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Home , Isotopes of magnesium, Nonmetal

103 – Help Sheet #2 Atoms, , and Do the topics appropriate for your course Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resources page) Nuggets: Atomic structure; ; Counting neutrons, , electrons; Average ; Mass Spectroscopy (Part I); Ionic vs molecular compounds; Naming chemicals; Writing formulas from names; Polyatomic ions; Coulomb’s law

ATOMIC STRUCTURE Rutherford's Au Foil Experiment: Mass of atom is concentrated in a highly dense nucleus that was positively charged; the rest of the atom has only a small amount of mass; alpha particles are positively charged so they are repelled by the nucleus comprised of protons (positively charged) and are small compared to the size of the atom allowing them to pass through the atom

Nucleus: contains nearly all the atom’s mass; neutrons and protons located in the nucleus and the nucleus is a very small volume of the atom; extremely dense; atom is mostly empty space Elements: determined by the number of protons they contain Masses of all atoms are compared to 12C which is the standard reference; 12C is defined as 12.0000... atomic mass units (exact value) = 12.0000...amu; (amu = atomic mass units) 1 23 -24 1 amu = /6.022 x 10 = 1.6605 x 10 g Neutron (n): has a neutral charge; mass ≈ 1amu ≈ 1 x 10-24g (1.6749 x 10-24g) (p+): has a charge of +1; mass ≈ 1amu ≈ 1 x 10-24g (1.6726 x 10-24g) Electron (e-): has a charge of -1; mass ≈ 0.0005amu ≈ 1 x 10-27g (9.1094 x 10-28g) Relative masses: n > p+ >> e-; mass of n ≈ p+ ≈ 1 with e- ≈ 1/2000

ISOTOPES: elements with the same #p+ (same element) but different #n (different mass) A + + Symbol: ZElement Symbol A = mass number = #p + #n; Z = atomic number = #p 12 13 14 + C isotopes: 6C , 6C , and 6C; the 3 isotopes of C each have Z = 6 (6p ) but different #n (6, 7, and 8) Example 1: What is mass of one atom of 16O-2 in kg? ⎛ −24 ⎞ -26 16 -2 16 -2 16 −2 1.6605 x 10 g ⎛ 1kg ⎞ −26 Answer 1:€ 2.7 x 10 kg { O = 16amu; mass of 1 atom O = 16u; O = 16amu⎜ ⎟ ⎜ ⎟ = 2.6568 x 10 kg } € € € ⎝⎜ 1amu ⎠⎟ ⎝ 1000g⎠

AVERAGE ATOMIC MASS (AAM): weighted average of all isotopes of one element taking into account the abundance of each isotope; units = amu; e.g., average atomic mass Br = 79.90amu Abundance: percent of one isotope as compared to all atoms of that given element AVERAGE ATOMIC MASS (AAM) = S(mass of isotope) x (fractional abundance) AAM = (massiso1)(fractional abundanceiso1) + (massiso2)(fractional abundanceiso2) + … = AAM = (M1)(FA1) + (M2)(FA2) + … Sum of fractional abundances = 1 (i.e., 100%); for elements with only 2 isotopes: FA1 + FA2 = 1

Example 2: Calculate the average atomic mass of an element given following data: isotope 1: 27.977amu, 92.21% abundance; isotope 2: 28.976amu, 4.70% abundance; and isotope 3: 29.974amu, 3.09% abundance. Answer 2: 28.09amu {AAM = (M1)(FA1) + (M2)(FA2) + (M3)(FA3) = (27.977)(0.9221) + (28.976)(0.0470) + (29.974)(0.0309) = 28.09amu}

Example 3: What are the abundances of the 2 B isotopes (10B mass = 10.0129u, 11B mass = 11.0093u), if the average atomic mass of B is 10.811amu? Answer 3: 10B = 19.90% and 11B = 80.10% AAM = (M1)(FA1) + (M2)(FA2); 10.811 = 10.0129x + 11.0093y; since there are 2 unknowns 2 equations are needed; second equation: FA1 + FA2 = 1 (which is 100%); x + y = 1 ® y = 1 - x; substitute this into the first equation: 10.811 = 10.0129x + 11.0093(1-x); solve for x: 10.811 = 10.0129x +11.0093 – 11.0093x; 0.1983 = 0.9964x; x = 0.1990 = 19.90%; y = 1-x = 1-0.1990 = 0.8010 = 80.10%; since x was the abundance for 10B ® abundance of 10B = 19.90% and abundance of 11B = y = 80.10%} MASS SPECTROSCOPY: single atoms experimental technique that yields a plot called a mass spectrum of mass to charge ratio, (m/z), versus abundance (or intensity); see the class video M2Q3 for a wonderful description of a mass spectrometer Example 4: Mg has 3 isotopes: 24Mg (79% abundance), 25Mg (10% abundance), 26Mg (11% abundance). Draw the mass spectrum for this element?

