CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS


 Kory Carpenter
 11 months ago
 Views:
Transcription
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 31
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 32
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 ) 33
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 34 ) 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 34
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 wholenumber 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 Wholenumber 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 wholenumber 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 36
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 wholenumber multiple of the empirical formula. molar mass of compound g/mol Wholenumber 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 Wholenumber 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. 37
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 wholenumber multiple of the empirical formula. molar mass of compound g/mol Wholenumber 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. 38
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 carboncontaining 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 39
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 310
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 (followup 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. 311
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 (followup 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) 312
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 315
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 316
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 317
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. 318
19 a) (62.0 mg Mn) 1 g 1 mol Mn 10 3 mg g Mn = 1.13 x 103 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. 319
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
Chapter 3 Mass Relationships in Chemical Reactions
Chapter 3 Mass Relationships in Chemical Reactions Student: 1. An atom of bromine has a mass about four times greater than that of an atom of neon. Which choice makes the correct comparison of the relative
More informationChem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses
Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses B. Calculations of moles C. Calculations of number of atoms from moles/molar masses 1. Avagadro
More informationChapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGrawHill 2009 1
Chapter 3 Insert picture from First page of chapter Stoichiometry: Ratios of Combination Copyright McGrawHill 2009 1 3.1 Molecular and Formula Masses Molecular mass  (molecular weight) The mass in amu
More informationChemical Reactions. Chemical Equations. Mole as Conversion Factor: To convert between number of particles and an equivalent number of moles:
Quantities of Reactants and Products CHAPTER 3 Chemical Reactions Stoichiometry Application of The Law of Conservation of Matter Chemical bookkeeping Chemical Equations Chemical equations: Describe proportions
More informationMOLE CONVERSION PROBLEMS. 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles]
MOLE CONVERSION PROBLEMS 1. What is the molar mass of MgO? [40.31 g/mol] 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles] 3. How many moles are present in 2.5 x 10 23 molecules of CH
More informationChapter 3: Stoichiometry
Chapter 3: Stoichiometry Key Skills: Balance chemical equations Predict the products of simple combination, decomposition, and combustion reactions. Calculate formula weights Convert grams to moles and
More informationStoichiometry. Unit Outline
3 Stoichiometry Unit Outline 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis
More informationChapter 3 Stoichiometry
Chapter 3 Stoichiometry 31 Chapter 3 Stoichiometry In This Chapter As you have learned in previous chapters, much of chemistry involves using macroscopic measurements to deduce what happens between atoms
More informationChapter 5, Calculations and the Chemical Equation
1. How many iron atoms are present in one mole of iron? Ans. 6.02 1023 atoms 2. How many grams of sulfur are found in 0.150 mol of sulfur? [Use atomic weight: S, 32.06 amu] Ans. 4.81 g 3. How many moles
More informationChapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT
Lecture Presentation Chapter 3 Chemical Reactions and Reaction James F. Kirby Quinnipiac University Hamden, CT The study of the mass relationships in chemistry Based on the Law of Conservation of Mass
More informationStoichiometry. What is the atomic mass for carbon? For zinc?
Stoichiometry Atomic Mass (atomic weight) Atoms are so small, it is difficult to discuss how much they weigh in grams We use atomic mass units an atomic mass unit (AMU) is one twelfth the mass of the catbon12
More informationMolar Mass Worksheet Answer Key
Molar Mass Worksheet Answer Key Calculate the molar masses of the following chemicals: 1) Cl 2 71 g/mol 2) KOH 56.1 g/mol 3) BeCl 2 80 g/mol 4) FeCl 3 162.3 g/mol 5) BF 3 67.8 g/mol 6) CCl 2 F 2 121 g/mol
More informationFormulae, stoichiometry and the mole concept
3 Formulae, stoichiometry and the mole concept Content 3.1 Symbols, Formulae and Chemical equations 3.2 Concept of Relative Mass 3.3 Mole Concept and Stoichiometry Learning Outcomes Candidates should be
More informationChemical Proportions in Compounds
Chapter 6 Chemical Proportions in Compounds Solutions for Practice Problems Student Textbook page 201 1. Problem A sample of a compound is analyzed and found to contain 0.90 g of calcium and 1.60 g of
More informationStudy Guide For Chapter 7
Name: Class: Date: ID: A Study Guide For Chapter 7 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The number of atoms in a mole of any pure substance
More informationChapter 1 The Atomic Nature of Matter
Chapter 1 The Atomic Nature of Matter 6. Substances that cannot be decomposed into two or more simpler substances by chemical means are called a. pure substances. b. compounds. c. molecules. d. elements.
