Section 64: Molar Mass Tutorial 1 Practice, pae 275 1 (a) Given: substance with formula S 8 Required: ar mass of S 8, n NaCl2 Step 1 Look up the ar mass of the element M S = 3207 Step 2 Multiply the ar mass of the element by the number of atoms of that element in the substance M S8 = 8M S = 8 3207 $ % M S8 = 25656 Statement: The ar mass of S 8 is 25656 / (b) Given: compound with formula H 2 S Required: ar mass of H 2 S, M H2 S Step 1 Look up the ar masses of the elements M H = 101 ; M = 3207 S Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M = 2M + M H2 S H S = 2 101 $ % + 3207 $ % M = 3409 H2 S Statement: The ar mass of H 2 S is 3409 / (c) Given: compound with formula NaH Required: ar mass of NaH, M NaH Step 1 Look up the ar masses of the elements M Na = 2299 ; M = 1600 ; M = 101 H Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-1
Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M NaH = M Na + M + M H = 2299 $ % + 1600 $ % + 101 $ % M NaH = 4000 Statement: The ar mass of NaH is 4000 / (d) Given: compound with formula Fe(H) 3 Required: ar mass of Fe(H) 3, M Fe(H)3 Step 1 Look up the ar masses of the elements M Fe = 5585 ; M = 1600 ; M = 101 H Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M Fe(H)3 = M Fe + 3( M + M H ) = 5585 $ % + 3 1600 + 101 $ % M Fe(H)3 = 10688 Statement: The ar mass of Fe(H) 3 is 10688 / (e) Given: compound with formula (NH 4 ) 2 S Required: ar mass of (NH 4 ) 2 S, M (NH4 ) 2 S Step 1 Look up the ar masses of the elements M N = 1401 ; M = 101 H ; M = 3207 S Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M = 2( M + 4M ) + M (NH4 ) 2 S N H S = 2 1401 $ + 4 101 % + 3207 % $ M (NH4 ) 2 S = 6817 Statement: The ar mass of (NH 4 ) 2 S is 6817 / Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-2
(f) Given: compound with formula Ca 3 (P 4 ) 2 Required: ar mass of Ca 3 (P 4 ) 2, M Ca3 (P 4 ) 2 Step 1 Look up the ar masses of the elements M Ca = 4008 ; M = 3097 P ; M = 1600 Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M Ca3 = 3M (P 4 ) 2 Ca + 2( M P + 4M ) = 3 4008 $ % + 2 3097 + 4 1600 $ % M Ca3 = 31018 (P 4 ) 2 Statement: The ar mass of Ca 3 (P 4 ) 2 is 31018 / () Given: compound with formula MS 4 7H 2 Required: ar mass of MS 4 7H 2, M MS4 7H 2 Step 1 Look up the ar masses of the elements M M = 2431 ; M = 3207 S ; M = 1600 ; M = 101 H Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M = M + M + 4M + 7(2M + M ) MS4 7H 2 M S H = 2431 $ % + 3207 $ % + 4 1600 $ % + 7 2 101 + 1600 $ % M MS4 7H 2 = 24652 Statement: The ar mass of MS 4 7H 2 is 24652 / Mini Investiation: The Mole Exhibit, pae 275 A Answers may vary Answers are determined from experimental data Molar mass of a compound is calculated by addin the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound Sample answer: Given: amount of calcium chloride, n CaCl2 = 100 Required: ar mass of calcium chloride, Step 1 Look up the ar masses of the elements that are in the substance M Ca = 4008 /; M Cl = 3545 / Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-3
