 (SO 4 ) 3 + H 2 a. Balance the equation. b. For a 8.1 g samples of Al, 1) Calculate the moles of Al present. 12
13 2) Calculate the moles and grams of H 2 SO 4 required for complete reaction. 3) Calculate the moles and grams of Al 2 (SO 4 ) 3 and of H 2 that would be formed. 4) Show that mass is conserved in this reaction. 20. For the following reaction: HCl + O > H 2 O + Cl 2 a. Balance the equation. b. For a 0.16 g sample of O 2, 1) Calculate the moles of O 2 present 2) Calculate the moles and grams of HCl required for complete reaction. 3) Calculate the moles and grams of H 2 O and of Cl 2 formed. 4) Show that mass is conserved in this reaction. 21. Butane, C 4 H 10, reacts with O 2 to give CO 2 and H 2 O. a. Write the balanced equation for the reaction. b. Suppose 17.4 g of C 4 H 10 and 64.0 g of O 2 are mixed and the reaction allowed to take place. 1) Which reactant will be in excess? How many moles of this reactant will be left? How may grams? 2) How many grams of CO 2 is formed? 3) How many grams of H 2 O is formed? 22. Zinc sulfide, ZnS, reacts with O 2 to give ZnO and SO 2. a. Write the balanced equation for the reaction. b. Suppose 20.0 g of ZnS and 15.0 g of O 2 are mixed and the reaction allowed to take place. 1) What reactant will be in excess? How many moles of this reactant will be left? How many grams will be left? 2) How many grams of SO 2 will be formed? c. SO 2 reacts with CaO according to the equation: CaO + SO > CaSO 3 13
14 How many grams of ZnS would be required to produce enough SO 2 to react with 8.4 g of CaO? 23. Aluminum carbonate reacts with HBr according to the equation: Al 2 (CO 3 ) 3 (s) + 6 HBr(aq) 2 AlBr 3 (aq) + 3 H 2 O(l) + 3 CO 2 (g) Suppose that 5.84 g of Al 2 (CO 3 ) 3 reacts., a. how many grams of AlBr 3 (FM = 267) is formed? b. what volume of CO 2 (g) would be collected over H 2 O at 24 C and torr total pressure? pressure of H 2 O at 24 C = 22.4 torr) c. how many ml of a 3.0 M HBr solution reaction? 24. Suppose that iron reacted with H 2 S to give Fe 2 S 3 and H 2 according to the equation: 2 Fe(s) + 3 H 2 S(aq) -----> Fe 2 S 3 (s) + 3 H 2 (g) (The vapor would be needed in this When 8.40 g of Fe reacts, a. what volume of H 2 (g) could be collected over water at 25 C and Torr atmospheric pressure from the reaction? (vapor pressure of water at 25 C, P H2 O = 23.8Torr ) b. how many Liters of a 0.20 M H 2 S solution would be required in the c. How many grams of Fe 2 S 3 would be produced in the reaction? Answers b. 1) 0.30; 2)0.45 mol, 44.1 g; 3) Al 2 (SO 4 ) 3, 0.15 mol, 51.3 g; H 2, 0.45 mol, 0.90 g 20. b. 1) 5.0x10 3 mol; 2) 0.02 mol, 0.73 g; 3) Cl 2, 0.01 mol, 0.71 g; H 2 O, 0.01 mol, 0.18 g 21. b. 1) O 2 in excess, 0.05 mol or 1.6 g remain; 2) 52.8 g; 3) 27.0 g 22. b. 1) O 2 in excess, 0.16 mol or 5.2 g remain; 2) 13.1 g; c g 23. a g; b. 1.9 L; c. 50 ml 24. a L; b ml; c reaction? 14
15 15
16 16
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 informationChemistry 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 informationChemical 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
More informationMoles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
More informationPart 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 informationCalculating 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 informationStoichiometry. 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 informationHonors 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
More informationChapter 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 informationProblem 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.
More informationStoichiometry. 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
More informationMole 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
More informationTutorial 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
More informationStoichiometry 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 informationChapter 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 informationChapter 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 informationF321 MOLES. Example If 1 atom has a mass of 1.241 x 10-23 g 1 mole of atoms will have a mass of 1.241 x 10-23 g x 6.02 x 10 23 = 7.
Moles 1 MOLES The mole the standard unit of amount of a substance (mol) the number of particles in a mole is known as Avogadro s constant (N A ) Avogadro s constant has a value of 6.02 x 10 23 mol -1.
