# CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS

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1 CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS FOLLOW UP PROBLEMS 3.1A Plan: The mass of carbon must be changed from mg to g. The molar mass of carbon can then be used to determine the number of moles g 1 mol C Moles of carbon = 315 mg C = x 10 1 mg 2 = 2.62 x 10 2 mol C g C Road map: Mass (mg) of C 10 3 mg = 1 g Mass (g) of C Divide by M (g/mol) Amount (moles) of C 3.1B Plan: The number of moles of aluminum must be changed to g. Then the mass of aluminum per can can be used to calculate the number of soda cans that can be made from 52 mol of Al g Al soda can Number of soda cans = 52 mol Al 1 = = 100 soda cans 1 mol Al 14 g Al Road map: Amount (mol) of Al Multiply by M (g/mol) (1 mol Al = g Al) Mass (g) of Al 14 g Al = 1 soda can Number of cans 3.2A Plan: Avogadro s number is needed to convert the number of nitrogen molecules to moles. Since nitrogen molecules are diatomic (composed of two N atoms), the moles of molecules must be multiplied by 2 to obtain moles of atoms mol N2 2 N atoms Moles of N atoms = ( 9.72 x 10 N2 molecules) x 10 N 1 mol N 2 molecules 2 = x 10 2 = 3.23 x 10 2 mol N 3-1

2 Road map: No. of N 2 molecules Divide by Avogadro s number (molecules/mol) Amount (moles) of N 2 Use chemical formula (1 mol N 2 = 2 mol N) Amount (moles) of N 3.2B Plan: Avogadro s number is needed to convert the number of moles of He to atoms. Number of He atoms = 325 mol He x 1023 He atoms = x = 1.96 x He atoms 1 mol He Road map: Amount (mol) of He Multiply by Avogadro s number (1 mol He = x He atoms) Number of He atoms 3.3A Plan: Avogadro s number is needed to convert the number of atoms to moles. The molar mass of manganese can then be used to determine the number of grams. Mass (g) of Mn = ( 20 1 mol Mn g Mn 3.22 x 10 Mn atoms) x 10 Mn atoms 1 mol Mn = x 10 2 = 2.94 x 10 2 g Mn Road map: No. of Mn atoms Divide by Avogadro s number (molecules/mol) Amount (moles) of Mn Multiply by M (g/mol) Mass (g) of Mn 3.3B Plan: Use the molar mass of copper to calculate the number of moles of copper present in a penny. Avogadro s number is then needed to convert the number of moles of Cu to Cu atoms mol Cu x 10 Cu atoms Number of Cu atoms = g Cu g Cu 1 mol Cu = x = 5.92 x Cu atoms 3-2

3 Road map: Mass (g) of Cu Divide by M (g/mol) (1 mol Cu = g Cu) Amount (moles) of Cu Multiply by Avogadro s number (1 mol Cu = x Cu atoms) No. of Cu atoms 3.4A Plan: Avogadro s number is used to change the number of formula units to moles. Moles may be changed to mass using the molar mass of sodium fluoride, which is calculated from its formula. The formula of sodium fluoride is NaF. M of NaF = (1 x M of Na) + (1 x M of F) = g/mol g/mol = g/mol Mass (g) of NaF = ( 19 1 mol NaF g NaF 1.19 x 10 NaF formula units) x 10 NaF formula units 1 mol NaF = x 10 4 = 8.30 x 10 4 g NaF Road map: No. of NaF formula units Divide by Avogadro s number (molecules/mol) Amount (moles) of NaF Multiply by M (g/mol) Mass (g) of NaF 3.4B Plan: Convert the mass of calcium chloride from pounds to g. Use the molar mass to calculate the number of moles of calcium chloride in the sample. Finally, use Avogadro s number to change the number of moles to formula units. M of CaCl 2 = (1 x M of Ca) + (2 x M of Cl) = g/mol + 2(35.45 g/mol) = g/mol Number of formula units of CaCl 2 = 400 lb g 1 lb 1 mol CaCl 2 23 x 10 formula units CaCl g CaCl 2 1 mol CaCl 2 = x = 1 x formula units CaCl 2 Road map: Mass (lb) of CaCl 2 1 lb = g Mass (g) of CaCl 2 Divide by M (g/mol) (1 mol CaCl 2 = g CaCl 2 ) 3-3

4 Amount (mol) of CaCl 2 Multiply by Avogadro s number (1 mol CaCl 2 = x CaCl 2 formula units) No. of formula units of CaCl 2 3.5A Plan: Avogadro s number is used to change the number of molecules to moles. Moles may be changed to mass by multiplying by the molar mass. The molar mass of tetraphosphorus decoxide is obtained from its chemical formula. Each molecule has four phosphorus atoms, so the total number of atoms is four times the number of molecules. a) Tetra = 4, and deca = 10 to give P 4 O 10. The molar mass, M, is the sum of the atomic weights, expressed in g/mol: P = 4(30.97) = g/mol O = 10(16.00) = g/mol = g/mol of P 4 O 10 Mass (g) of P 4 O 10 = ( 22 1 mol g 4.65 x 10 molecules P4 O10 ) x 10 molecules 1 mol = = 21.9 g P 4 O atoms P b) Number of P atoms = 4.65 x 10 molecules P4 O10 = 1.86 x P atoms 1 P 4 O 10 molecule 3.5B Plan: The mass of calcium phosphate is converted to moles of calcium phosphate by dividing by the molar mass. Avogadro s number is used to change the number of moles to formula units. Each formula unit has two phosphate ions, so the total number of phosphate ions is two times the number of formula units. a) The formula of calcium phosphate is Ca 3 (PO 4 ) 2. The molar mass, M, is the sum of the atomic weights, expressed in g/mol: M = (3 x M of Ca) + (2 x M of P) + (8 x M of O) = (3 x g/mol Ca) + (2 x g/mol P) + (8 x g/mol O) = g/mol Ca 3 (PO 4 ) 2 No. of formula units Ca 3 (PO 4 ) 2 = 75.5 g Ca 3 (PO 4 ) 2 1 mol Ca 3(PO 4 ) x g Ca 3 (PO 4 ) 2 23 formula units Ca3 (PO 4 ) 2 1 mol Ca 3 (PO 4 ) 2 = x = 1.47 x formula units Ca 3 (PO 4 ) 2 3 b) No. of phosphate (PO 3-4 ) ions = 1.47 x PO formula units Ca 3 (PO 4 ) 2 4 ions 1 formula unit Ca 3 (PO 4 ) 2 = 2.94 x phosphate ions 3.6A Plan: Calculate the molar mass of glucose. The total mass of carbon in the compound divided by the molar mass of the compound, multiplied by 100% gives the mass percent of C. The formula for glucose is C 6 H 12 O 6. There are 6 atoms of C per each formula. Molar mass of C 6 H 12 O 6 = (6 x M of C) + (12 x M of H) + (6 x M of O) = (6 x g/mol) + (12 x g/mol) + (6 x g/mol) = g/mol 3-4

