Download CHEM 1411 EXAM I (Chapters 1, 2, 3): 25

CHEM 1411 PRACTICE EXAM I (Chapters 1, 2, 3): 25 questions. Q1‐7: Chapter 1; Q8‐13: Chapter 2; Q14‐25: Chapter 3. Multiple Choice: Select one best answer. 1. Which of the following is a physical change? (a) color of carpet faded (by sun light) (b) water evaporates (d) wine turns sour (c) zinc strip dissolves in vinegar (to produce hydrogen gas) (e) scramble an egg Hint: For 11th ed,: p.p. 10‐11. For both 9th and 10th editions: p.p. 14‐15. Physical change is the change using the physical properties, which are the ones you can get them back by changing the temperature or pressure. The procedure used by adapting physical properties is called the physical process. Chemical properties are the ones you cannot get them back. 2. Which of the following is a heterogeneous mixture? (a) benzene and hexane (b) salt, water, and sugar (c) oil and alcohol (d) 14‐K gold ring (e) air Hint: For 11th ed.: p.p. 6‐7. For both 9the and 10th editions: p. 11. Heterogeneous mixture indicates there are two or more phases in the mixture/solution. If the mixture is made by solid and liquid, seeing the solid indicates it’s the heterogeneous mixture. If the mixture is made by liquid and liquid, seeing drops (by shaking it) or layer indicating it’s the heterogeneous mixture. 3. Which of the following is inaccurate? (a) silicon, Si (b) mercury, Hg (c) silver, Ag (d) chromium, Cr (e) iron, Ir Hint: For 11th, 9th and 10th editions: Check with front cover or periodic table and memorize them 4. What is the equivalent temperature for 98.6 Fahrenheit in Kelvin? (a) 37 (b) 310 (c) 471.6 (d) 594.8 (e) ‐273.1 Hint: For 11th ed.: p.p. 15‐17. For both 9th and 10th editions: p.p. 19‐20. Example 1.3. There is no direct conversion between the K and oF. So in this question, you must convert oF to oC first and then convert o
C to K by applying two formulas: one is oC = (5/9) x (oF – 32) and the other is K = 273.15 + oC. 5. An unknown sample has a mass of 13.9 g and a volume of 17.4 mL. What is its density (g/mL)? (a) 0.798 (b) 1.04 (c) 3.16 (d) 4.62 (e) 5.07 Hint: For 11th ed.: p.p. 14‐15. For both 9th and 10th editions: p. 18. Example 1.1. Density = mass/volume 6. How many significant figures are there in the measurement of 6.124 x 10‐5? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Hint: For 11th ed.: p.p. 19‐21, also Example 1.4(e). For both 9th and 10th editions: p.p. 23‐26. Memorize the rules. Example 1.4. 7. How many significant figures are appropriate to show in the result after carrying out the operation below? (223.7 + 0.27) ÷ 4.21 =? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 th
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Hint: For 11 ed.: p. 21. also p. 31: 1.35 (d) & 1.36(d). For both 9 and 10 editions: p. 25. Example 1.5. You need to memorize all the rules concerning the significant figures and the rules related to addition or subtraction and multiplication or division. Many students do not know the differences between addition/subtraction and multiplication/division. See two examples below: 1
Remember that the answers obtained from calculators are always WRONG when considering the significant figures. So when the question combining addition/subtraction and multiplication/division together, each follows its own rule. Thus, 223.7 + 0.27 = 223.97 from calculator, which must be corrected with the least decimal points of the components, and thus it is 224.0. So the question turns to 224.0 ÷ 4.21 = 53.20665083 from calculator, which must be corrected with the lease digit of the significant figures of the components, and thus it is 53.2. Since this question only asks how many significant figures in the result, thus from the components, 224.0 and 4.21, we know one has four significant figures and one has three significant figures. Thus the number of significant figures of the result should follow the one with the least significant figures, that is, three significant figures. ‐