More informationChem 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 informationUnit 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 informationFormulas, 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 informationChapter 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 informationIB 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
More informationCalculations 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 informationAppendix 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
More informationOther 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
More informationAS1 MOLES. oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol -1
Moles 1 MOLES The mole the standard unit of amount of a substance the number of particles in a mole is known as Avogadro s constant (L) Avogadro s constant has a value of 6.023 x 10 23 mol -1. Example
More informationChapter 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 informationMoles. 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 informationName 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
More informationUnit 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
More informationChapter 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 informationChemical 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 informationAtomic 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 informationUnit 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?
More informationThe 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
More informationW1 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 informationFormulae, 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 informationneutrons 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
More informationChemical 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.
More informationChem 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 informationName 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.
More informationLiquid phase. Balance equation Moles A Stoic. coefficient. Aqueous phase
STOICHIOMETRY Objective The purpose of this exercise is to give you some practice on some Stoichiometry calculations. Discussion The molecular mass of a compound is the sum of the atomic masses of all
More informationBalance 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 informationSCH 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 informationChapter 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,
More informationUnit 2: Quantities in Chemistry
Mass, Moles, & Molar Mass Relative quantities of isotopes in a natural occurring element (%) E.g. Carbon has 2 isotopes C-12 and C-13. Of Carbon s two isotopes, there is 98.9% C-12 and 11.1% C-13. Find
More informationCalculation 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 informationConcept 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 informationstoichiometry = 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
More informationElement of same atomic number, but different atomic mass o Example: Hydrogen
Atomic mass: p + = protons; e - = electrons; n 0 = neutrons p + + n 0 = atomic mass o For carbon-12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine-35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass
More information1. 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?
More informationStoichiometry. 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
More informationStoichiometry. 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 informationChemistry: 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,
More information7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.
CHATER 3. The atmosphere is a homogeneous mixture (a solution) of gases.. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. have volumes that depend on their conditions,
More informationChapter 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 informationThe Mole. Chapter 10. Dimensional Analysis. The Mole. How much mass is in one atom of carbon-12? Molar Mass of Atoms 3/1/2015
The Mole Chapter 10 1 Objectives Use the mole and molar mass to make conversions among moles, mass, and number of particles Determine the percent composition of the components of a compound Calculate empirical
More information1. 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 informationName Date Class CHEMICAL QUANTITIES. SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296)
Name Date Class 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
More information1. 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 informationLimiting Reagent Worksheet #1
Limiting Reagent Worksheet #1 1. Given the following reaction: (Balance the equation first!) C 3 H 8 + O 2 -------> CO 2 + H 2 O a) If you start with 14.8 g of C 3 H 8 and 3.44 g of O 2, determine the
More informationChemistry 65 Chapter 6 THE MOLE CONCEPT
THE MOLE CONCEPT Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present. In order to relate the mass and number of
More information87 16 70 20 58 24 44 32 35 40 29 48 (a) graph Y versus X (b) graph Y versus 1/X
HOMEWORK 5A Barometer; Boyle s Law 1. The pressure of the first two gases below is determined with a manometer that is filled with mercury (density = 13.6 g/ml). The pressure of the last two gases below
More informationCP Chemistry Review for Stoichiometry Test
CP Chemistry Review for Stoichiometry Test Stoichiometry Problems (one given reactant): 1. Make sure you have a balanced chemical equation 2. Convert to moles of the known substance. (Use the periodic
More informationChapter 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
More informationThe Mole and Molar Mass
The Mole and Molar Mass 1 Molar mass is the mass of one mole of a substance. Molar mass is numerically equal to atomic mass, molecular mass, or formula mass. However the units of molar mass are g/mol.
More informationChapter 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= 1.038 atm. 760 mm Hg. = 0.989 atm. d. 767 torr = 767 mm Hg. = 1.01 atm
Chapter 13 Gases 1. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. Gases have volumes that depend on their conditions, and can be compressed or expanded by
More informationSample 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 informationCHEM 105 HOUR EXAM III 28-OCT-99. = -163 kj/mole determine H f 0 for Ni(CO) 4 (g) = -260 kj/mole determine H f 0 for Cr(CO) 6 (g)
CHEM 15 HOUR EXAM III 28-OCT-99 NAME (please print) 1. a. given: Ni (s) + 4 CO (g) = Ni(CO) 4 (g) H Rxn = -163 k/mole determine H f for Ni(CO) 4 (g) b. given: Cr (s) + 6 CO (g) = Cr(CO) 6 (g) H Rxn = -26
More informationCh. 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
More informationEXPERIMENT 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
More informationChem 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
More informationCHEMISTRY II FINAL EXAM REVIEW
Name Period CHEMISTRY II FINAL EXAM REVIEW Final Exam: approximately 75 multiple choice questions Ch 12: Stoichiometry Ch 5 & 6: Electron Configurations & Periodic Properties Ch 7 & 8: Bonding Ch 14: Gas
More informationChapter 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
More informationChapter 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 informationSample Problem: STOICHIOMETRY and percent yield calculations. How much H 2 O will be formed if 454 g of. decomposes? NH 4 NO 3 N 2 O + 2 H 2 O
STOICHIOMETRY and percent yield calculations 1 Steps for solving Stoichiometric Problems 2 Step 1 Write the balanced equation for the reaction. Step 2 Identify your known and unknown quantities. Step 3
More informationCHEMICAL 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 informationChapter 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 informationAPPENDIX B: EXERCISES
BUILDING CHEMISTRY LABORATORY SESSIONS APPENDIX B: EXERCISES Molecular mass, the mole, and mass percent Relative atomic and molecular mass Relative atomic mass (A r ) is a constant that expresses the ratio
More information2. The percent yield is the maximum amount of product that can be produced from the given amount of limiting reactant.