5 total mass of C 6 x g/mol Mass % of C = (100) = (100) = = 40.00% C molar mass of C 6 H 12 O g/mol 3.6B Plan: Calculate the molar mass of CCl 3 F. The total mass of chlorine in the compound divided by the molar mass of the compound, multiplied by 100% gives the mass percent of Cl. The formula is CCl 3 F. There are 3 atoms of Cl per each formula. Molar mass of CCl 3 F = (1 x M of C) + (3 x M of Cl) + (1 x M of F) = (1 x g/mol) + (3 x g/mol) + (1 x g/mol) = g/mol total mass of Cl 3 x g/mol Mass % of Cl = (100) = (100) = = 77.42% Cl molar mass of CCl 3 F g/mol 3.7A Plan: Multiply the mass of the sample by the mass fraction of C found in the preceding problem g C Mass (g) of C = g C 6 H 12 O 6 = = g C g C 6 H 12 O 6 3.7B Plan: Multiply the mass of the sample by the mass fraction of Cl found in the preceding problem g Cl Mass (g) of Cl = 112 g CCl 3 F = = 86.7 g Cl g CCl 3 F 3.8A Plan: We are given fractional amounts of the elements as subscripts. Convert the fractional amounts to whole numbers by dividing each number by the smaller number and then multiplying by the smallest integer that will turn both subscripts into integers. Divide each subscript by the smaller value, 0.170: B0.170O = B 1 O Multiply the subscripts by 2 to obtain integers: B 1 x 2 O 1.5 x 2 = B 2 O 3 3.8B Plan: We are given fractional amounts of the elements as subscripts. Convert the fractional amounts to whole numbers by dividing each number by the smaller number and then multiplying by the smallest integer that will turn both subscripts into integers. Divide each subscript by the smaller value, 6.80: C H 18.1 = C 1 H Multiply the subscripts by 3 to obtain integers: C 1x3 H 2.66x3 = C 3 H 8 3.9A Plan: Calculate the number of moles of each element in the sample by dividing by the molar mass of the corresponding element. The calculated numbers of moles are the fractional amounts of the elements and can be used as subscripts in a chemical formula. Convert the fractional amounts to whole numbers by dividing each number by the smallest subscripted number. Moles of H = 1.23 g H 1 mol H = 1.22 mol H g H Moles of P = g P 1 mol P = mol P g P Moles of O = g O 1 mol O = 1.63 mol O g O Divide each subscript by the smaller value, 0.408: H 1.22 P0.408O 1.63 = H 3 PO

6 3.9B Plan: The moles of sulfur may be calculated by dividing the mass of sulfur by the molar mass of sulfur. The moles of sulfur and the chemical formula will give the moles of M. The mass of M divided by the moles of M will give the molar mass of M. The molar mass of M can identify the element. 1 mol S Moles of S = ( 2.88 g S) = mol S g S 2 mol M Moles of M = ( mol S) = mol M 3 mol S 3.12 g M Molar mass of M = = = 52.1 g/mol mol M The element is Cr (52.00 g/mol); M is Chromium and M 2 S 3 is chromium(iii) sulfide. 3.10A Plan: If we assume there are 100 grams of this compound, then the masses of carbon and hydrogen, in grams, are numerically equivalent to the percentages. Divide the atomic mass of each element by its molar mass to obtain the moles of each element. Dividing each of the moles by the smaller value gives the simplest ratio of C and H. The smallest multiplier to convert the ratios to whole numbers gives the empirical formula. To obtain the molecular formula, divide the given molar mass of the compound by the molar mass of the empirical formula to find the whole-number by which the empirical formula is multiplied. Assuming 100 g of compound gives g C and 4.79 g H: 1 mol C Moles of C = g C = mol C g C 1 mol H Mole of H = 4.79 g H = mol H g H Divide each of the moles by , the smaller value: C H = C H The value is 5/3, so the moles of C and H must each be multiplied by 3. If it is not obvious that the value is near 5/3, use a trial and error procedure whereby the value is multiplied by the successively larger integer until a value near an integer results. This gives C 5 H 3 as the empirical formula. The molar mass of this formula is: (5 x g/mol) + (3 x g/mol) = g/mol molar mass of compound g/mol Whole-number multiple = = = 4 molar mass of empirical formula g/mol Thus, the empirical formula must be multiplied by 4 to give 4(C 5 H 3 ) = C 20 H 12 as the molecular formula of benzo[a]pyrene. 3.10B Plan: If we assume there are 100 grams of this compound, then the masses of carbon, hydrogen, nitrogen, and oxygen, in grams, are numerically equivalent to the percentages. Divide the atomic mass of each element by its molar mass to obtain the moles of each element. Dividing each of the moles by the smaller value gives the simplest ratio of C, H, N, and O. To obtain the molecular formula, divide the given molar mass of the compound by the molar mass of the empirical formula to find the whole-number by which the empirical formula is multiplied. Assuming 100 g of compound gives g C, 5.19 g H, g N, and g O: Moles of C = g C 1 mol C = mol C g C Moles of H = 5.19 g H 1 mol H = 5.15 mol H g H Moles of N = g N 1 mol N = mol N g N 3-6