More informationChapter 6 OxidationReduction Reactions. Section 6.1 2. Which one of the statements below is true concerning an oxidationreduction reaction?
Chapter 6 OxidationReduction Reactions 1. Oxidation is defined as a. gain of a proton b. loss of a proton c. gain of an electron! d. loss of an electron e. capture of an electron by a neutron 2. Which
More informationSample Exercise 3.1 Interpreting and Balancing Chemical Equations
Sample Exercise 3.1 Interpreting and Balancing Chemical Equations The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms.
More informationBalance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O
Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Ans: 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO 2 + 11 H 2 O 3.2 Chemical Symbols at Different levels Chemical symbols represent
More informationMoles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
More informationMass and Moles of a Substance
Chapter Three Calculations with Chemical Formulas and Equations Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows
More informationW1 WORKSHOP ON STOICHIOMETRY
INTRODUCTION W1 WORKSHOP ON STOICHIOMETRY These notes and exercises are designed to introduce you to the basic concepts required to understand a chemical formula or equation. Relative atomic masses of
More information1. How many hydrogen atoms are in 1.00 g of hydrogen?
MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x 1024 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?
More informationWriting and Balancing Chemical Equations
Name Writing and Balancing Chemical Equations Period When a substance undergoes a chemical reaction, chemical bonds are broken and new bonds are formed. This results in one or more new substances, often
More informationProblem Solving. Stoichiometry of Gases
Skills Worksheet Problem Solving Stoichiometry of Gases Now that you have worked with relationships among moles, mass, and volumes of gases, you can easily put these to work in stoichiometry calculations.
More information0.786 mol carbon dioxide to grams g lithium carbonate to mol
1 2 Convert: 2.54 x 10 22 atoms of Cr to mol 4.32 mol NaCl to grams 0.786 mol carbon dioxide to grams 2.67 g lithium carbonate to mol 1.000 atom of C 12 to grams 3 Convert: 2.54 x 10 22 atoms of Cr to
More informationStoichiometry. Lecture Examples Answer Key
Stoichiometry Lecture Examples Answer Key Ex. 1 Balance the following chemical equations: 3 NaBr + 1 H 3 PO 4 3 HBr + 1 Na 3 PO 4 2 C 3 H 5 N 3 O 9 6 CO 2 + 3 N 2 + 5 H 2 O + 9 O 2 2 Ca(OH) 2 + 2 SO 2
More informationMolecular Formula: Example
Molecular Formula: Example A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? 1 CHAPTER 3 Chemical
More informationIB Chemistry. DP Chemistry Review
DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount
More informationMoles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:
Like a recipe: Balancing Eqns Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) coefficients subscripts Balancing Eqns Balancing Symbols (s) (l) (aq) (g) or Yields or Produces solid liquid (pure liquid)
More informationChemical calculations
Chemical calculations Stoichiometry refers to the quantities of material which react according to a balanced chemical equation. Compounds are formed when atoms combine in fixed proportions. E.g. 2Mg +
More informationFormulas, Equations and Moles
Chapter 3 Formulas, Equations and Moles Interpreting Chemical Equations You can interpret a balanced chemical equation in many ways. On a microscopic level, two molecules of H 2 react with one molecule
More informationHow much does a single atom weigh? Different elements weigh different amounts related to what makes them unique.
How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique. What units do we use to define the weight of an atom? amu units of atomic weight. (atomic
More informationIB Chemistry 1 Mole. One atom of C12 has a mass of 12 amu. One mole of C12 has a mass of 12 g. Grams we can use more easily.