8. How many protons, electrons, and neutrons are there in 72Br ? (a) 35p, 35e, 72n (b) 34p, 35e, 37n (c) 35p, 36e, 72n (d) 36p, 34e, 72n (e) 35p, 36e, 37n Hint: For 11th ed.: p.p. 46‐47, p.p. 50‐51 and p. 71: 2.73 and 2.83. For 9th edition: p. 53; 10th edition: p. 54. Also p. 50: Example 2.1 to calculate numbers of protons, neutrons, and electrons. For Positive charge indicates there are more protons than electrons; while the negative charge indicates there are more electrons than protons. This is because neutron does not carry charge, a proton carries one positive charge, and an electron carries one negative charge. The ion in this question carries one negative charge indicating that the electron is one more than the protons. From periodic table, the atomic number (that is, the proton number) of Br is 35, and thus there are 35‐(‐1) = 36 electrons. 9. Which of the following pairs would have similar chemical and physical properties? (a) Ni, Mg (b) H, Na (c) C, Si (d) Cu, Ca (e) Cr, Cl Hint: For 11th ed.: p.p. 48‐50. For 9th edition: p.p. 50‐51. For 10th edition: p. 51. The elements from the same group (i.e. the vertical column) will have similar physical and chemical properties. Be careful that H is a nonmetal while Na is a metal. 10. Which of the following is a molecular compound? (a) KCl (b) CsF (c) HCN (d) AlBr3 (e) NaOH Hint: For 11th ed.: p.p. 59‐63; For 9th and 10th editions: p. Starting 62. Molecular compound is composed by nonmetal reacts with nonmetal atoms; whle the ionic compound is composed by metal and nonmetal atoms. However, there is an exception: For compounds that contain ammonium ion, NH4+ , they are the ionic compounds as they are formed by acid, HCl, reacts with base, NH3. 11. Which of the following is the empirical formula? (b) CH3COOH (c) C4H10O2 (d) NH4NO3 (e) B2H6 (a) C8H18 Hint: For 11th ed.: p.p. 53‐55; 9th edition: p.p. 55‐57; 10th edition: p. 56‐57. Both editions: Example 2.3. Empirical formula is the simplest integral ratio among each atom. For (b), the molecular formula can be written as C2H4O2 and that for (d) as N2H4O3. 12. Which of the following is not a correct match? (b) H2O dihydrogen monoxide (a) AlCl3, aluminum chloride (d) HNO3, nitrous acid (c) CrF3, chromium (III) fluoride (e) CuSO4.5H2O, copper (II) sulfate pentahydrate Hint: For 11th ed.: p.p. 56‐64; for both 9th and 10th editions: p.p. 59‐68. Follow the rules of naming the compounds. Be able to differentiate the rules between ionic and molecular compounds. Prefix must be included in the name of molecular compound except the first element with subscript as one. For 2
ionic compound whose metal is a transition metal or with multiple charges must use Roman numeral to specify the charge on the metal ion. 13. Which of the following pairs is a correct match? (a) Ba3(PO4)2, barium (II) phosphate (b) (NH4)2SO4, diammonium sulfate (c) Na2O2, sodium oxide (e) CH3COOH, carbonic acid (d) Ca(NO3)2, calcium nitrate Hint: For 11th ed.: p.p. 56‐64; for both 9th and 10th editions: p.p. 59‐68. Follow the rules of naming the compounds. Be able to differentiate the rules between ionic and molecular compounds. 14. The atomic masses of 10B and 11B are 10.0129 amu (natural abundance 19.78%) and 11.0093 amu (natural abundance 80.22%), respectively. What is the average atomic mass of B? (a) 9.467 (b) 9.966 (c) 10.042 (c) 10.504 (e) 10.810 th