Step 2 Determine the sum of the ar masses of the elements in the substance M CaCl2 = M Ca + 2M Cl = 4008 + 2 3545 $ % M CaCl2 = 11098 Statement: The ar mass of 100 of calcium chloride is 111 / I will have to measure approximately 111 of calcium chloride for 100 of the substance B Each ar quantity contains 602 10 23 entities Therefore there 602 10 23 entities of calcium chloride in the beaker Tutorial 2: Practice, pae 277 1 (a) Given: m NaCl = 500 Required: amount of table salt, n NaCl Step 1 Calculate the ar mass of table salt, M NaCl M NaCl = 2299 $ % + 3545 $ % M NaCl = 5844 Step 2 Use the mass of table salt and its ar mass to calculate the amount of table salt 1 $ n NaCl = (500 ) 5844 % n NaCl = 856 Statement: The amount of sodium chloride in a 500 of table salt is 856 (b) Given: m Al = 142 Required: amount of aluminum, n Al Use the ar mass of aluminum to calculate the amount of aluminum 1 $ n Al = (142 ) 2698 % n Al = 0526 Statement: The amount of aluminum in a 142 sample is 0526 Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-4
(c) Given: m Ca = 100 k Required: amount of calcium oxide, n Ca Step 1 Calculate the ar mass of calcium oxide, M Ca M Ca = 4008 $ % + 1600 $ % M Ca = 5608 Step 2 Use the mass of calcium oxide and its ar mass to calculate the amount of calcium oxide m Ca = 100 k = 100 10 3 1 % n Ca = (100 10 3 ) $ 5608 n Ca = 178 Statement: The amount of calcium oxide in a 100 k sample is 178 (d) Given: m HCl = 175 k Required: amount of hydroen chloride, n HCl Step 1 Calculate the ar mass of hydroen chloride, M HCl M HCl = 101 $ % + 3545 $ % M HCl = 3646 Step 2 Use the mass of hydroen chloride and its ar mass to calculate the amount of hydroen chloride m HCl = 175 k = 175 10 3 1 % n HCl = (175 10 3 ) $ 3646 n HCl = 480 Statement: The amount of hydroen chloride in a 175 k sample is 480 Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-5
(e) Given: m C13 H 18 2 = 2000 m Required: amount of ibuprofen, n C13 H 18 2 Step 1 Calculate the ar mass of ibuprofen, M C13 H 18 2 M C13 = 13 1201 H 18 2 $ % + 18 101 $ % + 2 1600 $ % M C13 = 20631 H 18 2 Step 2 Use the mass of ibuprofen and its ar mass to calculate the amount of ibuprofen m C13 = 2000 m H 18 2 m C13 H 18 2 = 2000 10 3 = 2000 10 1 1 n C13 = (2000 10 1 ) H 18 2 % $ 20631 ( n C13 H 18 2 = 9694 10 4 Statement: The amount of ibuprofen in a 2000 m tablet is 9694 10 4 2 (a) Given: n H2 2 = 080 Required: mass of hydroen peroxide, m H2 2 Step 1 Calculate the ar mass of hydroen peroxide, M H2 2 M H2 = 2 101 2 $ % + 2 1600 $ % M H2 = 3402 2 Step 2 Use the amount of hydroen peroxide and its ar mass to calculate the mass of hydroen peroxide 3402 $ m H2 = (080 ) 2 1 % m H2 2 = 27 Statement: The mass of 080 of hydroen peroxide is 27 Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-6
(b) Given: n NaHS4 = 325 Required: mass of sodium hydroen sulfate, m NaHS4 Step 1 Calculate the ar mass of sodium hydroen sulfate, M NaHS4 M NaHS4 = 2299 $ % + 101 $ % + 3207 $ % + 4 1600 $ % M NaHS4 = 12007 Step 2 Use the amount of sodium hydroen sulfate and its ar mass to calculate the mass of sodium hydroen sulfate m NaHS4 = (325 ) 12007 $ 1 % m NaHS4 = 390 10 2 Statement: The mass of 325 of sodium hydroen sulfate is 390 10 2 (c) Given: n CaC3 = 50 m Required: mass of calcium carbonate, m CaC3 Step 1 Calculate the ar mass of calcium carbonate, M CaC3 M CaC3 = 4008 $ % + 1201 $ % + 3 1600 $ % M CaC3 = 10009 Step 2 Use the amount of calcium carbonate and its ar mass to calculate the mass of calcium carbonate n CaC3 = 50 m = 50 10 3 m CaC3 = (50 10 3 ) 10009 % ( $ 1 m CaC3 = 050 Statement: The mass of 50 m of calcium carbonate is 050 Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-7
