HL First-Year Chemistry

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Susan Baker
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1 1. Atomic Structure and Stoichiometry Solutions to further problems: 1. (a) Fe (b) Rb (c) Tl (d) Cr HL First-Year Chemistry. Introduction of the source, vaporization of the sample, ionization, acceleration, velocity selection, deflection, and detection A R (Ti) = X 6 + X 7 + X 8 + X 9 + X = 7.95 = 7.9 or 8.. Relative atomic masses of some elements are so far off from whole numbers because it represents the average mass of multiple isotopes that make up the element. 5. (a) protons = 55, neutrons = 77, electrons = 55, (b) protons = 8, neutrons = 67, electrons = 6, (c) protons = 81, neutrons = 11, electrons = 81, (d) protons = 7, neutrons = 58, electrons = 6, (e) protons =, neutrons =, electrons = 6 6. The sample is first vaporized (into gaseous atoms) and then they are ionized, and then accelerated by a strong electric field into an area where they are further exposed to a varying magnetic or electric field ultimately reaching the detector where the charged ions are detected as electric current. 7. (a) atoms of Na, 1 atom of C, and atoms of O (b) atoms of N, 1 of H, 1 of P and of O (c) atoms of Na, 1 of Ag, of S, 6 of O 8. (a) moles of Na atoms, 1 mole of C atoms, and moles of O atoms (b) moles of N atoms, 1 moles of H atoms, 1 mole of P atoms and moles of O atoms (c) moles of Na atoms, 1 mole of Ag atoms, moles of S atoms, 6 moles of O atoms 9. Fe and CO are the products. The equation is NOT balanced. 10. (a) Products: CaCl, CO and H O, reactants: CaCO and HCl (b) equation is unbalanced. (c) CaCO = solid, HCl = aqueous, CaCl = aqueous, CO = gas, and H O = liquid 11. HgS (s) + O (g) Hg (l) + SO (g) 1. (a) there is no unit for relative molecular mass (b) g 1. (a) g (b) 17.6 g (c) 7.10 g g Fe 1mol Fe atoms 1. a) = 9.7 X 10 g Fe/atom of Fe 1mol Fe 6.0 X10 atoms of Fe b) (5 X mol N molecules N O) 6.0 X10 molecules N = 9.7 X 10 g Fe/atom of Fe ( sig figs).00 g N O 1mol N =.65 X 10 1 g N O = X 10 1 g N O (1 sig fig) 15. Percent composition of: a) NH (ammonia): A R (N) = 1.01, A R (H) = 1.01 M R (NH ) = (1.01) = = 17.0 [ decimal places] 1.01 % N = X 100 = 8.18 = 8. [ sig. figs] 17.0

2 .0 % H = X 100 = = 17.8 [ sig. figs.] 17.0 [NB: The total of the percent compositions add up to 100.0, which when rounded off to one decimal place to be consistent with the percent composition of hydrogen it comes out to ] OR considering a mole of ammonia: Mass of N in a mole of ammonia = 1.01 g Mass of H in a mole of ammonia = (1.01 g) =.0 g Molar mass of ammonia = 17.0 g 1.01g % N = X 100 = 8.18 = 8. [ sig. figs] 17.0 g.0 g % H = X 100 = = 17.8 [ sig. figs.] 17.0 g [NB: It makes no difference whether you determine percent compositions using relative mass or molar mass!] b) C H O (ethanoic acid): A R (C) = 1.01, A R (H) = 1.01, A R (O) = M R (C H O ) = (1.01) + (1.01) + (16.00) = [ decimal places].0 % C = X 100 = 9.99 [ sig. figs.] % H = X 100 = 6.7 [ sig. figs.] % O = X 100 = 5.8 [ sig. figs.] c) K SO (potassium sulfate): A R (K) = 9.10, A R (S) =.06, A R (O) = formula mass (K SO ) = (9.10) (16.00) = 17.6 [ decimal places] 78.0 % K = X 100 =.88 [ sig. figs.] % S = X 100 = 18.0 [ sig. figs.] % O = X 100 = 6.7 [ sig. figs.] 17.6 d) Ca(OH) : A R (Ca) = 0.08, A R (O) = 16.00, A R (H) = 1.01 formula mass (Ca(OH) ) = 7.10 [ decimal places] 0.08 % Ca = X 100 = 5.09 [ sig. figs.] % O = X 100 =.18 [ sig. figs.] % H = X 100 =.7 [ sig. figs.] 7.10 (e) Zn(NO ) 6H O: formula mass = [ decimal places] 65.7 % Zn = X 100 = 1.97 [ sig. figs.] % N = X 100 = 9.18 [ sig. figs.] 97.51