7 Moles of O = g O 1 mol O = mol O g O Divide each subscript by the smaller value, 1.030: C4.119H N O1.030= C 4 H 5 N 2 O This gives C 4 H 5 N 2 O as the empirical formula. The molar mass of this formula is: (4 x g/mol) + (5 x g/mol) + (2 x g/mol) + (1 x g/mol) = g/mol The molar mass of caffeine is g/mol, which is larger than the empirical formula mass of g/mol, so the molecular formula must be a whole-number multiple of the empirical formula. molar mass of compound g/mol Whole-number multiple = = molar mass of empirical formula g/mol = 2 Thus, the empirical formula must be multiplied by 2 to give 2(C 4 H 5 N 2 O) = C 8 H 10 N 4 O 2 as the molecular formula of caffeine. 3.11A Plan: The carbon in the sample is converted to carbon dioxide, the hydrogen is converted to water, and the remaining material is chlorine. The grams of carbon dioxide and the grams of water are both converted to moles. One mole of carbon dioxide gives one mole of carbon, while one mole of water gives two moles of hydrogen. Using the molar masses of carbon and hydrogen, the grams of each of these elements in the original sample may be determined. The original mass of sample minus the masses of carbon and hydrogen gives the mass of chlorine. The mass of chlorine and the molar mass of chlorine will give the moles of chlorine. Once the moles of each of the elements have been calculated, divide by the smallest value, and, if necessary, multiply by the smallest number required to give a set of whole numbers for the empirical formula. Compare the molar mass of the empirical formula to the molar mass given in the problem to find the molecular formula. Determine the moles and the masses of carbon and hydrogen produced by combustion of the sample. 1 mol CO2 1 mol C g C g CO2 = mol C = g C g CO2 1 mol CO2 1 mol C 1 mol H2O 2 mol H g H g H2O = mol H = g H g H2O 1 mol H2O 1 mol H The mass of chlorine is given by: g sample ( g C g H) = g Cl The moles of chlorine are: 1 mol Cl g Cl = mol Cl. This is the smallest number of moles g Cl Divide each mole value by the lowest value, : C H Cl = C 3 H 2 Cl The empirical formula has the following molar mass: (3 x g/mol) + (2 x g/mol) + (35.45 g/mol) = g/mol C 3 H 2 Cl molar mass of compound g/mol Whole-number multiple = = molar mass of empirical formula g/mol = 2 Thus, the molecular formula is two times the empirical formula, 2(C 3 H 2 Cl) = C 6 H 4 Cl B Plan: The carbon in the sample is converted to carbon dioxide, the hydrogen is converted to water, and the remaining material is oxygen. The grams of carbon dioxide and the grams of water are both converted to moles. One mole of carbon dioxide gives one mole of carbon, while one mole of water gives two moles of hydrogen. Using the molar masses of carbon and hydrogen, the grams of each of these elements in the original sample may be determined. The original mass of sample minus the masses of carbon and hydrogen gives the mass of oxygen. The mass of oxygen and the molar mass of oxygen will give the moles of oxygen. Once the moles of each of the elements have been calculated, divide by the smallest value, and, if necessary, multiply by the smallest number required to give a set of whole numbers for the empirical formula. Compare the molar mass of the empirical formula to the molar mass given in the problem to find the molecular formula. Determine the moles and the masses of carbon and hydrogen produced by combustion of the sample. 3-7

8 3.516 g CO 2 1 mol CO 2 1 mol C = mol C g C g CO 2 1 mol CO 2 1 mol C g H 2 O 1 mol H 2O 2 mol H g H = mol H g H 2 O 1 mol H 2 O 1 mol H = g C = g H The mass of oxygen is given by: g sample ( g C g H) = g O The moles of C and H are calculated above. The moles of oxygen are: g O 1 mol O = mol O. This is the smallest number of moles g O H Divide each subscript by the smallest value, : C O = C 10 H 14 O This gives C 10 H 14 O as the empirical formula. The molar mass of this formula is: (10 x g/mol) + (14 x g/mol) + (1 x g/mol) = g/mol The molar mass of the steroid is g/mol, which is larger than the empirical formula mass of g/mol, so the molecular formula must be a whole-number multiple of the empirical formula. molar mass of compound g/mol Whole-number multiple = = molar mass of empirical formula g/mol = 2 Thus, the empirical formula must be multiplied by 2 to give 2(C 10 H 14 O) = C 20 H 28 O 2 as the molecular formula of the steroid. 3.12A Plan: In each part it is necessary to determine the chemical formulas, including the physical states, for both the reactants and products. The formulas are then placed on the appropriate sides of the reaction arrow. The equation is then balanced. a) Sodium is a metal (solid) that reacts with water (liquid) to produce hydrogen (gas) and a solution of sodium hydroxide (aqueous). Sodium is Na; water is H 2 O; hydrogen is H 2 ; and sodium hydroxide is NaOH. Na(s) + H 2 O(l) H 2 (g) + NaOH(aq) is the equation. Balancing will precede one element at a time. One way to balance hydrogen gives: Na(s) + 2H 2 O(l) H 2 (g) + 2NaOH(aq) Next, the sodium will be balanced: 2Na(s) + 2H 2 O(l) H 2 (g) + 2NaOH(aq) On inspection, we see that the oxygen is already balanced. b) Aqueous nitric acid reacts with calcium carbonate (solid) to produce carbon dioxide (gas), water (liquid), and aqueous calcium nitrate. Nitric acid is HNO 3 ; calcium carbonate is CaCO 3 ; carbon dioxide is CO 2 ; water is H 2 O; and calcium nitrate is Ca(NO 3 ) 2. The starting equation is HNO 3 (aq) + CaCO 3 (s) CO 2 (g) + H 2 O(l) + Ca(NO 3 ) 2 (aq) Initially, Ca and C are balanced. Proceeding to another element, such as N, or better yet the group of elements in NO 3 gives the following partially balanced equation: 2HNO 3 (aq) + CaCO 3 (s) CO 2 (g) + H 2 O(l) + Ca(NO 3 ) 2 (aq) Now, all the elements are balanced. c) We are told all the substances involved are gases. The reactants are phosphorus trichloride and hydrogen fluoride, while the products are phosphorus trifluoride and hydrogen chloride. Phosphorus trifluoride is PF 3 ; phosphorus trichloride is PCl 3 ; hydrogen fluoride is HF; and hydrogen chloride is HCl. The initial equation is: PCl 3 (g) + HF(g) PF 3 (g) + HCl(g) Initially, P and H are balanced. Proceed to another element (either F or Cl); if we will choose Cl, it balances as: PCl 3 (g) + HF(g) PF 3 (g) + 3HCl(g) The balancing of the Cl unbalances the H, this should be corrected by balancing the H as: PCl 3 (g) + 3HF(g) PF 3 (g) + 3HCl(g) Now, all the elements are balanced. 3-8