79% Mg

10%

orIntensity Abundance 11%

10 20 30 40 24 25 26 Answer 4: m/z

MASS SPECTROSCOPY: compounds (Not all instructors will be covering mass spectroscopy of compounds; instructors will be covering mass spectroscopy of single atom elements or molecules like Br2, Cl2, etc., but some sections may not cover compounds like C2H6O.) Hard Mass Spectroscopy: This technique breaks the into individual atoms Soft Mass Spectroscopy: This technique ionizes the molecule and the one peak shown represents the molar mass of the molecule (molar mass is the sum of the average atomic mass of the individual elements within the molecule; e.g., for C3H6O1 the molar mass = 3(mass of C) + 6(mass of H) + 1(mass of O) = 3(12) + 6(1) +1(16) = 58g/mol; this will be revisited in HelpSheet #4!) Empirical formula: the simplest formula that shows the ratio between atoms within the compound (this will be revisited in HelpSheet #4!) Molecular formula: the actual formula of the compound; (e.g., molecular formula = N2O4 while the empirical formula = NO2; i.e., the simplest ratio between N and O atoms in N2O4) Example 5: A was analyzed by mass spectroscopy and both “hard” and “soft” mass spectra were obtained. a. What is the empirical formula for the unknown compound? b. What is the molecular formula for the unknown compound?

Hard MS Soft MS 100% 60.0%

30.0%

Intensity orIntensity Abundance 10.0% orIntensity Abundance

1 12 16 116 m/z m/z Answer 5: a. empirical formula = C3H6O1 {from the hard mass spectrum (Hard MS) it can be determined that the compound contains H (mass = 1), C (mass = 12), and O (mass = 16). From the percent intensity or abundance the number of each element can be determined by dividing each element by the smallest percent; 10/0% in this example. 60% 30% 10% H = = 6 ; C = = 3 ; O = = 1 ; this yields an empirical formula = C H O } 10% 10% 10% 3 6 1 b. molecular formula = C6H12O2 {from the soft mass spectrum (Soft MS) the peak represents the compound’s molar mass; molar mass = 116g/mol; the empirical formula (C3H6O1) molar mass = 58g/mol [3(mass of C) + 6(mass of H) + 1(mass of O) = 3(12) + 6(1) +1(16) = 58g/mol] molar massmolecular formula 116g / mol to determine the molecular formula, determine ratio = ; ratio = = 2 ; molar massempiricalformula 58g / mol use the ratio just determined and multiply the empirical formula by 2: C(3 x 2)H(6 x 2)O(1 x 2) = C6H12O2}

IONIC COMPOUNDS: contain + (can substitute a polyatomic for either or both) Properties: extended ; not discrete molecules; high melting/boiling points; conducts electricity in molten

state; Examples: NaCl(s), Ca(NO3)2(s), NH4NO3(s), AgCl(s)

Name Formula - Ammonium + Ions: atoms or molecules that have lost/gained e (charges are not changed NH4 Hydronium + by gaining/losing protons!) H3O Cations: positively charged atoms (electrons have been lost) Hydroxide OH- - Anions: negatively charged atoms (electrons have been gained) CH3COO Polyatomic Ions: “many-atom ions”; these compounds are not broken up Nitrate - NO3 and are treated as a : Permanganate - MnO4 Carbonate - HCO3 Charges on elements in Ionic Compounds based on the column of Periodic Carbonate -2 Table the element resides in: CO3 Hydrogen Sulfate HSO - Group IA: +1 (H+, Li+, ...) Group IIA: +2 (Be+2, Mg+2, ...) 4 Sulfate -2 +3 +3 SO4 Group IIIA: +3 (B , Al , ...) Phosphate -3 -3 -3 -2 -2 PO4 Group VA: -3 (N , P , ...) Group VIA: -2 (O , S , ...) Group VIIA: -1 (F-, Cl-, ...)