3 19.00 % O = X 100 = 6.56 [5 sig. figs.] % H = X 100 =.07 [ sig. figs.] [NB: The sum of the percentages adds up to which when rounded off to decimal places, which is what you have to do since the percentage of zinc is correct only to decimal place, the sum does result in 100.] f) C H 0 O 1 N 7 M R (C H 0 O 1 N 7 ) = % C = X 100 = 5.99 [5 sig. figs.] % H = X 100 =.516 [ sig. figs.] % O = X 100 =.1 [5 sig. figs.] % N = X 100 = [ sig. figs.] kg NH 16. a) 1.00 lb NH. lb NH 1000 g NH 1kg NH 1mol NH 17.0 g NH = 6.7 mol NH [ sig figs] 1kg C H O b) 1.00 lb C H O 1000 g C H O 1mol C H O. lb C H O 1kg C H O = 7.50 mol C H O [ g C H O sig figs] 1kg K SO c) 1.00 lb K SO 1000 g K SO 1mol K SO. lb K SO 1kg K SO =.61 mol 17.6 g K SO K SO [ sig figs] 1kg Ca(OH) d) 1.00 lb Ca(OH) 1000 g Ca(OH) 1mol Ca(OH). lb Ca(OH) 1kg Ca(OH) = 6.1 mol 7.10 g Ca(OH) Ca(OH) [ sig figs] 1kg Zn(NO) H 1000 g Zn(NO) H e) 1.00 lb Zn(NO) 6H O. lb Zn(NO) H 1kg Zn(NO ) H 1mol Zn(NO) H = 1.5 mol Zn(NO ) H [ sig figs] 97.51g Zn(NO) H 1kg C H 0O1N g C H 0O1N 7 1mol C H 0O1N 7 f) 1.00 lb C H 0 O 1 N 7. lb C H 0O1N 7 1kg C H 0O1N g C H 0O1N 7 = mol C H O [ sig figs] 1mol Al 17. (a) (15. g Al) 6.98 g Al 6.0 X10 1mol Al atoms Al =. X 10 atoms Al ( sig figs)

4 1mol N X10 molecules N 5 (b) (1.8 g N O 5 ) g N 5 1mol N 5 = 8.5 X 10 molecules of N O 5 1mol Na CO mol Na ions 6.0 X10 Na ions (c) (8. g Na CO ) g Na CO 1mol Na CO 1mol Na ions 1mol CH8O mol O 18. (7 mol C) mol C = 7 mol O, which you should have been able to 1mol CH8O determine without even setting it up as shown above. Since in a mole of glycerol the number of moles of C and O are equal (:) any sample of glycerol would have equal number of moles of C and O. 19. Equation is: Cu (s) + S (aq) Cu S (s) 1mol Cu (100.0 g Cu) 1 mol S.06 g S 6.55 g Cu mol Cu = 5. = 5. g S 1 mol S Limiting reagent is copper. 1mol Cu (100.0 g Cu) 1 mol Cu S g Cu S 6.55 g Cu mol Cu = 15. = 15. g. ( sig figs) 1 mol Cu S 0. Equation is: C (s) + Cl (g) CCl (l) 1mol C (10.0 g C) mol Cl g Cl 1.01 g C 1 mol C = 59.0 Cl 1 mol Cl Limiting reagent is carbon. 1mol C (10.0 g C) 1 mol CCl g CCl 1.01 g C 1 mol C = = 118 g CCl ( sig figs) 1 mol CCl Mass of excess reagent left unreacted = g 59.0 g = 1.0 g 1. The equation for the reaction is: Fe (s) + O (g) Fe O (s) 1mol O (.80 g O ) mol Fe g Fe.00 g O mol O = g Fe = 11. g Fe (required) 1 mol Fe g Fe (0.150 mol Fe) = g Fe = 8.8 g Fe (available) 1 mol Fe Therefore, limiting reagent is Fe. mol Fe (0.150 mol Fe) g Fe mol Fe = g Fe O = 1.0 g Fe O (produced) 1 mol Fe Mass of Fe remaining = 0. mol O (0.150 mol Fe) g O mol Fe = 1.8 g O = 1.80 g O 1 mol O Mass of oxygen remaining = =.00 g ( dp). (a) NH SO, (b) CH, (c) C H 8 O, (d) Hg SO.. Molecular Empirical formula a) C H 7 O C H 7 O b)ch O CH O c) C H 8 O C H O d) C H 6 O C H 6 O

5 e) C H 7 N C H 7 N f) C H 6 O CH O. Tungsten Oxygen Mass ratio. g 5.. = 1.11 g 1 mol 1 mol Molar ratio. g = mol 1.11 g = mol g g Dividing by the smaller of the two numbers (to = 1 = get whole number ratios) Therefore the empirical formula is WO. 5. a) C O Percentage 7.% 7.7% Then in 100 g.9 g 57.1 g Molar ratio 1mol 1mol (.9 g) =.57 mol (57.1 g) 1.01g g Number of moles of O that = 1 = 1 combine with 1 mol of C =.57 mol Therefore the empirical formula is CO. b) CO c) NaSO d) Na S O. 6. There are a couple of different ways of tackling these kinds of questions. Method A. The following is the long version with explanations. And it uses the stoichiometric ratio between the anhydrous salt and the hydrated form. What the solution boils down to is determination of the number of moles of water produced by a mole of the hydrate knowing each of the moles of water and that of the hydrate in the sample of hydrate analyzed. a) NiSO xh O (s) NiSO (s) + xh O (g) residue = NiSO (s) 1mol NiSO 0.06 g NiSO = mol NiSO g NiSO Since 1 mole of NiSO xh O produces 1 mole of NiSO, number of moles of NiSO xh O in the original sample = number of moles of NiSO produced = Since the original mass was 0.50 g, then the mass of 1 mole of NiSO xh O is given by: 0.50 g NiSO xh = 6 g/mol NiSO xh O mol NiSO xh Mass of water in 1 mole of NiSO xh O = molar mass of NiSO xh O molar mass of NiSO (since molar ratio of NiSO xh O: NiSO = 1:1) = = 108 g 1mol H Number of moles of water in 108 g of water = 108 g H O = 5.99 = g H Therefore, x = 6. That is, in every mole of NiSO xh O, there are 6 moles of water molecules.