9 3.12B Plan: In each part it is necessary to determine the chemical formulas, including the physical states, for both the reactants and products. The formulas are then placed on the appropriate sides of the reaction arrow. The equation is then balanced. a) We are told that nitroglycerine is a liquid reactant, and that all the products are gases. The formula for nitroglycerine is given. Carbon dioxide is CO 2 ; water is H 2 O; nitrogen is N 2 ; and oxygen is O 2. The initial equation is: C 3 H 5 N 3 O 9 (l) CO 2 (g) + H 2 O(g) + N 2 (g) + O 2 (g) Counting the atoms shows no atoms are balanced. One element should be picked and balanced. Any element except oxygen will work. Oxygen will not work in this case because it appears more than once on one side of the reaction arrow. We will start with carbon. Balancing C gives: C 3 H 5 N 3 O 9 (l) 3CO 2 (g) + H 2 O(g) + N 2 (g) + O 2 (g) Now balancing the hydrogen gives: C 3 H 5 N 3 O 9 (l) 3CO 2 (g) + 5/2H 2 O(g) + N 2 (g) + O 2 (g) Similarly, if we balance N we get: C 3 H 5 N 3 O 9 (l) 3CO 2 (g) + 5/2H 2 O(g) + 3/2N 2 (g) + O 2 (g) Clear the fractions by multiplying everything except the unbalanced oxygen by 2: 2C 3 H 5 N 3 O 9 (l) 6CO 2 (g) + 5H 2 O(g) + 3N 2 (g) + O 2 (g) This leaves oxygen to balance. Balancing oxygen gives: 2C 3 H 5 N 3 O 9 (l) 6CO 2 (g) + 5H 2 O(g) + 3N 2 (g) + 1/2O 2 (g) Again clearing fractions by multiplying everything by 2 gives: 4C 3 H 5 N 3 O 9 (l) 12CO 2 (g) + 10H 2 O(g) + 6N 2 (g) + O 2 (g) Now all the elements are balanced. b) Potassium superoxide (KO 2 ) is a solid. Carbon dioxide (CO 2 ) and oxygen (O 2 ) are gases. Potassium carbonate (K 2 CO 3 ) is a solid. The initial equation is: KO 2 (s) + CO 2 (g) O 2 (g) + K 2 CO 3 (s) Counting the atoms indicates that the carbons are balanced, but none of the other atoms are balanced. One element should be picked and balanced. Any element except oxygen will work (oxygen will be more challenging to balance because it appears more than once on each side of the reaction arrow). Because the carbons are balanced, we will start with potassium. Balancing potassium gives: 2KO 2 (s) + CO 2 (g) O 2 (g) + K 2 CO 3 (s) Now all elements except for oxygen are balanced. Balancing oxygen by adding a coefficient in front of the O 2 gives: 2KO 2 (s) + CO 2 (g) 3/2O 2 (g) + K 2 CO 3 (s) Clearing the fractions by multiplying everything by 2 gives: 4KO 2 (s) + 2CO 2 (g) 3O 2 (g) + 2K 2 CO 3 (s) Now all the elements are balanced. c) Iron(III) oxide (Fe 2 O 3 ) is a solid, as is iron metal (Fe). Carbon monoxide (CO) and carbon dioxide (CO 2 ) are gases. The initial equation is: Fe 2 O 3 (s) + CO(g) Fe(s) + CO 2 (g) Counting the atoms indicates that the carbons are balanced, but none of the other atoms are balanced. One element should be picked and balanced. Because oxygen appears in more than one compound on one side of the reaction arrow, it is best not to start with that element. Because the carbons are balanced, we will start with iron. Balancing iron gives: Fe 2 O 3 (s) + CO(g) 2Fe(s) + CO 2 (g) Now all the atoms but oxygen are balanced. There are 4 oxygen atoms on the left hand side of the reaction arrow and 2 oxygen atoms on the right hand side of the reaction arrow. In order to balance the oxygen, we want to change the coefficients in front of the carbon-containing compounds (if we changed the coefficient in front of the iron(iii) oxide, the iron atoms would no longer be balanced). To maintain the balance of carbons, the coefficients in front of the carbon monoxide and the carbon dioxide must be the same. On the left hand side of the equation, there are 3 oxygens in Fe 2 O 3 plus 1X oxygen atoms from the CO (where X is the coefficient in the balanced equation). On the right hand side of the equation, there are 2X oxygen atoms. The number of oxygen atoms on both sides of the equation should be the same: 3 + 1X = 2X 3-9

10 3 = X Balancing oxygen by adding a coefficient of 3 in front of the CO and CO 2 gives: Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) Now all the elements are balanced. 3.13A Plan: Count the number of each type of atom in each molecule to write the formulas of the reactants and products. 6CO(g) + 3O 2 (g) 6CO 2 (g) or, 2CO(g) + O 2 (g) 2CO 2 (g) 3.13B Plan: Count the number of each type of atom in each molecule to write the formulas of the reactants and products. 6H 2 (g) + 2N 2 (g) 4NH 3 (g) or, 3H 2 (g) + N 2 (g) 2NH 3 (g) 3.14A Plan: The reaction, like all reactions, needs a balanced chemical equation. The balanced equation gives the molar ratio between the moles of iron and moles of iron(iii) oxide. The names and formulas of the substances involved are: iron(iii) oxide, Fe 2 O 3, and aluminum, Al, as reactants, and aluminum oxide, Al 2 O 3, and iron, Fe, as products. The iron is formed as a liquid; all other substances are solids. The equation begins as: Fe 2 O 3 (s) + Al(s) Al 2 O 3 (s) + Fe(l) There are 2 Fe, 3 O, and 1 Al on the reactant side and 1 Fe, 3 O, and 2 Al on the product side. Balancing aluminum: Fe 2 O 3 (s) + 2Al(s) Al 2 O 3 (s) + Fe(l) Balancing iron: Fe 2 O 3 (s) + 2Al(s) Al 2 O 3 (s) + 2Fe(l) Moles of Fe 2 O 3 = ( 3 1 mol Fe ) 2 O 3.60 x 10 mol Fe 3 2 mol Fe = 1.80 x 103 mol Fe 2 O 3 Road map: Amount (moles) of Fe Molar ratio (2 mol Fe = 1 mol Fe 2 O 3 ) Amount (moles) of Fe 2 O B Plan: The reaction, like all reactions, needs a balanced chemical equation. The balanced equation gives the molar ratio between the moles of aluminum and moles of silver sulfide. The names and formulas of the substances involved are: silver sulfide, Ag 2 S, and aluminum, Al, as reactants; and aluminum sulfide, Al 2 S 3, and silver, Ag, as products. All reactants and compounds are solids. The equation begins as: Ag 2 S(s) + Al(s) Al 2 S 3 (s) + Ag(s) There are 2 Ag, 1 S, and 1 Al on the reactant side and 2 Al, 3 S, and 1 Ag on the product side. Balancing sulfur: 3Ag 2 S(s) + Al(s) Al 2 S 3 (s) + Ag(s) Balancing silver: 3Ag 2 S(s) + Al(s) Al 2 S 3 (s) + 6Ag(s) Balancing aluminum: 3Ag 2 S(s) + 2Al(s) Al 2 S 3 (s) + 6Ag(s) 2 mol Al Moles of Al = mol Ag 2 S = = mol Al 3 mol Ag 2 S 3-10

11 Road map: Amount (moles) of Ag 2 S Molar ratio (3 mol Ag 2 S = 2 mol Al) Amount (moles) of Al 3.15A Plan: Divide the formula units of aluminum oxide by Avogadro s number to obtain moles of compound. The balanced equation gives the molar ratio between moles of iron(iii) oxide and moles of iron mol Fe2O3 2 mol Fe Moles of Fe = ( 1.85 x 10 Fe2O 3 formula units) x 10 Fe 1 mol Fe 2O 3 formula units 2O = 61.4 mol Fe Road map: No. of Fe 2 O 3 formula units Divide by Avogadro s number (6.022 x Fe 2 O 3 formula units = 1 mol Fe 2 O 3 ) Amount (moles) of Fe 2 O 3 Molar ratio (1 mol Fe 2 O 3 = 2 mol Fe) Amount (moles) of Fe 3.15B Plan: Divide the mass of silver sulfide by its molar mass to obtain moles of the compound. The balanced equation gives the molar ratio between moles of silver sulfide and moles of silver. Moles of Ag = 32.6 g Ag 2 S 1 mol Ag 2 S 6 mol Ag = = mol Ag g Ag 2 S 3 mol Ag 2 S Road map: Mass (g) of Ag 2 S Divide by M (g/mol) (247.9 g Ag 2 S = 1 mol Ag 2 S) Amount (moles) of Ag 2 S Molar ratio (3 mol Ag 2 S = 6 mol Ag) Amount (moles) of Ag 3.16A Plan: The mass of aluminum oxide must be converted to moles by dividing by its molar mass. The balanced chemical equation (follow-up problem 3.14A) shows there are two moles of aluminum for every mole of aluminum oxide. Multiply the moles of aluminum by Avogadro s number to obtain atoms of Al. 3-11