UNIT 6 stoichiometry practice test True/False Indicate whether the statement is true or false. moles F 1. The mole ratio is a comparison of how many grams of one substance are required to participate in
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Chapter 10 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A gas at a pressure of 10.0 Pa exerts a force of N on an area of 5.5 m2. A) 1.8 B) 0.55
More informationHow To Calculate Mass In Chemical Reactions
We have used the mole concept to calculate mass relationships in chemical formulas Molar mass of ethanol (C 2 H 5 OH)? Molar mass = 2 x 12.011 + 6 x 1.008 + 1 x15.999 = 46.069 g/mol Mass percentage of
More informationCHEM 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 informationCHEMICAL REACTIONS AND REACTING MASSES AND VOLUMES
CHEMICAL REACTIONS AND REACTING MASSES AND VOLUMES The meaning of stoichiometric coefficients: 2 H 2 (g) + O 2 (g) 2 H 2 O(l) number of reacting particles 2 molecules of hydrogen react with 1 molecule
More informationName 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
More informationChemical 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
More information4.1 Stoichiometry. 3 Basic Steps. 4. Stoichiometry. Stoichiometry. Butane Lighter 2C 4 H 10 + 13O 2 10H 2 O + 8CO 2
4. Stoichiometry 1. Stoichiometric Equations 2. Limiting Reagent Problems 3. Percent Yield 4. Limiting Reagent Problems 5. Concentrations of Solutes 6. Solution Stoichiometry 7. ph and Acid Base Titrations
More informationCalculations with Chemical Formulas and Equations
Chapter 3 Calculations with Chemical Formulas and Equations Concept Check 3.1 You have 1.5 moles of tricycles. a. How many moles of seats do you have? b. How many moles of tires do you have? c. How could
More informationThe Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10
Our Atmosphere The Gas Laws 99% N 2 and O 2 78% N 2 80 70 Nitrogen Chapter 10 21% O 2 1% CO 2 and the Noble Gases 60 50 40 Oxygen 30 20 10 0 Gas Carbon dioxide and Noble Gases Pressure Pressure = Force
More informationPART 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 informationExam 4 Practice Problems false false
Exam 4 Practice Problems 1 1. Which of the following statements is false? a. Condensed states have much higher densities than gases. b. Molecules are very far apart in gases and closer together in liquids
More informationMolar Mass Worksheet Answer Key
Molar Mass Worksheet Answer Key Calculate the molar masses of the following chemicals: 1) Cl 2 71 g/mol 2) KOH 56.1 g/mol 3) BeCl 2 80 g/mol 4) FeCl 3 162.3 g/mol 5) BF 3 67.8 g/mol 6) CCl 2 F 2 121 g/mol
More informationConcentration of a solution
Revision of calculations Stoichiometric calculations Osmotic pressure and osmolarity MUDr. Jan Pláteník, PhD Concentration of a solution mass concentration: grams of substance per litre of solution molar
More informationCHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS
CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS This chapter reviews the mole concept, balancing chemical equations, and stoichiometry. The topics covered in this chapter are: Atomic mass and average
More informationWriting 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
More informationBalancing 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.
More informationUnit 6 The Mole Concept
Chemistry Form 3 Page 62 Ms. R. Buttigieg Unit 6 The Mole Concept See Chemistry for You Chapter 28 pg. 352-363 See GCSE Chemistry Chapter 5 pg. 70-79 6.1 Relative atomic mass. The relative atomic mass
More informationStudy 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
More informationGas Laws. The kinetic theory of matter states that particles which make up all types of matter are in constant motion.
Name Period Gas Laws Kinetic energy is the energy of motion of molecules. Gas state of matter made up of tiny particles (atoms or molecules). Each atom or molecule is very far from other atoms or molecules.
More information