Naming • Metal + Nonmetal ® ionic compound; metal name followed by nonmetal “root”+"ide" Example 6: What is the name for CaCl2? ® Answer 6: chloride (note: not dichloride!) • Contains one or two Polyatomic Ions ® ionic compound; use name Example 7: What is the name for Ca3(PO4)2? ® Answer 7: calcium phosphate (note: not tricalcium nor diphosphate!) Example 8: What is the name for (NH4)2SO4? ® Answer 8: ammonium sulfate (note: not diammonium!) • Contains ® ionic compound; metal name (charge in Roman numerals) Example 9: What is the name for Fe2(SO4)3? ® Answer 9: (III) sulfate (see below for process to follow) 1. Transition have charges that vary so the metal charge needs to be determined and specified in the name; the charge is written in Roman numerals in parentheses after the metal name -2 2. Find charge on Fe. First, SO4 charge is -2 (memorized); total negative charge: 3(-2) = -6 from the 3 SO4 ; since the compound is neutral the 2 Fe must total +6 ® therefore, each Fe = +3 = Fe+3; the Roman numeral for 3 = III

IONIC COMPOUNDS (continued): Writing Formulas 1. Take metal name or polyatomic ion and write formula including charge • If metal has a Roman numeral, then charge is determined from that numeral (e.g., (II) = Cu+2) + • If it does not have a Roman numeral, it is a memorized polyatomic ion (e.g., ammonium = NH4 ) or determined from the column in the it resides in (e.g., = Mg+2 since it is in column IIA) 2. Take nonmetal name or polyatomic ion and write formula including charge • If nonmetal name ends with “ide” it is a single atom ion with a negative charge with the charge determined from column in Periodic Table it resides in (e.g., chloride = Cl- since column VIIA has a -1 charge); exceptions: hydroxide (OH-) and cyanide (CN-) are polyatomic ions with a name that ends in “ide” • If name ends with “ite” or “ate”, or starts with “hypo” or “per” it is a memorized polyatomic ion 3. Balance charges within formula Example 10: Write formula for (III) carbonate. +3 -2 Answer 10: charge on Cr is ® Cr from Roman numeral; carbonate = CO3 (memorized) ® Crx(CO3)y; choose x and y to balance the charge: x = 2 ® [2(+3) = +6] and y = 3 ® [3(-2) = -6]; +6 balances -6 ® Cr2(CO3)3

MOLECULAR COMPOUNDS: contain 2 nonmetal elements; Properties: low melting/boiling points; discrete molecules; do not conduct electricity Examples: CO(g), CO2(g), N2O4(g)

Naming: Prefix (if greater than 1)+1st nonmetal name & Prefix(always included)+2nd nonmetal root+"ide” Prefixes: mono (1), di (2), tri (3), tetra (4), penta (5), hexa (6), hepta (7), octa (8), nona (9), deca (10) Example 11: What is the name for NO? (Has 2 ® molecular compound) Answer 11: Has one N ® no prefix ® ; has one O ® prefix always used = mono; ® answer: nitrogen monoxide Example 12: What is the name for N2O? (Has 2 nonmetals ® molecular compound) Answer 12: Has two N ® use prefix = di ® dinitrogen; has one O ® prefix always used = mono; monoxide ® dinitrogen monoxide

Writing Formulas - translate name with prefixes Example 13: What is the formula for disulfur trioxide? Answer 13: disulfur ® 2 S atoms; trioxide ® 3 O atoms; answer: S2O3

ACIDS (not emphasized) -2 -2 Naming – chemicals named as are neutral compounds (e.g., SO4 and HSO4 are not named as acids but H2SO4 is named as an ) Acids that contain anions ending in “ide”: change “ide” to “ic”, add “hydro” in front and “acid” at end; e.g., Cl- = chloride; HCl = hydrochloric acid -2 “ate”: change “ate” to “ic” and add “acid” at the end; e.g., SO4 = sulfate; H2SO4 = sulfuric acid Common acids: Strong acids (best to memorize): HCl (hydrochloric acid), HBr (hydrobromic acid), HI (hydroiodic acid), HNO3 (nitric acid), H2SO4 (sulfuric acid), HClO4 (perchloric acid); Weak acids: HF (), HCN (hydrocyanic acid), H3PO4 (phosphoric acid; memorize), CH3COOH (acetic acid; memorize), H2CO3 (carbonic acid; memorize)

COULOMB’S LAW: Describes the attractive force between a positive (cation) and negative (anion) ion (Q1)(Q2) Force of Attraction = k where Q1, Q2 are ionic charges; d is distance between the nuclei centers d2 (Q )(Q ) Potential Energy = U = Force x distance; Potential Energy = U = k 1 2 d ionic charges (Q) ­ -- or -- ionic radii ¯ ® d ¯ Þ Force or Energy ­ Þ mp ­ (ionic radii ¯ as you go up a column); Note: melting pointmolecular compounds << melting pointionic compounds