12 Atoms of Al = ( 1.00 g Al O ) Road map: Mass (g) of Al 2 O mol Al2O3 2 mol Al x 10 atoms Al g Al2O3 1 mol Al2O3 1 mol Al = x = 1.18 x atoms Al Divide by M (g/mol) Amount (moles) of Al 2 O 3 Molar ratio Amount (moles) of Al Multiply by Avogadro s number Number of Al atoms 3.16B Plan: The mass of aluminum sulfide must be converted to moles by dividing by its molar mass. The balanced chemical equation (follow-up problem 3.14B) shows there are two moles of aluminum for every mole of aluminum sulfide. Multiply the moles of aluminum by its molar mass to obtain the mass (g) of aluminum. Mass (g) of Al = 12.1 g Al 2 S 3 1 mol Al 2S 3 2 mol Al g Al = = 4.35 g Al 1 mol Al 2 S 3 1 mol Al g Al 2 S 3 Road map: Mass (g) of Al 2 S 3 Divide by M (g/mol) ( g Al 2 S 3 = 1 mol Al 2 S 3 ) Amount (moles) of Al 2 S 3 Molar ratio (1 mol Al 2 S 3 = 2 mol Al) Amount (moles) of Al Multiply by M (g/mol) (1 mol Al = g Al) Mass (g) of Al 3.17A Plan: Write the balanced chemical equation for each step. Add the equations, canceling common substances. Step 1 2SO 2 (g) + O 2 (g) 2SO 3 (g) Step 2 SO 3 (g) + H 2 O(l) H 2 SO 4 (aq) 3-12

13 Adjust the coefficients since 2 moles of SO 3 are produced in Step 1 but only 1 mole of SO 3 is consumed in Step 2. We have to double all of the coefficients in Step 2 so that the amount of SO 3 formed in Step 1 is used in Step 2. Step 1 2SO 2 (g) + O 2 (g) 2SO 3 (g) Step 2 2SO 3 (g) + 2H 2 O(l) 2H 2 SO 4 (aq) Add the two equations and cancel common substances. Step 1 2SO 2 (g) + O 2 (g) 2SO 3 (g) Step 2 2SO 3 (g) + 2H 2 O(l) 2H 2 SO 4 (aq) 2SO 2 (g) + O 2 (g) + 2SO 3 (g) + 2H 2 O(l) 2SO 3 (g) + 2H 2 SO 4 (aq) Or 2SO 2 (g) + O 2 (g) + 2H 2 O(l) 2H 2 SO 4 (aq) 3.17B Plan: Write the balanced chemical equation for each step. Add the equations, canceling common substances. Step 1 N 2 (g) + O 2 (g) 2NO(g) Step 2 NO(g) + O 3 (g) NO 2 (g) + O 2 (g) Adjust the coefficients since 2 moles of NO are produced in Step 1 but only 1 mole of NO is consumed in Step 2. We have to double all of the coefficients in Step 2 so that the amount of NO formed in Step 1 is used in Step 2. Step 1 N 2 (g) + O 2 (g) 2NO(g) Step 2 2NO(g) + 2O 3 (g) 2NO 2 (g) + 2O 2 (g) Add the two equations and cancel common substances. Step 1 N 2 (g) + O 2 (g) 2NO(g) Step 2 2NO(g) + 2O 3 (g) 2NO 2 (g) + 2O 2 (g) N 2 (g) + O 2 (g) + 2NO(g) + 2O 3 (g) 2NO(g) + 2NO 2 (g) + 2O 2 (g) Or N 2 (g) + 2O 3 (g) 2NO 2 (g) + O 2 (g) 3.18A Plan: Count the molecules of each type, and find the simplest ratio. The simplest ratio leads to a balanced chemical equation. The substance with no remaining particles is the limiting reagent. 4 AB molecules react with 3 B 2 molecules to produce 4 molecules of AB 2, with 1 B 2 molecule remaining unreacted. The balanced chemical equation is 4AB(g) + 2B 2 (g) 4AB 2 (g) or 2AB(g) + B 2 (g) 2AB 2 (g) The limiting reagent is AB since there is a B 2 molecule left over (excess). 3.18B Plan: Write a balanced equation for the reaction. Use the molar ratios in the balanced equation to find the amount (molecules) of SO 3 produced when each reactant is consumed. The reactant that gives the smaller amount of product is the limiting reagent. 5 SO 2 molecules react with 2 O 2 molecules to produce molecules of SO 3. The balanced chemical equation is 2SO 2 (g) + O 2 (g) 2SO 3 (g) Amount (molecules) of SO 3 produced from the SO 2 = 5 molecules SO 2 2 molecules SO 3 2 molecules SO 2 = 5 molecules SO 3 Amount (molecules) of SO 3 produced from the O 2 = 2 molecules SO 2 2 molecules SO 3 1 molecule O 2 = 4 molecules SO 3 O 2 is the limiting reagent since it produces less SO 3 than the SO 2 does. 3.19A Plan: Use the molar ratios in the balanced equation to find the amount of AB 2 produced when 1.5 moles of each reactant is consumed. The smaller amount of product formed is the actual amount. 2 mol AB2 Moles of AB 2 from AB = ( 1.5 mol AB) 2 mol AB = 1.5 mol AB