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Hint: For 11 ed.: p.p. 76‐77 and Example 3.1; for 9 edition: p.p. 78‐79. Example 3.1; 10th edition: p.p. 80‐81 Example 3.1. You must convert the % into decimals first and then apply the formula and memorize it: M = M1X1 + M2X2 = 10.0129x0.1978 + 11.0093x0.8022 = 10.810. Note that the average atomic mass (with decimal) can be found in the periodic table, while the mass number (a whole number) can not be found. For three isotopes: Example Exercise 5.3: 28Si (27.977 amu; 92.21% abundance), 29Si (28.976 amu; 4.70% abundance) and 30Si (29.974 amu; 3.09% abundance). The atomic mass M = M1X1 + M2X2 + M3X3 = 27.977x0.9221 + 28.976x0.0470 + 29.974x0.0309 = 28.09 amu by rounding to two decimals. 15. How many moles of S are there in 64.2 g of S? (a) 1.325 (b) 1.764 (c) 1.968 (d) 2.003 (e) 2.475 Hint: For 11th ed.: Example 3.2 and Example 3.6; for 9th edition: p. 81. Examples 3.2 and 3.6. For 10th edition: p.p. 82‐83: Example 3.2. Mole = mass (g) / molar mass (g/mole) = 64.2/32 = 2.003 16. What is the molar mass for calcium nitrate, Ca(NO3)2? (a) 44 (b) 56 (c) 87 (d) 93 (e) 164 Hint: For 11th ed.: p.p. 81‐82 and Example 3.5; for 9th edition: p.p. 83‐84. Example 3.5. For 10th edition: p.p. 85‐87 Example 3.5. In the formula, Ca(NO3)2, it indicates there are 1 Ca, 1x2 = 2 N and 3x2 = 6 O. Go to the periodic table and locate the atomic mass for each atom. Thus the formula mass of Ca(NO3)2 = 1x40 + 2x14 + 6x16 = 164 17. How many hydrogen atoms are there in 48.0 g of CH4? (b) 7.22x1024 (c) 6.02x1023 (d) 1.20x1025 (e) 4.70x1025 (a) 1.81x1023 th
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Hint: For 11 ed.: p. 83 and Example 3.7; for 9 edition: p. 85 and Example 3.7. For 10th edition: p. 87 and Example 3.7. According to the chemical formula, one mole of CH4 contains 1 mole of C atoms and 4 moles of hydrogen atoms. Thus, the mole of H = 4 x {mass of CH4/molar mass of CH4}. When converting moles into atoms, it needs to multiply the Avogadro’s number, that is, 6.02x1023. Thus the number of H atoms = moles of H atoms x Avogadro’s number = (6.02x1023)x{4x(48.0/16.0)} = 7.22x1024 18. What is the mass percent (%) for O in SO2? (a) 38.09 (b) 45.41 (c) 50.00 (d) 53.86 (e) 56.43 Hint: For 11th ed.: p.p. 85‐86 and Example 3.8; for 9th edition: p. 87: Equation (3.1); for 10th edition: p.p. 88‐89: Example 3.8. % S = (mass of 2 O/ mass of SO2) x 100% = {2x16/(1x32+2x16)}x100% = 50.0% 3
19. Chemical analysis shows the composition of a compound containing carbon, hydrogen, chlorine, and oxygen, to be 37.84% C, 2.12% H, 55.84% Cl, and 4.20% O. What is its empirical formula? (c) C12H8Cl6O (d) C12H8Cl6O4 (a) CHClO (b) C2HClO4 Hint: For 11th ed.: p.p. 86‐89: Examples 3.9 and 3.11. For 9th edition: p. 88. Example 3.9 (Vitamin C): very important or the most difficult for empirical formula calculation. Figure 3.5. or p. 109 (3.50). For 10th edition: p.p. 92‐94: Example 3.9 (p. 90). The empirical formula is the simplest integral ratio of moles among each atom. Here, there are four different kinds of atoms, C, H, Cl and O. Thus mole of C = 37.84/12 = 3.15; mole of H = 2.12/1 = 2.12; mole of Cl = 55.84/35.45 = 1.58; mole of O = 4.20/16 = 0.26. Note that as long as one of the moles is not an integer, we have to divide the smallest value among them: here the smallest value is 0.26. So C : H : Cl : O = 3.15/0.26 : 2.12/0.26 : 1.58/0.26 : 0.26/0.26 = 12.1: 8.1 : 6.1: 1 = 12: 8: 6: 1, which indicates that the empirical formula contains 12 C, 8 H, 6 Cl and 1 O. Thus, the empirical formula is written as C12H8Cl6O as 1 is usually not written in the formula. *****What is the empirical formula for methyl benzoate, a compound used in the manufacture of perfumes, contains 70.57% carbon, 5.94% hydrogen, and 23.49% oxygen? Note: subscripts must be integers. B) C2H2O0.5 C) C8H8O2 D) CHO A) C4H4O Hint: Section 9.8. This is a very, very important question. The empirical formula is the simplest integral ratio of moles among each atom. Here, there are three different kinds of atoms, C, H and O. Thus mole of C = 70.57/12 = 5.88; mole of H = 5.94/1 = 5.94; mole of O = 23.49/16 = 1.47. Note that as long as one of the moles is not an integer, we have to divide the smallest value among them: here the smallest value is 1.47. So C : H : O = 5.88/1.47 : 5.94/1.47 : 1.47/1.47 = 4 : 4.04 : 1. Since 4.04 is very close to 4.00 and thus we can round it to 4.00. So C : H : O = 4 : 4 : 1, which indicates that the empirical formula contains four C, four H and one O. Thus, the empirical formula is written as C4H4O as 1 is usually not written in the formula. 