(d) Given: n NaCl = 12 10 3 Required: mass of sodium hypochlorite, m NaCl Step 1 Calculate the ar mass of sodium hypochlorite, M NaCl M NaCl = 2299 $ % + 1600 $ % + 3545 $ % M = 7444 NaCl Step 2 Use the amount of sodium hypochlorite and its ar mass to calculate the mass of sodium hypochlorite 7444 % m NaCl = (12 10 3 ) $ 1 m NaCl = 89 10 4 Statement: The mass of 12 10 3 of sodium hypochlorite is 89 10 4 (e) Given: n He = 45 m Required: mass of helium, m He Use the ar mass of helium to calculate the mass of helium n He = 45 m = 45 10 2 400 m He = (45 10 2 )% ( $ 1 m He = 18 10 1 Statement: The mass of 45 m of helium is 18 10 1 Section 64 Questions, pae 277 1 ne e samples of different compounds have different masses because the ar mass of the compounds are different 2 A precise estimate of the number of entities in a sample is determined by dividin the mass of the sample by the ar mass of the substance in the sample In this way, the mass of the sample is an indirect way to count the number of entities present 3 The amount obtained usin this equation is always an estimate because mass is a measured quantity There is some deree of uncertainty in all measured quantities Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-8
4 (a) Given: iron(iii) oxide, Fe 2 3 Required: ar mass of Fe 2 3, M Fe2 3 Step 1 Look up the ar masses of the elements M Fe = 5585 ; M = 1600 Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M Fe2 = 2M 3 Fe + 3M = 2 5585 $ % + 3 1600 $ % M Fe2 = 15970 3 Statement: The ar mass of iron(iii) oxide is 15970 / (b) Given: calcium carbonate, CaC 3 Required: ar mass of CaC 3, M CaC3 Step 1 Look up the ar masses of the elements M Ca = 4008 ; M = 1201 C ; M = 1600 Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M CaC3 = M Ca + M C + 3M = 4008 $ % + 1201 $ % + 3 1600 $ % M CaC3 = 10009 Statement: The ar mass of calcium carbonate is 10009 / (c) Given: octane, C 8 H 18 Required: ar mass of C 8 H 18, M C8 H 18 Step 1 Look up the ar masses of the elements M C = 1201 ; M = 101 H Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-9
Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M C8 = 8M H 18 C + 18M H = 8 1201 $ % + 18 101 $ % M C8 = 11426 H 18 Statement: The ar mass of octane is 11426 / (d) Given: calcium chlorate, Ca(Cl 3 ) 2 Required: ar mass of Ca(Cl 3 ) 2, M Ca(Cl3 ) 2 Step 1 Look up the ar masses of the elements M Ca = 4008 ; M = 3545 Cl ; M = 1600 Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M Ca(Cl3 = M ) 2 Ca + 2( M Cl + 3M ) = 4008 $ % + 2 3545 + 3 1600 $ % M Ca(Cl3 = 20698 ) 2 Statement: The ar mass of calcium chlorate is 20698 / (e) Given: ammonium carbonate, (NH 4 ) 2 C 3 Required: ar mass of (NH 4 ) 2 C 3, M (NH4 ) 2 C 3 Step 1 Look up the ar masses of the elements M N = 1401 ; M = 101 H ; M = 1201 C ; M = 1600 Step 2 Add the ar masses of the elements, multiplyin the ar mass of each element by the number of atoms of that element in the compound M (NH4 = 2( M ) 2 C 3 N + 4M H ) + M C + 3M = 2 1401 $ + 4 101 % + 1201 % $ + 3 1600 % $ M (NH4 = 9611 ) 2 C 3 Statement: The ar mass of ammonium carbonate is 9611 / Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-10