The Mole Atomic mass units and atoms are not convenient units to work with. The concept of the mole was invented. This was the number of atoms of carbon12 that were needed to make 12 g of carbon. 1 mole
More informationCh. 10 The Mole I. Molar Conversions
Ch. 10 The Mole I. Molar Conversions I II III IV A. What is the Mole? A counting number (like a dozen) Avogadro s number (N A ) 1 mole = 6.022 10 23 representative particles B. Mole/Particle Conversions
More informationSolution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent
Solution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent Water a polar solvent: dissolves most ionic compounds as well as many molecular compounds Aqueous solution:
More informationChapter 6 Chemical Calculations
Chapter 6 Chemical Calculations 1 Submicroscopic Macroscopic 2 Chapter Outline 1. Formula Masses (Ch 6.1) 2. Percent Composition (supplemental material) 3. The Mole & Avogadro s Number (Ch 6.2) 4. Molar
More informationUnit 10A Stoichiometry Notes
Unit 10A Stoichiometry Notes Stoichiometry is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations
More informationSCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGrawHill Ryerson, 2001
SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGrawHill Ryerson, 2001 1. A small pin contains 0.0178 mol of iron. How many atoms of iron are in the pin? 2. A sample
More informationStoichiometry Review
Stoichiometry Review There are 20 problems in this review set. Answers, including problem setup, can be found in the second half of this document. 1. N 2 (g) + 3H 2 (g) > 2NH 3 (g) a. nitrogen
More informationChapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations
CHE11 Chapter Chapter Stoichiometry: Calculations with Chemical Formulas and Equations 1. When the following equation is balanced, the coefficients are. NH (g) + O (g) NO (g) + H O (g) (a). 1, 1, 1, 1
More informationMole Notes.notebook. October 29, 2014
1 2 How do chemists count atoms/formula units/molecules? How do we go from the atomic scale to the scale of everyday measurements (macroscopic scale)? The gateway is the mole! But before we get to the
More informationChapter 3. Molecules, Compounds, and Chemical Composition
Chapter 3 Molecules, Compounds, and Chemical Composition Elements and Compounds Elements combine together to make an almost limitless number of compounds. The properties of the compound are totally different
More informationConcept 1. The meaning and usefulness of the mole. The mole (or mol) represents a certain number of objects.
Chapter 3. Stoichiometry: MoleMass Relationships in Chemical Reactions Concept 1. The meaning and usefulness of the mole The mole (or mol) represents a certain number of objects. SI def.: the amount of
More informationChapter 3 Calculation with Chemical Formulas and Equations
Chapter 3 Calculation with Chemical Formulas and Equations Practical Applications of Chemistry Determining chemical formula of a substance Predicting the amount of substances consumed during a reaction
More informationGHW#9. Louisiana Tech University, Chemistry 100. POGIL Exercise on Chapter 4. Quantities of Reactants and Products: Equations, Patterns and Balancing
GHW#9. Louisiana Tech University, Chemistry 100. POGIL Exercise on Chapter 4. Quantities of Reactants and Products: Equations, Patterns and Balancing Why? In chemistry, chemical equations represent changes
More informationCHAPTER 3 COMPOUNDS AND MOLECULES
Chapter 3 Compounds and Molecules Page 1 CHAPTER 3 COMPOUNDS AND MOLECULES 31. Octane, a component of gasoline, has eight carbon atoms and eighteen hydrogen atoms per molecule. Its formula is written
More informationChemical Equations. Chemical Equations. Chemical reactions describe processes involving chemical change
Chemical Reactions Chemical Equations Chemical reactions describe processes involving chemical change The chemical change involves rearranging matter Converting one or more pure substances into new pure
More informationOther Stoich Calculations A. mole mass (mass mole) calculations. GIVEN mol A x CE mol B. PT g A CE mol A MOLE MASS :
Chem. I Notes Ch. 12, part 2 Using Moles NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics. 1 MOLE = 6.02 x 10 23 representative particles (representative particles
More informationUnit 2: Quantities in Chemistry
Mass, Moles, & Molar Mass Relative quantities of isotopes in a natural occurring element (%) E.g. Carbon has 2 isotopes C12 and C13. Of Carbon s two isotopes, there is 98.9% C12 and 11.1% C13. Find
More informationReactions. Balancing Chemical Equations uses Law of conservation of mass: matter cannot be lost in any chemical reaction
Reactions Chapter 8 Combustion Decomposition Combination Chapter 9 Aqueous Reactions Exchange reactions (Metathesis) Formation of a precipitate Formation of a gas Formation of a week or nonelectrolyte
More informationWriting, Balancing and Predicting Products of Chemical Reactions.