Solubility: Why do ionic compounds dissolve? There is a balance of two energies: Lattice Energy: The energy of attraction between the ions within an ionic compound keeps the ionic compound as a . Solvation Energy: The energy between the ions and the solvent molecules (usually ). Soluble (dissolves): solvation energy > lattice energy Insoluble (doesn’t dissolve): lattice energy > solvation energy

1. Determine the number of protons, neutrons, and electrons present in each of the following atoms. 53 55 9 40 3 a. 24 Cr b. 25 Mn c. 4 Be d. 20 Ca e. 1H

+ - 28 +3 41 +2 59 +2 36 − 2. How many p , n, and e are in each of the following ions? a. 13Al b. 20 Ca c. 28 Ni d. 17 Cl

3. Use the following list of isotopes to answer the questions. I. 60 Mn+2 II. 62 Co+2 III. 40 Ca+2 IV. 40 S−2 V. 58 Mn+5 a. Which of these isotopes have the same number of protons? b. Which of these isotopes have the same number of neutrons? c. Which of these isotopes have the same number of electrons? d. Which of these isotopes have the same charge? e. Which of these isotopes have the same mass number?

4. State whether the statement is true or false. a. The number of neutrons distinguishes from nitrogen. b. Elements are neutral because they have equal numbers of protons and electrons. c. Elements that have different numbers of neutrons are called isotopes.

5. Write the complete isotopic symbol with mass and atomic number for each question. a. a neutral atom with 28 protons and 30 neutrons b. a neutral atom contains 22 protons and 21 neutrons c. an atom with 10 neutrons and the same number of electrons as protons d. chromium atom with a mass number of 54 and no charge e. a with 35 protons, 35 electrons, and 44 neutrons f. an metal with the fewest protons and 4 neutrons and no charge g. an iron atom with mass number of 56 and 3 more protons than electrons h. sulfide ion with 17 neutrons

6. Which of the following statements is false? a. All atoms contain neutrons. b. All atoms contain protons. c. The mass of the neutron is greater than the mass of the proton? d. The electrons form a cloud around the nucleus. e. Atoms can easily gain or lose electrons but cannot easily gain or lose neutrons.

7. Describe the Rutherford experiment and include a diagram.

18 17 16 8. Oxygen exists as three isotopes: 8 O , 8 O , and 8 O . If the average atomic weight of oxygen is 15.9994, 18 what do you expect the abundance of 8 O to be approximately? (Hint: No calculation needed.) <5% 25% 50% 75% 90%

9. Calcium chloride dissolves in water; draw a picture of the ions and water in and their orientation to one another.

10. I. What is mass of one atom of 18O in kg? II. What is mass of one atom of 52Cr in kg?

11. An atom has a mass of 4.48 x 10-26kg. Which isotope is it? a. 27Al b. 14N c. 59Co d. 19F e. 23Na

12. An element has two naturally occurring isotopes with the following abundances and masses: Isotope Atomic Mass (amu) Fractional Abundance 1 84.912 0.7215 2 86.909 0.2785 What is the average atomic mass of this element? What is the identity of the element?

13. is found commonly as two isotopes: 107Ag (mass = 106.91amu) and 109Ag (mass = 108.90amu). The average atomic mass of silver is 107.869amu. What are the percent abundances of the two isotopes, 107Ag and 109Ag?

14. There are 3 isotopes of magnesium: 24Mg (mass = 23.985amu), 25Mg (mass = 24.986amu), and 26Mg (mass = 25.983amu). If the abundance of 25Mg is 10.00% what are the percent abundances of the other two isotopes of magnesium? 73.7%

Intensity orIntensity Abundance 8.3% 7.4% 5.4% 5.2%

46 47 48 49 50 15. Shown is the mass spectrum of Ti. What is the average atomic mass of Ti? m/z

35 37 16. , Cl2, exists as a diatomic gas and is found as two isotopes: Cl and Cl with approximate abundances: 35Cl = 75% and 37Cl = 25%. A sample of chlorine gas was run on a mass spectrometer and analyzed. a. Draw the expected hard mass spectrum for Cl2. Include on the x-axis mass values, on the y-axis the peak intensity, and label the axes. Label the percent intensities for each peak drawn. Assume the isotopic mass number is the mass of that isotope (i.e., 35Cl has a mass of 35amu). b. Draw the expected soft mass spectrum for Cl2. Again, include values on the axes, label the axes, and label the percent intensities for each peak drawn.