5 Given: m CaCl2 = 100 Required: amount of calcium chloride, n CaCl2 Step 1 Calculate the ar mass of calcium chloride, M CaCl2 M CaCl2 = 4008 $ % + 2 3545 $ % M CaCl2 = 11098 Step 2 Use the mass of calcium chloride and its ar mass to calculate the amount of calcium chloride 1 $ n CaCl2 = (100 ) 11098 % n CaCl2 = 00901 Statement: The amount of calcium chloride in a 100 sample is 00901 6 Given: m C8 H 10 N 4 2 = 800 m Required: amount of caffeine, n C8 H 10 N 4 2 Step 1 Calculate the ar mass of caffeine, M C8 H 10 N 4 2 M C8 = 8 1201 H 10 N 4 2 $ % + 10 101 $ % + 4 1401 $ % + 2 1600 $ % M C8 = 19422 H 10 N 4 2 Step 2 Use the mass of caffeine and its ar mass to calculate the amount of caffeine m C8 = 800 m = 800 10 2 H 10 N 4 2 1 n C8 = (800 10 2 ) H 10 N 4 2 % $ 19422 ( n C8 H 10 N 4 2 = 412 10 4 Statement: The amount of caffeine in an 800 m sample is 412 10 4 Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-11
7 Given: n H2 = 122 103 Required: mass of water, m H2 Step 1 Calculate the ar mass of water, M H2 M = 2 101 $ H2 % + 1600 $ % M = 1802 H2 Step 2 Use the amount of water and its ar mass to calculate the mass of water m H2 = (122 103 ) 1802 % ( $ 1 m H2 = 220 102 Statement: The mass of 122 10 3 of water is 220 10 2 8 Given: m Si = 58 m Required: amount of silicon, n Si Use the ar mass of silicon to calculate the amount of silicon 1 n Si = (58 10 3 ) % $ 2809 ( n Si = 21 10 4 Statement: The amount of silicon in a 58 m sample is 21 10 4 9 (a) Given:V H2 S 4 = 34 000 L ; d H2 S 4 = 184 k/l Required: mass of sulfuric acid, m H2 S 4 Use the formula d = m V to calculate the mass of sulfuric acid d = m V m = Vd m H2 = V S 4 H2 d S 4 H2 S 4 184 k$ = (34 000 L) L % m H2 S 4 = 62 560 k [2 extra diits carried] Statement: The mass of sulfuric acid in the truck is 63 000 k Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-12
(b) Given: m H2 S 4 = 62 560 k Required: amount of sulfuric acid, n H2 S 4 Step 1 Calculate the ar mass of sulfuric acid, M H2 S 4 M H2 = 2 101 S 4 $ % + 3207 $ % + 4 1600 $ % M H2 = 9809 S 4 Step 2 Use the mass of sulfuric acid and its ar mass to calculate the amount of sulfuric acid m H2 = 62 560 k S 4 m H2 S 4 = 62 560 10 3 = 6256 10 7 1 % n H2 = (6256 10 7 ) S 4 $ 9808 n H2 S 4 = 64 10 5 Statement: The amount of sulfuric acid in the 34 000 L sample is 64 10 5 10 (a) Given: n NaCl = 0154 Required: mass of sodium chloride, m NaCl Step 1 Calculate the ar mass of sodium chloride, M NaCl M NaCl = 2299 $ % + 3545 $ % M NaCl = 5844 Step 2 Use the amount of sodium chloride and its ar mass to calculate the mass of sodium chloride m NaCl = (0154 ) 5844 $ 1 % m NaCl = 900 Statement: The mass amount of 0154 of sodium chloride is 900 Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-13
11 Given: m compound = 29 ; n compound = 0050 Required: ar mass of compound, M compound Use the formula n = m M to calculate the ar mass of the compound n = m M M = m n M compound = m compound n compound = 29 0050 M compound = 58 Statement: The ar mass of the compound is 58 / 12 Given: n Na = 100 ; m atom = 2299 u ; conversion factor = 11661 10 24 u ; atoms N A = 602 10 23 Required: mass of 100 sodium atom, m Na Calculate the number of atoms in 100, then the mass of this number of atoms in atomic mass units, u, then convert the mass of this number of atoms from atomic mass units, u, to rams, atoms % u % m Na = (100 ) $ 602 10 23 $ 2299 atom 11661 % $ 10(24 u m Na = 230 Statement: The mass of 100 sodium atoms is 230 Copyriht 2011 Nelson Education Ltd Chapter 6: Quantities in Chemical Formulas 64-14