14 2 mol AB2 Moles of AB 2 from B 2 = ( 1.5 mol B2 ) = 3.0 mol AB 2 1 mol B2 Thus AB is the limiting reagent and only 1.5 mol of AB 2 will form. 3.19B Plan: Use the molar ratios in the balanced equation to find the amount of SO 3 produced when 4.2 moles of SO 2 are consumed and, separately, the amount of SO 3 produced when 3.6 moles of O 2 are consumed. The smaller amount of product formed is the actual amount. The balanced chemical equation is 2SO 2 (g) + O 2 (g) 2SO 3 (g) Amount (mol) of SO 3 produced from the SO 2 = 4.2 mol SO 2 2 mol SO 3 2 mol SO 2 = 4.2 mol SO 3 Amount (mol) of SO 3 produced from the O 2 = 3.6 mol SO 2 2 mol SO 3 1 mol O 2 = 7.2 mol SO mol of SO 3 (the smaller amount) will be produced. 3.20A Plan: First, determine the formulas of the materials in the reaction and write a balanced chemical equation. Using the molar mass of each reactant, determine the moles of each reactant. Use molar ratios from the balanced equation to determine the moles of aluminum sulfide that may be produced from each reactant. The reactant that generates the smaller number of moles is limiting. Change the moles of aluminum sulfide from the limiting reactant to the grams of product using the molar mass of aluminum sulfide. To find the excess reactant amount, find the amount of excess reactant required to react with the limiting reagent and subtract that amount from the amount given in the problem. The balanced equation is 2Al(s) + 3S(s) Al 2 S 3 (s) Determining the moles of product from each reactant: 1 mol Al Moles of Al 2 S 3 from Al = (10.0 g Al) g Al 1 mol Al 2S 3 = mol Al 2 S 3 2 mol Al Moles of Al 2 S 3 from S = (15.0 g S) 1 mol S g S 1 mol Al 2S 3 3 mol S = mol Al 2S 3 Sulfur produces less product so it is the limiting reactant. Mass (g) of Al 2 S 3 = ( mol Al 2 S 3 ) g Al 2S 3 1 mol Al 2 S 3 = = 23.4 g Al 2 S 3 The mass of aluminum used in the reaction is now determined: Mass (g) of Al= (15.0 g S) 1 mol S mol Al g Al = g Al used g S 3 mol Al 1 mol Al Subtracting the mass of aluminum used from the initial aluminum gives the mass remaining. Excess Al = Initial mass of Al mass of Al reacted = 10.0 g g = = 1.6 g Al 3.20B Plan: First, determine the formulas of the materials in the reaction and write a balanced chemical equation. Using the molar mass of each reactant, determine the moles of each reactant. Use molar ratios and the molar mass of carbon dioxide from the balanced equation to determine the mass of carbon dioxide that may be produced from each reactant. The reactant that generates the smaller mass of carbon dioxide is limiting. To find the excess reactant amount, find the amount of excess reactant required to react with the limiting reagent and subtract that amount from the amount given in the problem. The balanced equation is: 2C 4 H 10 (g) + 13O 2 (g) 8CO 2 (g) + 10H 2 O(g) Determining the mass of product formed from each reactant: CO 2 Mass (g) of CO 2 from C 4 H 10 = 4.65 g C 4 H 10 1 mol C 4H g C 4 H 10 8 mol CO 2 2 mol C 4 H g CO 2 1 mol CO 2 = = 14.1 g Mass (g) of CO 2 from O 2 = 10.0 g O 2 1 mol O g O 2 8 mol CO 2 13 mol O g CO 2 1 mol CO 2 = = 8.46 g CO

15 Oxygen produces the smallest amount of product, so it is the limiting reagent, and 8.46 g of CO 2 are produced. The mass of butane used in the reaction is now determined: Mass (g) of C 4 H 10 = 10.0 g O 2 1 mol O g O 2 2 mol C 4H mol O g C 4H 10 1 mol C 4 H 10 = = 2.79 g C 4 H 10 used Subtracting the mass of butane used from the initial butane gives the mass remaining. Excess butane = Initial mass of butane mass of butane reacted = 4.65 g 2.79 g = 1.86 g butane 3.21A Plan: Determine the formulas, and then balance the chemical equation. The mass of marble is converted to moles, the molar ratio (from the balanced equation) gives the moles of CO 2, and finally the theoretical yield of CO 2 is determined from the moles of CO 2 and its molar mass. To calculate percent yield, divide the given actual yield of CO 2 by the theoretical yield, and multiply by 100. The balanced equation: CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) Find the theoretical yield of carbon dioxide. 1 mol CaCO3 1 mol CO g CO2 Mass (g) of CO 2 = ( 10.0 g CaCO3 ) g CaCO3 1 mol CaCO3 1 mol CO2 = g CO 2 The percent yield: actual yield theoretical yield ( 100% ) 3.65 g CO g CO 2 = ( 100% ) 2 = = 83.0% 3.21B Plan: Determine the formulas, and then balance the chemical equation. The mass of sodium chloride is converted to moles, the molar ratio (from the balanced equation) gives the moles of sodium carbonate, and finally the theoretical yield of sodium carbonate is determined from the moles of sodium carbonate and its molar mass. To calculate percent yield, divide the given actual yield of sodium carbonate by the theoretical yield, and multiply by 100. The balanced equation: 2NaCl + CaCO 3 CaCl 2 + Na 2 CO 3 Find the theoretical yield of sodium carbonate. 1 mol NaCl Mass (g) of Na 2 CO 3 = 112 g NaCl g NaCl 1 mol Na 2CO g Na 2CO 3 = = 102 g Na 2 CO 3 2 mol NaCl 1 mol Na 2 CO 3 The percent yield: actual yield % yield of Na 2 CO 3 = theoretical yield (100%) = 92.6 g Na 2CO 3 (100%) = = 90.8% 102 g Na 2 CO 3 END OF CHAPTER PROBLEMS 3.1 Plan: The atomic mass of an element expressed in amu is numerically the same as the mass of 1 mole of the element expressed in grams. We know the moles of each element and have to find the mass (in g). To convert moles of element to grams of element, multiply the number of moles by the molar mass of the element. Al amu g/mol Al g Al Mass Al (g) = ( 3 mol Al) 1 mol Al = g Al Cl amu g/mol Cl g Cl Mass Cl (g) = ( 2 mol Cl) 1 mol Cl = g Cl 3-15

16 3.2 Plan: The molecular formula of sucrose tells us that 1 mole of sucrose contains 12 moles of carbon atoms. Multiply the moles of sucrose by 12 to obtain moles of carbon atoms; multiply the moles of carbon atoms by Avogadro s number to convert from moles to atoms. 12 mol C a) Moles of C atoms = ( 1 mol C12H22O11) = 12 mol C 1 mol C12H22O mol C x10 C atoms b) C atoms = ( 2 mol C12H22O11) = 1.445x10 25 C atoms 1 mol C12H22O11 1 mol C 3.3 Plan: Review the list of elements that exist as diatomic or polyatomic molecules. 1 mol of chlorine could be interpreted as a mole of chlorine atoms or a mole of chlorine molecules, Cl 2. Specify which to avoid confusion. The same problem is possible with other diatomic or polyatomic molecules, e.g., F 2, Br 2, I 2, H 2, O 2, N 2, S 8, and P 4. For these elements, as for chlorine, it is not clear if atoms or molecules are being discussed. 3.4 The molecular mass is the sum of the atomic masses of the atoms or ions in a molecule. The molar mass is the mass of 1 mole of a chemical entity. Both will have the same numeric value for a given chemical substance but molecular mass will have the units of amu and molar mass will have the units of g/mol. 3.5 A mole of a particular substance represents a fixed number of chemical entities and has a fixed mass. Therefore the mole gives us an easy way to determine the number of particles (atoms, molecules, etc.) in a sample by weighing it. The mole maintains the same mass relationship between macroscopic samples as exist between individual chemical entities. It relates the number of chemical entities (atoms, molecules, ions, electrons) to the mass. 3.6 Plan: The mass of the compound is given. Divide the given mass by the molar mass of the compound to convert from mass of compound to number of moles of compound. The molecular formula of the compound tells us that 1 mole of compound contains 2 moles of phosphorus atoms. Use the ratio between P atoms and P 4 molecules (4:1) to convert moles of phosphorus atoms to moles of phosphorus molecules. Finally, multiply moles of P 4 molecules by Avogadro s number to find the number of molecules. Roadmap Mass (g) of Ca 3 (PO 4 ) 2 Divide by M (g/mol) Amount (mol) of Ca 3 (PO 4 ) 2 Molar ratio between Ca 3 (PO 4 ) 2 and P atoms Amount (moles) of P atoms Molar ratio between P atoms and P 4 molecules Amount (moles) of P 4 molecules Multiply by 6.022x10 23 formula units/mol Number of P 4 molecules 3-16