Writing, Balancing and Predicting Products of Chemical Reactions. A chemical equation is a concise shorthand expression which represents the relative amount of reactants and products involved in a chemical
More informationChapter 7: Chemical Equations. Name: Date: Period:
Chapter 7: Chemical Equations Name: Date: Period: 71 What is a chemical reaction? Read pages 232237 a) Explain what a chemical reaction is. b) Distinguish between evidence that suggests a chemical reaction
More informationChemistry B11 Chapter 4 Chemical reactions
Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl
More informationPart One: Mass and Moles of Substance. Molecular Mass = sum of the Atomic Masses in a molecule
CHAPTER THREE: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Part One: Mass and Moles of Substance A. Molecular Mass and Formula Mass. (Section 3.1) 1. Just as we can talk about mass of one atom of
More informationChapter 3: ex. P2O5 molecular mass = 2(30.97 amu) + 5(16.00 amu) = amu
Molecular Mass and Formula Mass for molecular compounds: the molecular mass is the mass (in amu) of one molecule of the compound molecular mass = atomic masses of elements present Chapter 3: ex. P2O5 molecular
More information2 Stoichiometry: Chemical Arithmetic Formula Conventions (1 of 24) 2 Stoichiometry: Chemical Arithmetic Stoichiometry Terms (2 of 24)
Formula Conventions (1 of 24) Superscripts used to show the charges on ions Mg 2+ the 2 means a 2+ charge (lost 2 electrons) Subscripts used to show numbers of atoms in a formula unit H 2 SO 4 two H s,
More informationAtomic mass is the mass of an atom in atomic mass units (amu)
Micro World atoms & molecules Laboratory scale measurements Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00
More informationChemical Equations & Stoichiometry
Chemical Equations & Stoichiometry Chapter Goals Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of the term
More informationMoles, Molecules, and Grams Worksheet Answer Key
Moles, Molecules, and Grams Worksheet Answer Key 1) How many are there in 24 grams of FeF 3? 1.28 x 10 23 2) How many are there in 450 grams of Na 2 SO 4? 1.91 x 10 24 3) How many grams are there in 2.3
More informationChapter 9. Answers to Questions
Chapter 9 Answers to Questions 1. Word equation: Silicon Tetrachloride + Water Silicon Dioxide + Hydrogen Chloride Formulas: Next, the chemical formulas are needed. As these are all covalent compounds,
More informationCHEMICAL EQUATIONS and REACTION TYPES
31 CHEMICAL EQUATIONS and REACTION TYPES The purpose of this laboratory exercise is to develop skills in writing and balancing chemical equations. The relevance of this exercise is illustrated by a series
More information1. P 2 O 5 2. P 5 O 2 3. P 10 O 4 4. P 4 O 10
Teacher: Mr. gerraputa Print Close Name: 1. A chemical formula is an expression used to represent 1. mixtures, only 3. compounds, only 2. elements, only 4. compounds and elements 2. What is the total number
More informationChapter 5. Chemical Reactions and Equations. Introduction. Chapter 5 Topics. 5.1 What is a Chemical Reaction
Introduction Chapter 5 Chemical Reactions and Equations Chemical reactions occur all around us. How do we make sense of these changes? What patterns can we find? 1 2 Copyright The McGrawHill Companies,
More informationCalculating Atoms, Ions, or Molecules Using Moles
TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary
More informationBALANCING CHEMICAL EQUATIONS
BALANCING CHEMICAL EQUATIONS The Conservation of Matter states that matter can neither be created nor destroyed, it just changes form. If this is the case then we must account for all of the atoms in a
More informationMOLES AND MOLE CALCULATIONS
35 MOLES ND MOLE CLCULTIONS INTRODUCTION The purpose of this section is to present some methods for calculating both how much of each reactant is used in a chemical reaction, and how much of each product
More informationChapter 3! Stoichiometry: Calculations with Chemical Formulas and Equations. Stoichiometry
Chapter 3! : Calculations with Chemical Formulas and Equations Anatomy of a Chemical Equation CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2
More informationBalancing Chemical Equations
Balancing Chemical Equations Academic Success Center Science Tutoring Area Science Tutoring Area Law of Conservation of Mass Matter cannot be created nor destroyed Therefore the number of each type of
More informationAP Chem Unit 1 Assignment 3 Chemical Equations
Symbols used in chemical equations: Symbol Meaning + used to separate one reactant or product from another used to separate the reactants from the products  it is pronounced "yields" or "produces" when
More informationAnswers and Solutions to Text Problems
Chapter 7 Answers and Solutions 7 Answers and Solutions to Text Problems 7.1 A mole is the amount of a substance that contains 6.02 x 10 23 items. For example, one mole of water contains 6.02 10 23 molecules
More informationChemistry I: Using Chemical Formulas. Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu.