17. Which of the following will have the lowest ? (Note: Ionic radii increase as you go down a column in the Periodic Table) a. MgF2 b. LiF c. RbF d. BaF2

18. Which of the following will have the highest melting point? (Note: Ionic radii increase as you go down a column in the Periodic Table) a. BeO b. NH3 c. MgS d. KBr

19. Name the following molecules. a. NO b. PCl3 c. KBr d. Na2CO3 e. N2O4 f. K2SO4 g. Ca3(PO4)2 h. Cu2O i. Mn3(PO4)2 j. Co(CH3COO)3 k. CoSO4 l. KMnO4 m. CH4

20. Write the chemical formula for the following names. a. trioxide b. calcium fluoride c. chlorine monobromide d. sulfate e. ammonium nitrate f. carbonate g. calcium nitrate h. (III) sulfate i. chromium(VI) j. (IV) sulfide k. calcium acetate

21. Which of the following when placed in an ionic compound, will commonly have a charge of –2? I. Cl II. O III. SO4 IV. Ca V. PO4 VI. NO3 a. II b. II and III c. II and IV d. II, III, and IV e. II and VI

(For more practice naming/writing polyatomic ion formulas see Practice Sheet #3 or for more practice naming/writing formulas for compounds Practice Sheet #4.

Mass Spectroscopy of Compounds (skip if not covered) 22. A chemical sample was analyzed by mass spectroscopy and both a “hard” and “soft” mass spectra were obtained. a. What is the empirical formula for the unknown compound? b. What is the molecular formula for the unknown compound?

66.6% 100%

22.2%

Intensity orIntensity Abundance 11.1% orIntensity Abundance

1 12 14 88 m/z m/z Hard Mass Spectrum Soft Mass Spectrum

23. A chemical sample was analyzed by mass spectroscopy and both a “hard” and “soft” mass spectra were obtained. a. What is the empirical formula for the unknown compound? b. What is the molecular formula for the unknown compound?

42.8% 100%

28.6% 28.6% Intensity orIntensity Abundance orIntensity Abundance

1 12 16 148 m/z m/z Hard Mass Spectrum Soft Mass Spectrum

24. Draw the “hard” and “soft” mass spectra for ethanol, C2H6O. Label the axes and explain any reasoning needed. Assume the dominant H isotope is 1H, dominant C isotope is 12C, and dominant O isotope is 16O.

35 25. A compound, C2H5Cl, was analyzed by mass spectroscopy. There are 2 : Cl (75% abundance) and 37Cl (25% abundance). Assume the dominant isotope of C is 12C and 1H for H. a. Draw the soft mass spectrum for this compound. b. Draw the hard mass spectrum for this compound.

ANSWERS 1. a. 24 p+, 24 e-, 29 n {bottom number = atomic number = #p+ = 24; top number = mass number = #n + #p+; 53 = #n + 24; #n = 29; #e = #p+ because the isotope is neutral (charge = 0); #e- = 24} b. 25 p+, 25 e-, 30 n {bottom number = atomic number = #p+ = 25; top number = mass number = #n + #p+; 55 = #n + 25; #n = 30; #e = #p+ because the isotope is neutral (charge = 0); #e- = 25} c. 4 p+, 4 e-, 5 n {bottom number = atomic number = #p+ = 4; top number = mass number = #n + #p+; 9 = #n + 4; #n = 5; #e = #p+ because the isotope is neutral (charge = 0); #e- = 4} d. 20 p+, 20 e-, 20 n {bottom number = atomic number = #p+ = 20; top number = mass number = #n + #p+; 40 = #n + 20; #n = 20; #e = #p+ because the isotope is neutral (charge = 0); #e- = 20} e. 1 p+, 1 e-, 2 n {bottom number = atomic number = #p+ = 1; top number = mass number = #n + #p+; 3 = #n + 1; #n = 2; #e = #p+ because the isotope is neutral (charge = 0); #e- = 1}

2. a. 13 p+, 10 e-, 15 n {bottom number = atomic number = #p+ = 13; top number = mass number = #n + #p+; 28 = #n + 13; #n = 15; charge = +3; +3 = #p(+1) + #e(-1); 3 = 13 – x; x = #e = 10} b. 20 p+, 18 e-, 21 n {bottom number = atomic number = #p+ = 20; top number = mass number = #n + #p+; 41 = #n + 20; #n = 21; charge = +2; +2 = #p(+1) + #e(-1); 2 = 20 – x; x = #e = 18} c. 28 p+, 26 e-, 31 n {bottom number = atomic number = #p+ = 28; top number = mass number = #n + #p+; 59 = #n + 28; #n = 31; charge = +2; +2 = #p(+1) + #e(-1); 2 = 28 – x; x = #e = 26} d. 17 p+, 18 e-, 19 n {bottom number = atomic number = #p+ = 17; top number = mass number = #n + #p+; 36 = #n + 17; #n = 19; charge = -1; -1 = #p(+1) + #e(-1); -1 = 17 – x; x = #e = 18}