17 3.7 Plan: The relative atomic masses of each element can be found by counting the number of atoms of each element and comparing the overall masses of the two samples. a) The element on the left (green) has the higher molar mass because only 5 green balls are necessary to counterbalance the mass of 6 yellow balls. Since the green ball is heavier, its atomic mass is larger, and therefore its molar mass is larger. b) The element on the left (red) has more atoms per gram. This figure requires more thought because the number of red and blue balls is unequal and their masses are unequal. If each pan contained 3 balls, then the red balls would be lighter. The presence of 6 red balls means that they are that much lighter. Because the red ball is lighter, more red atoms are required to make 1 g. c) The element on the left (orange) has fewer atoms per gram. The orange balls are heavier, and it takes fewer orange balls to make 1 g. d) Neither element has more atoms per mole. Both the left and right elements have the same number of atoms per mole. The number of atoms per mole (6.022x10 23 ) is constant and so is the same for every element. 3.8 Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. a) M = (1 x M of Sr) + (2 x M of O) + (2 x M of H) = (1 x g/mol Sr) + (2 x g/mol O) + (2 x g/mol H) = g/mol of Sr(OH) 2 b) M = (2 x M of N) + (3 x M of O) = (2 x g/mol N) + (3 x g/mol O) = g/mol of N 2 O 3 c) M = (1 x M of Na) + (1 x M of Cl) + (3 x M of O) = (1 x g/mol Na) + (1 x g/mol Cl) + (3 x g/mol O) = g/mol of NaClO 3 d) M = (2 x M of Cr) + (3 x M of O) = (2 x g/mol Cr) + (3 x g/mol O) = g/mol of Cr 2 O Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. a) M = (3 x M of N) + (12 x M of H) + (1 x M of P) + (4 x M of O) = (3 x g/mol N) + (12 x g/mol H) + (1 x g/mol P) + (4 x g/mol O) = g/mol of (NH 4 ) 3 PO 4 b) M = (1 x M of C) + (2 x M of H) + (2 x M of Cl) = (1 x g/mol C) + (2 x g/mol H) + (2 x g/mol Cl) = g/mol of CH 2 Cl 2 c) M = (1 x M of Cu) + (1 x M of S) + (9 x M of O) + (10 x M of H) = (1 x g/mol Cu) + (1 x g/mol S) + (9 x g/mol O) + (10 x g/mol H) = g/mol of CuSO 4 5H 2 O d) M = (1 x M of Br) + (3 x M of F) = (1 x g/mol Br) + (3 x g/mol F) = g/mol of BrF Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of 3-17

18 the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. a) M = (1 x M of Sn) + (1 x M of O) = (1 x g/mol Sn) + (1 x g/mol O) = g/mol of SnO b) M = (1 x M of Ba) + (2 x M of F) = (1 x g/mol Ba) + (2 x g/mol F) = g/mol of BaF 2 c) M = (2 x M of Al) + (3 x M of S) + (12 x M of O) = (2 x g/mol Al) + (3 x g/mol S) + (12 x g/mol O) = g/mol of Al 2 (SO 4 ) 3 d) M = (1 x M of Mn) + (2 x M of Cl) = (1 x g/mol Mn) + (2 x g/mol Cl) = g/mol of MnCl Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element multiplied by the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance expressed in g/mol. a) M = (2 x M of N) + (4 x M of O) = (2 x g/mol N) + (4 x g/mol O) = g/mol of N 2 O 4 b) M = (4 x M of C) + (10 x M of H) + (1 x M of O) = (4 x g/mol C) + (10 x g/mol H) + (1 x g/mol O) = g/mol of C 4 H 9 OH c) M = (1 x M of Mg) + (1 x M of S) + (11 x M of O) + (14 x M of H) = (1 x g/mol Mg) + (1 x g/mol S) + (11 x g/mol O) + (14 x g/mol H) = g/mol of MgSO 4 7H 2 O d) M = (1 x M of Ca) + (4 x M of C) + (6 x M of H) + (4 x M of O) = (1 x g/mol Ca) + (4 x g/mol C) + (6 x g/mol H) + (4 x g/mol O) = g/mol of Ca(C 2 H 3 O 2 ) Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. To find the mass in part a), multiply the number of moles by the molar mass of Zn. In part b), first multiply by Avogadro s number to obtain the number of F 2 molecules. The molecular formula tells us that there are 2 F atoms in each molecule of F 2 ; use the 2:1 ratio to convert F 2 molecules to F atoms. In part c), convert mass of Ca to moles of Ca by dividing by the molar mass of Ca. Then multiply by Avogadro s number to obtain the number of Ca atoms g Zn a) (0.346 mol Zn) = 22.6 g Zn 1 mol Zn b) (2.62 mol F 2 ) x 1023 F 2 molecules 1 mol F 2 2 F atoms 1 F 2 molecule = 3.16 x 1024 F atoms 23 1 mol Ca x 10 Ca atoms c) 28.5 g Ca = 4.28 x Ca atoms g Ca 1 mol Ca 3.13 Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. In part a), convert mg units to g units by dividing by 10 3 ; then convert mass of Mn to moles of Mn by dividing by the molar mass of Mn. In part b) convert number of Cu atoms to moles of Cu by dividing by Avogadro s number. In part c) divide by Avogadro s number to convert number of Li atoms to moles of Li; then multiply by the molar mass of Li to find the mass. 3-18