6 Method B. This method uses the relationship between the products: What the solution boils down to is determination of the number of moles of water produced for every mole of the anhydrous salt produced knowing each of the moles of water and that of the anhydrous salt in the sample of hydrate analyzed. From the equation, the ratio of the products NiSO to H O = 1 to x moles of NiSO That is, 1 mole of NiSO = moles of H x moles of NiSO The ratio of in the experiment must also be the same. moles of H Therefore, experimental moles of NiSO 1 mole of NiSO = (Equation I) experimental moles of H x Moles of NiSO produced in the experiment 1mol NiSO 0.06 g NiSO = mol NiSO g NiSO Mass of water in the experiment = ( ) g = 0.1 g Moles of H O produced in the experiment: 1mol H 0.1 g H O = mol H O 18.0 g H Substituting in equation I, and solving for x, mole NiSO 1 mole of NiSO = mole H x mole H x = (1 mole of NiSO ) = 5.95 mol H O = 6 mole H O mole of NiSO b) MnI xh O (s) MnI (s) + xh O (g) moles of H produced x = (1 mole of MnI ) moles of MnI produced Mass of water produced = ( ) g = g 1mol H g H O = mol H O 18.0 g H Moles of MnI produced 1mol MnI 0.76 g MnI = mol MnI 08.7 g MnI moles of H x = (1 mole of MnI ) =.98 mol H O = mol H O moles of MnI c) similar to the ones above. MgSO xh O (s) MgSO (s) + xh O (g) d) similar to the ones above. CdSO xh O (s) CdSO (s) + xh O (g) e) Similar to the ones above. KAl(SO ) xh O (s) KAl(SO ) (s) + xh O (g) 7. Compound + O (g) H O (g) + CO (g) Mass of compound = 1.1 mg

7 Mass of H O produced = 5.6 mg Mass of hydrogen in the sample of water produced 1g H 5.6 mg H O 1mol H mol H 1.01g H 1000 mg H 18.0 g H = g H 1mol H 1mol H Mass of CO produced = 0.6 mg Mass of carbon in the carbon dioxide produced 1g CO 0.6 mg CO 1mol CO 1000 mg CO.01g CO 1mol C 1.01g C = g C 1mol CO 1mol C Since all the hydrogen in water and carbon in carbon dioxide came from the original organic compound, the mass of hydrogen in the original compound = g, and mass of carbon in the original compound = g. Therefore, mass of oxygen in the original compound = g ( ) g = g g = g C H O Mass ratio g g 1mol Molar ratio ( g) 1.01g 1mol g) g Moles of C and H that combines with 1mol of O 1mol ( g) 1.01g ( = mol = mol = mol =.98 =.99 = Multiplying by.98 X = X = X = Therefore, the empirical formula is C 7 H 6 O. 1 mol N 8. Out of g compound: 0. g N =.17 mol N 1.01 g N % O = = 69.6% 1 mol O 69.6 g O =.5 mol O g O Number of moles of O per mol of N = = Empirical formula = NO Empirical formula mass = (16.00) = 6.01 g/mol

8 9 g g = ; Therefore, the molecular formula is N O. 9. Assuming g of compound Mass of hydrogen = = 6.90 g 1 mol C 1 mol O 9.1 g C =.106 mol C;.79 g O =.77 mol O; 1.01 g C g O 1 mol H 6.90 g H = 6.8 mol H 1.01 g H Dividing all mole values by.77 to get the the number of moles of C and H per mole of O = C; =.95 H Multiplying the ratio by yields whole number ratios givening an empirical formula C H 5 O. Empirical formula mass = (1.01) + 5(1.01) + (16.00) = 7.08 g/mol molar mass 16.1 = = =.00 empirical formula mass 7.08 Therefore, the molecular formula is C 6 H 10 O. 0. Assuming g of compound Mass of oxygen = = 55.1 g 1 mol C 1 mol O 1.9 g C =.6 mol C; 55.1 g O =.6 mol O; 1.01 g C g O 1 mol H.7 g H =.5 mol H 1.01 g H All three are the same mole values, therefore the empirical formula is CHO. Empirical formula mass = = 9.0 g/mol 15.0 g Molar mass of compound = 116 g/mol 0.19 mol = molar mass 116 = =.00 empirical formula mass 9.0 Therefore, the molecular formula is C H O.

IGCSE (9-1) Edexcel - Chemistry

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