20. Chemical analysis shows the composition of a compound containing carbon and hydrogen, to be 80.00% carbon and 20% hydrogen and the molar mass is 30 g. What is its molecular formula? (c) C2H6 (d) C6H12 (e) C10H22 (a) CH (b) C2H4 th
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Hint: For 11 ed.: p.p. 89‐90 and Example 3.11; for 9 edition: p.p. 91‐92 and Example 3.11 or p. 109 (3.54). For 10th edition: p. 93: Example 3.11. The molecular formula is an integral multiple of empirical formula. That is, the molar mass = empirical molar mass x integer. From C: H = 80.00/12 : 20.00/1 = 6.66: 20 = 1: 3. So the empirical formula is CH3 and the empirical molar mass of CH3 = 12x1+1x3 =15. So the integer = 30/15 = 2. Thus there are two empirical formulas in a molecular formula. Therefore, the molecular formula is C2H6. 21. What are the coefficients respectively when the equation __PH3 + __O2  __P2O5 + __H2O is balanced? (a) 2, 2, 1, 3 (b) 2, 1, 3, 4 (c) 2, 3, 1, 2 (d) 2, 4, 1, 3 (e) 1, 1, 3, 3 Hint: For 11th ed.: 90‐95 and Example 3.12; For 9th edition: p.p. 94‐97. Example 3.12 or p. 109 (3.60). 10th edition: p.p. 94‐99: Example 3.12. Balancing an equation is a process of trial an error. Sometimes it requires more than one trial. Usually start visual examination and select the most bulky species, that is, the one with the most different kinds of atoms and number of atoms as the reference and set its coefficient as one. In this question, P2O5 is the most bulky one, we put 1 in front of it to remind us we have done examining P2O5. Now the equation is updated to __PH3 + __O2  1 P2O5 + __H2O 4
Since P2O5 contains 2 phosphorus atoms, so we need two phosphorus atoms at the left side, which leads us to put 2 (called coefficient) in front of the PH3. Now the equation is updated to 2 PH3 + __O2  1 P2O5 + __H2O As there are 6 hydrogen atoms in 2 PH3, thus we need to balance the hydrogen atoms at the right side, which lead us to put 3 in front of the H2O. Now the equation is updated to 2 PH3 + __O2  1 P2O5 + 3 H2O Now we need to balance the oxygen atoms. Since there are 1x5+3x1 = 8 oxygen atoms at the right side, and thus the left side must have the same number of oxygen atoms. That is to say, __x2 = 8. So __ = 4. Now the equation is updated to 2 PH3 + 4 O2  1 P2O5 + 3 H2O Since all the atoms of each type has the same amount, this equation is balanced. 22. In the reaction of Al(OH)3 with H2SO4, how many moles of water can be produced If the reaction is begun with 5.500 mole of Al(OH)3? 2Al(OH)3 + 3H2SO4  Al2(SO4)3 + 6H2O (a) 2.50 (b) 4.75 (c) 6.32 (d) 7.58 (e) 16.50 Hint: For 11th ed.: p.p. 95‐97 and Examples 3.13 and 3.14 (application of Figure 3.8); for 9th edition: p.p. 97‐101. Examples 3.13. and 3.14. Figure 3.8. For 10th edition: p.p.99‐103: Examples 3.13. &3.14. Stoichiometrey. Method One (ratio approach): From the equation, the involving species, 2 Al(OH)3 and 6 H2O with coefficients 2 and 6 respectively, tell us that for 2 moles of Al(OH)3 it produces 6 moles of H2O. According to this proportion or ratio, 5.500 mole of Al(OH)3 requires 5.500 x (6/2) = 16.50 moles of H2O. Method Two (road map approach; figure 3.8): Apply the road map or say the dimensional analysis. The road map is grams of substance A  moles