19 a) (62.0 mg Mn) 1 g 1 mol Mn 10 3 mg g Mn = 1.13 x 10-3 mol Mn b) (1.36 x mol Cu Cu atoms) x = mol Cu Cu atoms c) (8.05 x mol Li g Li Li atoms) x = 92.8 g Li Li atoms 1 mol Li 3.14 Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. To find the mass in part a), multiply the number of moles by the molar mass of the substance. In part b), first convert mass of compound to moles of compound by dividing by the molar mass of the compound. The molecular formula of the compound tells us that 1 mole of compound contains 6 moles of oxygen atoms; use the 1:6 ratio to convert moles of compound to moles of oxygen atoms. In part c), convert mass of compound to moles of compound by dividing by the molar mass of the compound. Since 1 mole of compound contains 6 moles of oxygen atoms, multiply the moles of compound by 6 to obtain moles of oxygen atoms; then multiply by Avogadro s number to obtain the number of oxygen atoms. a) M of KMnO 4 = (1 x M of K) + (1 x M of Mn) + (4 x M of O) = (1 x g/mol K) + (1 x g/mol Mn) + (4 x g/mol O) = g/mol of KMnO g KMnO4 Mass of KMnO 4 = ( 0.68 mol KMnO4 ) = = 1.1x10 2 g KMnO 4 1 mol KMnO4 b) M of Ba(NO 3 ) 2 = (1 x M of Ba) + (2 x M of N) + (6 x M of O) = (1 x g/mol Ba) + (2 x g/mol N) + (6 x g/mol O) = g/mol Ba(NO 3 ) 2 1 mol Ba(NO 3) 2 Moles of Ba(NO 3 ) 2 = ( 8.18 g Ba(NO 3) 2 ) = mol Ba(NO 3 ) g Ba(NO 3) 2 6 mol O atoms Moles of O atoms = ( mol Ba(NO 3) 2 ) = = mol O atoms 1 mol Ba(NO 3) 2 c) M of CaSO 4 2H 2 O = (1 x M of Ca) + (1 x M of S) + (6 x M of O) + (4 x M of H) = (1 x g/mol Ca) + (1 x g/mol S) + (6 x g/mol O) + (4 x g/mol H) = g/mol (Note that the waters of hydration are included in the molar mass.) 3 1 mol CaSO4 2H2O Moles of CaSO 4 2H 2 O = ( 7.3x10 g CaSO4 2H2O) = x10 5 mol g CaSO 4 2H 2 O 5 6 mol O atoms Moles of O atoms = ( x10 mol CaSO4 2H2O) 1 mol CaSO 4 2H 2 O = x10 5 mol O atoms x10 O atoms Number of O atoms = ( x10 mol O atoms) 1 mol O atoms = x10 20 = 1.5x10 20 O atoms 3.15 Plan: Determine the molar mass of each substance, then perform the appropriate molar conversions. To find the mass in part a), divide the number of molecules by Avogadro s number to find moles of compound and then multiply the mole amount by the molar mass in grams; convert from mass in g to mass in kg. In part b), first convert mass of compound to moles of compound by dividing by the molar mass of the compound. The molecular formula of the compound tells us that 1 mole of compound contains 2 moles of chlorine atoms; use the 1:2 ratio to convert moles of compound to moles of chlorine atoms. In part c), convert mass of compound to moles of compound by dividing by the molar mass of the compound. Since 1 mole of compound contains 2 moles of H ions, multiply the moles of compound by 2 to obtain moles of H ions; then multiply by Avogadro s number to obtain the number of H ions. 3-19

20 a) M of NO 2 = (1 x M of N) + (2 x M of O) = (1 x g/mol N) + (2 x g/mol O) = 46.01g/mol of NO mol NO 2 Moles of NO 2 = ( 4.6x10 molecules NO2 ) 23 = x x10 molecules NO 3 mol NO g NO2 1 kg Mass (kg) of NO 2 = ( x10 mol NO2 ) 3 = x10 4 = 3.5x10 4 kg NO 2 1 mol NO2 10 g b) M of C 2 H 4 Cl 2 = (2 x M of C) + (4 x M of H) + (2 x M of Cl) = (2 x 12.01g/mol C) + (4 x g/mol H) + (2 x g/mol Cl) = g/mol of C 2 H 4 Cl 2 1 mol C2H4Cl2 Moles of C 2 H 4 Cl 2 = ( g C2H4Cl2 ) = x10 4 mol C 2 H 4 Cl g C2H4Cl2 Moles of Cl atoms = ( 4 2 mol Cl atoms x10 mol C2H4Cl2) = x mol C2H4Cl2 = 1.24x10 3 mol Cl atoms c) M of SrH 2 = (1 x M of Sr) + (2 x M of H) = (1 x g/mol Sr) + (2 x g/mol H) = g/mol of SrH 2 1 mol SrH2 Moles of SrH 2 = ( 5.82 g SrH2 ) = mol SrH g SrH2 Moles of H 2 mol H ions = ( mol SrH2 ) = mol H 1 mol SrH ions 2 23 Number of H 6.022x10 H ions ions = ( mol H ions) = x10 1 mol H 22 = 7.82x10 22 H ions 3.16 Plan: Determine the molar mass of each substance; then perform the appropriate molar conversions. To find the mass in part a), multiply the number of moles by the molar mass of the substance. In part b), first convert the mass of compound in kg to mass in g and divide by the molar mass of the compound to find moles of compound. In part c), convert mass of compound in mg to mass in g and divide by the molar mass of the compound to find moles of compound. Since 1 mole of compound contains 2 moles of nitrogen atoms, multiply the moles of compound by 2 to obtain moles of nitrogen atoms; then multiply by Avogadro s number to obtain the number of nitrogen atoms. a) M of MnSO 4 = (1 x M of Mn) + (1 x M of S) + (4 x M of O) = (1 x g/mol Mn) + (1 x g/mol S) + (4 x g/mol O) = g/mol of MnSO g MnSO4 Mass (g) of MnSO 4 = ( 6.44x10 mol MnSO4 ) = = 9.72 g MnSO 4 1 mol MnSO4 b) M of Fe(ClO 4 ) 3 = (1 x M of Fe) + (3 x M of Cl) + (12 x M of O) = (1 x g/mol Fe) + (3 x g/mol S) + (12 x g/mol O) = g/mol of Fe(ClO 4 ) g Mass (g) of Fe(ClO 4 ) 3 = ( 15.8 kg Fe(ClO 4 ) 3) = 1.58 x 10 1 kg 4 kg Fe(ClO 4 ) mol Fe(ClO 4) 3 Moles of Fe(ClO 4 ) 3 = ( 1.58x10 g Fe(ClO 4) 3) = = 44.6 mol Fe(ClO 4 ) g Fe(ClO 4) 3 c) M of NH 4 NO 2 = (2 x M of N) + (4 x M of H) + (2 x M of O) = (2 x g/mol N) + (4 x g/mol H) + (2 x g/mol O) = g/mol NH 4 NO g Mass (g) of NH 4 NO 2 = ( 92.6 mg NH4NO2) = g NH 1 mg 4 NO

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