3. a. I and V {for this question, same element means same #p+, hence, choices I and V; for the rest of the questions it is best to determine the number of p+, n, e-, mass number and charge of each; in the isotopic symbols below the atomic number has been added} Element #p+ #n #e- Mass number Charge 60 +2 I. 25 Mn 25 35 23 60 +2 62 +2 II. 27Co 27 35 25 62 +2 40 +2 III. 20Ca 20 20 18 40 +2 40 −2 IV. 16S 16 24 18 40 -2 58 +5 V. 25 Mn 25 33 20 58 +5 b. I and II c. III and IV d. I, II, and III e. III and IV

4. a. F {elements are distinguished by protons not neutrons} b. T c. T

58 + + 5. a. 28 Ni {28p = Ni; bottom number = atomic number = #p = 28; neutral = charge = 0; + 58 top number = mass number = #n + #p = 30 + 28 = 58; 28 Ni } 43 + + b. 22Ti {22p = Ti; bottom number = atomic number = #p = 22; neutral = charge = 0; + 43 top number = mass number = #n + #p = 21 + 22 = 43; 22Ti } 18 + + c. 8 O {O = 8p = bottom number = atomic number = #p = 8; neutral = charge = 0; + 18 top number = mass number = #n + #p = 10 + 8 = 18; 8O } 54 + + d. 24 Cr {Cr = 24p = bottom number = atomic number = #p = 24; neutral = charge = 0; + 54 top number = mass number = #n + #p = 54; 24Cr } 79 + + + - e. 35 Br {35p = Br; bottom number = atomic number = #p = 35; #p = #e = charge = 0; + 79 top number = mass number = #n + #p = 44 + 35 = 79; 35 Br } 7 + + + - f. 3Li { = Group IA; fewest protons = 3p = Li; bottom number = atomic number = #p = 3; #p = #e = charge = 0; + 7 top number = mass number = #n + #p = 4 + 3 = 7; 3Li } 56 +3 + + + - g. 26 Fe {iron = Fe = 26p = bottom number = atomic number = #p = 26; 3 extra p than e : + 56 +3 charge = #p(+1) + #e(-1) = 26(+1) + 23(-1) = +3; charge = +3; top number = mass number = #n + #p = 56; 26 Fe } 33 -2 + + h. 16 S {sulfide = S = 16p = bottom number = atomic number = #p = 16; ion means charged; Group VIA has a charge of -2 + 33 -2 (memorized); top number = mass number = #n + #p = 17 + 16 = 33; 16S }

6. a {hydrogen, 1H, does not contain neutrons}

7. Rutherford targeted alpha particles at thin Au foil and Au Foil

placed a photographic detector all around the foil. empty space nucleus Nearly all the a particles went through the foil but a α particles small number bounced back. He deduced that most of the atom is "empty" (hence most a particles pass through the foil) but there is a highly dense, positively charged, small "nucleus" (hence, a small number of a particles bounced back at sharp angles).

8. <5% {since the average atomic weight of O = 15.9994 ≈ 16, then the abundance of 16O must be close to 100% since the average weight is close to 16 and therefore the abundance of 18O must be very close to zero otherwise the average atomic mass would be higher than 16}

9.

⎛ −24 ⎞ -26 18 18 18 1.6605 x 10 g ⎛ 1kg ⎞ −26 10. I. 2.99 x 10 kg { O = 18u; mass of 1 atom O = 18u; O = 18amu⎜ ⎟ ⎜ ⎟ = 2.9889 x 10 kg } ⎝⎜ 1amu ⎠⎟ ⎝ 1000g⎠ ⎛ −24 ⎞ -26 52 52 52 1.6605 x 10 g ⎛ 1kg ⎞ −26 II. 8.63 x 10 kg { Cr = 52u; mass of 1 atom Cr = 52u; Cr = 52amu⎜ ⎟ ⎜ ⎟ = 8.6346 x 10 kg } ⎝⎜ 1amu ⎠⎟ ⎝ 1000g⎠

⎛ −24 ⎞ 26 1.6605 x 10 g ⎛ 1kg ⎞ ⎛ 1000g⎞ ⎛ 1u ⎞ 11. a { 4.48 x 10− kg = X⎜ ⎟ ; solve for X: X = (4.48 x 10−26 kg) = 26.98 ; X = 27 ® Al} ⎜ ⎟ ⎜ ⎟ ⎜ 24 ⎟ ⎝⎜ 1amu ⎠⎟ ⎝ 1000g⎠ ⎝ 1kg ⎠ ⎝ 1.6605 x 10− g⎠