of substance A moles of substance B  grams of substance B. Note: (1) substance A is the one with given (or known) information of mass (or mole); substance B is the one needed to be calculated. (2) grams of substance A  molar mass of substance A = moles of substance A. (3) moles of substance A  coefficient of substance B  coefficient of substance A = moles of substance B. (4) grams of substance B = moles of substance B  molar mass of substance B. Since this question starts at moles and thus we only apply (2): 5.500 moles Al(OH)3 x {6 H2O/2 Al(OH)3} = 16.50 moles H2O 23. How many grams of H2O could be formed by the reaction of 16.0 g of CH4 with 48.0 g of O2? CH4 + 2O2  CO2 + 2H2O (a) 27.0 (b) 37.3 (c) 46.8 (d) 54.1 (e) 58.7 Hint: For 11th ed.: p.p. 99‐102 and Example 3.15; for 9th edition: p.p. 97‐101. Examples 3.13. and 3.14. Figure 3.8. 10th edition: p.p. 99‐103. Examples 3.13 & 3.14. Stoichiometrey. This question provides two known quantities of substance A and thus this is the limiting reagent question. So we need to apply two times of all‐four‐step‐road map to figure out what is the true mass of water. Note that the limiting reagent limits the (maximum) of product(s) that can be produced. In the road map 5
application, the one produces the smallest amount of the product is the true limiting reagent. The one with the left over is the excess reagent. Assume CH4 (i. e. substance A # 1) is the limiting reagent: x 18 g/mol ÷ 16 g/mol x 2H2O/1CH4 16 g CH4 ‐‐‐‐‐‐‐‐‐‐‐‐‐ 1 mole CH4 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 2 mole H2O ‐‐‐‐‐‐‐‐‐‐‐‐‐ 36 g H2O Assume O2 is the limiting reagent: x 18 g/mol ÷ 32 g/mol x 2H2O/2O2 48 g O2 ‐‐‐‐‐‐‐‐‐‐‐‐‐ 1.5 mole O2 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 1.5 mole H2O‐‐‐‐‐‐‐‐‐‐‐‐‐ 27 g H2O Because O2 produces the least amount of H2O (can compare by using mole, 1.5 < 2, or grams, 27 < 36), it is the true limiting reagent and the maximum amount of water produced is 27 grams. The CH4 is the excess reagent. 24. What is the excess reagent in the above reaction (Q.23)of CH4 and O2? How many grams of the excess reagent were consumed? (b) O2, 12 grams (c) CH4, 4 grams (a) CH4, 12 grams (e) CH4, 8 grams (d) O2, 4 grams Hint: 9th edition: p.p. 101‐103. Example 3.15. 10th edition: p.p. 103‐106: Example 3.15. Now the substance A is O2 and CH4 is the substance B because we need to calculate its amount consumed stoichiometrically. Apply the road map: x 16 g/mol ÷ 32 g/mol x 1CH4/2O2 48 g O2 ‐‐‐‐‐‐‐‐‐‐‐‐‐ 1.5 mole O2 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 0.75 mole CH4‐‐‐‐‐‐‐‐‐‐‐‐‐ 12 g CH4 Thus, there are (16 – 12) = 4 grams of CH4 left. This is because # grams excess = # grams original – # grams used. Or say # moles excess = # moles original – # moles used. 25. A 15.6 grams of benzene (C6H6) is mixed with excess HNO3 to prepare nitrobenzene (C6H5NO2). After the reaction there are 15.6 grams of nitrobenzene produced. What is the percent yield of nitrobenzene? C6H6 + HNO3  C6H5NO2 + H2O (a) 34.3% (b) 47.6% (c) 58.9% (d) 63.4% (e) 71.2% Hint: For 11th ed.: p.p. 103‐104; for 9th edition: p.p. 103‐106. Example 3.16. 10th edition: p.p. 106‐107: Example 3.16. The amount of product calculated according to the road map is the theoretical yield and the one obtained by weighing is the actual yield. % yield = (actual yield/ theoretical yield) x 100% Now from the road map, we calculate the theoretical yield (note that since the actual yield given is in unit of grams, so we need to complete the entire road map). x 123g/mol ÷ 78 g/mol x 1C6H5NO2 /1C6H6 15.6 gC6H6‐‐‐‐‐‐‐‐‐‐‐‐ 0.2mole C6H6 ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 0.2mole C6H5NO2 ‐‐‐‐‐‐‐‐‐ 24.6g C6H5NO2 Thus the percent yield = {15.6 g / 24.6 g} x 100% = 63.41%
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