12. 85.47amu; Rb {(84.912)(0.7215) + (86.909)(0.2785) = 85.468amu}

13. 107Ag: 51.8%, and 109Ag: 48.2% {107.869 = 106.91(x) + 108.90(1 - x); x = abundance 107Ag; 107.869 = 106.91x + 108.90 – 108.90x; -1.031 = -1.990x; x = 0.5181 = 107Ag abundance; 109Ag abundance = 1 – x = 1 – 0.5181 = 0.4819}

14. 24Mg = 78.99%; 26Mg = 11.01% {x = abundance 24Mg; y = abundance 26Mg; sum of fractional abundances: 1.000 = x + 0.1000 + y; y = 1.000 – 0.1000 – x; 24.3050 = 23.985(x) + 24.986(0.1000) + 25.983(1 – 0.1000 – x); 24.3050 = 23.985x + 2.4986 + 23.3847 – 25.983x; 24.3050 = 25.8833 – 1.998x; –1.5783 = –1.998x; x = 0.7899 = 24Mg abundance = 78.99%; 26Mg abundance = 100.00 – 10.00 – 78.99 = 11.01%}

15. 47.9amu {Average atomic mass = (M1)(FA1) + (M2)(FA2) + . . . ; AAM = (46)(0.083) + (47)(0.074) + (48)(0.737) + (49)(0.054) + (50)(0.052) = 47.92amu}

75%

25% Intensity orIntensity Abundance

35 36 37 16. a. m/z {Since there are 2 isotopes 2 peaks are expected in the hard mass spectrum. The mass of each peak corresponds to the mass number; hence, peaks at mass = 35 and mass = 37 are expected. The intensity of each peak reflects the relative abundance of each isotope so 35Cl would be 75% and 37Cl would be 25%. [The peaks need to be in a ratio of 3:1 for 35 35 37 Cl ⎛ 75%⎞ 3 the Cl and Cl isotopes which are in a ratio of 3:1 from = ⎜ ⎟ = } 37Cl ⎝ 25%⎠ 1

56%

38%

Intensity orIntensity Abundance 6%

70 72 74 b. m/z {Generate all the possible isotopic pairs for Cl2. The masses of each are determined by adding the masses of the isotopes. The probability is determined by multiplying their respective abundances. Versions of Cl2 Mass of Cl2 Probability of this isotopic version Peak (mass) Peak Height 35Cl35Cl 70 (0.75)(0.75) = 0.5625 = 56.3% 70 56.3% 35Cl37Cl 72 (0.75)(0.25) = 0.1875 = 18.8% 72 18.8% + 18.8% = 37.6% 37Cl35Cl 72 (0.25)(0.75) = 0.1875 = 18.8% ------37Cl37Cl 74 (0.25)(0.25) = 0.0625 = 6.3% 74 6.3% Draw the three peaks with their masses at 70, 72, and 74 reflecting the possible isotopic versions. The heights should reflect their abundances of 56.3%, 37.6%, and 6.3%, respectively. 17. c {LiF and RbF have +1/-1 charges while MgF2 and BaF2 have +2/-1 charges; lower charges will yield lower melting points so +1 -1 +1 -1 drop MgF2 and BaF2; since LiF and RbF have the same charges: Li F and Rb F to differentiate between these the larger the ions the lower the melting point; ions lower in a column on the Periodic Table are larger, and therefore since Rb+ is larger than Li+, RbF should have the lowest melting point}

18. a {NH3 is a molecular compound (2 nonmetals) and molecular compounds have much lower melting points than ionic compounds; KBr has +1/-1 charges while BeO and MgS have +2/-2 charges; higher charges will yield higher melting points so drop KBr; the other ionic compounds have the same charges: Be+2O-2, Mg+2S-2; to differentiate between these the smaller the ions the higher the melting point; ions higher in a column on the Periodic Table are smaller, and therefore since Be+2 is smaller than Mg+2 and likewise O-2 is smaller than S-2, BeO will have the highest melting point}

19. a. nitrogen monoxide b. trichloride c. bromide d. sodium carbonate e. dinitrogen tetroxide f. potassium sulfate g. calcium phosphate h. copper(I) oxide i. manganese(II) phosphate j. (III) acetate k. cobalt(II) sulfate l. potassium permanganate m. methane (common name; memorize)

20. a. SO3 b. CaF2 c. ClBr d. Na2SO4 e. NH4NO3 f. Li2CO3 g. Ca(NO3)2 h. Mn2(SO4)3 i. CrO3 j. TiS2 k. Ca(C2H3O2)2

21. b {memorized}

22. a. empirical formula = C2H6N1 {from the hard mass spectrum it can be determined that the compound contains H (mass = 1), C (mass = 12), and N (mass = 14). From the percent intensity or abundance the number of each element can be determined by dividing each element by the smallest percent. 66.6% 22.2% 11.1% H = = 6 ; C = = 2 ; N = = 1; this yields an empirical formula = C H N } 11.1% 11.1% 11.1% 2 6 1 b. molecular formula = C4H12N2 {from the soft mass spectrum the parent peak describes the overall molar mass of the compound; the molar mass = 88g/mol; the empirical formula’s molar mass = 44g/mol; molar massmolecular formula 88g / mol determine ratio = ; ratio = = 2 ; multiply the empirical formula by 2: molar massempiricalformula 44g / mol C(2 x 2)H(6 x 2)N(1 x 2) = C4H12N2}

23. a. empirical formula = C2H2O3 {from the hard mass spectrum it can be determined that the compound contains H (mass = 1), C (mass = 12), and O (mass = 16). From the percent intensity or abundance the number of each element can be determined by dividing each element by the smallest percent. 28.6% 28.6% 42.8% H = = 1 ; C = = 1 ; O = = 1.5 ; this yields a formula = C H O ; deal with the fraction: 28.6% 28.6% 28.6% 1 1 1.5 C(1 x 2)H(1 x 2)O(1.5 x 2) = C2H2O3 = empirical formula} b. molecular formula = C4H4O6 {from the soft mass spectrum the parent peak describes the overall molar mass of the compound; the molar mass = 88g/mol; the empirical formula’s molar mass = 44g/mol; molar massmolecular formula 148g / mol determine ratio = ; ratio = = 2 ; multiply the empirical formula by 2: molar massempiricalformula 74g / mol C(2 x 2)H(2 x 2)O(3 x 2) = C4H4O6}

24.

67% 100%

22%

Intensity orIntensity Abundance 11% orIntensity Abundance

1 12 16 46 m/z m/z Hard Mass Spectrum Soft Mass Spectrum

{For the soft spectrum determine the molar mass of the chemical; in this case it is 46g/mol. One peak denoted with 100% at this m/z will be the soft mass spectrum. For the hard mass spectrum there will be a peak at mass = 1 for H, a peak at mass = 12 for C, and a peak at mass = 16 for O. To determine each peak’s intensity in the hard mass spectrum divide the number of atoms of that element by the total number of atoms present in the compound. ⎛ 2 ⎞ ⎛ 6 ⎞ ⎛ 1 ⎞ C = x 100% = 22.2% ; H = x 100% = 66.7% ; O = x 100% = 11.1% .} ⎝⎜ 2 + 6 +1⎠⎟ ⎝⎜ 2 + 6 +1⎠⎟ ⎝⎜ 2 + 6 +1⎠⎟

25. See below. a. To determine the soft spectrum, determine the molar mass of the compound. Since there are 2 isotopes of Cl there would be 2 35 37 35 37 different versions of the compound: C2H5 Cl and C2H5 Cl. The mass of C2H5 Cl = 64 and the mass of C2H5 Cl = 66. Since the Cl isotopes are 75% and 25% in abundance, the peak at 64 would be 75% and the one at 66 would be 25%. b. To determine the hard spectrum, there will be unique peaks for each element/isotope. Hence, there will be a peak at mass = 1 for H, a peak at mass = 12 for C, and two peaks for Cl one at mass = 35 and one at mass = 37. To determine each element’s peak’s intensity in the mass spectrum divide the number of atoms of that element by the total of atoms in the compound. The Cl is ⎛ 2 ⎞ ⎛ 5 ⎞ counted as 1 initially; we’ll take care of the 2 isotopes next. C = x 100% = 25.0% ; H = x 100% = 62.5% ; ⎝⎜ 2 + 5 +1⎠⎟ ⎝⎜ 2 + 5 +1⎠⎟ ⎛ 1 ⎞ Cl = x 100% = 12.5% . Since there are 2 isotopes of Cl, their total is 12.5%, and one is 75% abundant while the other is ⎝⎜ 2 + 5 +1⎠⎟ 25% abundant this to: 35Cl = (0.75)(12.5%) = 9.4% and 37Cl = (0.25)(12.5%) = 3.1%.

62.5% 75%

25% 25% 9.4% Intensity orIntensity Abundance Intensity orIntensity Abundance 3.1%

64 66 1 12 35 37 m/z m/z Soft mass spectrum Hard mass spectrum