Chemistry I: Using Chemical Formulas Formula Mass The sum of the average atomic masses of all elements in the compound. Units are amu. Molar Mass  The mass in grams of 1 mole of a substance. Substance
More informationSolution. Practice Exercise. Concept Exercise
Example Exercise 9.1 Atomic Mass and Avogadro s Number Refer to the atomic masses in the periodic table inside the front cover of this textbook. State the mass of Avogadro s number of atoms for each of
More information1. What is the molecular formula of a compound with the empirical formula PO and a grammolecular mass of 284 grams?
Name: Tuesday, May 20, 2008 1. What is the molecular formula of a compound with the empirical formula PO and a grammolecular mass of 284 grams? 2 5 1. P2O 5 3. P10O4 2. P5O 2 4. P4O10 2. Which substance
More informationstoichiometry = the numerical relationships between chemical amounts in a reaction.
1 REACTIONS AND YIELD ANSWERS stoichiometry = the numerical relationships between chemical amounts in a reaction. 2C 8 H 18 (l) + 25O 2 16CO 2 (g) + 18H 2 O(g) From the equation, 16 moles of CO 2 (a greenhouse
More informationChemistry 65 Chapter 6 THE MOLE CONCEPT
THE MOLE CONCEPT Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present. In order to relate the mass and number of
More informationUnit 9 Stoichiometry Notes (The Mole Continues)
Unit 9 Stoichiometry Notes (The Mole Continues) is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations
More informationChemical Calculations: Formula Masses, Moles, and Chemical Equations
Chemical Calculations: Formula Masses, Moles, and Chemical Equations Atomic Mass & Formula Mass Recall from Chapter Three that the average mass of an atom of a given element can be found on the periodic
More informationHonors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C4.4)
Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1 1. 2. 3. 4. 5. 6. Question What is a symbolic representation of a chemical reaction? What 3 things (values) is a mole of a chemical
More informationChapter 1: Moles and equations. Learning outcomes. you should be able to:
Chapter 1: Moles and equations 1 Learning outcomes you should be able to: define and use the terms: relative atomic mass, isotopic mass and formula mass based on the 12 C scale perform calculations, including
More informationAppendix D. Reaction Stoichiometry D.1 INTRODUCTION
Appendix D Reaction Stoichiometry D.1 INTRODUCTION In Appendix A, the stoichiometry of elements and compounds was presented. There, the relationships among grams, moles and number of atoms and molecules
More informationTopic 4 National Chemistry Summary Notes. Formulae, Equations, Balancing Equations and The Mole
Topic 4 National Chemistry Summary Notes Formulae, Equations, Balancing Equations and The Mole LI 1 The chemical formula of a covalent molecular compound tells us the number of atoms of each element present
More information1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) 3 (s) + H 2 (g)
1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) A) 1 B) 2 C) 4 D) 5 E) Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) Al (s) + H 2 O (l)? Al(OH)
More informationChemical Equations and Chemical Reactions. Chapter 8.1
Chemical Equations and Chemical Reactions Chapter 8.1 Objectives List observations that suggest that a chemical reaction has taken place List the requirements for a correctly written chemical equation.
More informationChapter 6: Chemical Composition
C h e m i s t r y 1 2 C h 6 : C h e m i c a l C o m p o s i t i o n P a g e 1 Chapter 6: Chemical Composition Bonus: 17, 21, 31, 39, 43, 47, 55, 61, 63, 69, 71, 77, 81, 85, 91, 95, 97, 99 Check the deadlines
More informationChapter 3 Molecules, Moles, and Chemical Equations. Chapter Objectives. Warning!! Chapter Objectives. Chapter Objectives
Larry Brown Tom Holme www.cengage.com/chemistry/brown Chapter 3 Molecules, Moles, and Chemical Equations Jacqueline Bennett SUNY Oneonta 2 Warning!! These slides contains visual aids for learning BUT they
More informationIntroductory Chemistry Fourth Edition Nivaldo J. Tro
Introductory Chemistry Fourth Edition Nivaldo J. Tro Chapter 6 Chemical Composition Dr. Sylvia Esjornson Southwestern Oklahoma State University Weatherford, OK 6.1 How Much Sodium? Sodium is an important
More informationPART I: MULTIPLE CHOICE (30 multiple choice questions. Each multiple choice question is worth 2 points)
CHEMISTRY 12307 Midterm #1 Answer key October 14, 2010 Statistics: Average: 74 p (74%); Highest: 97 p (95%); Lowest: 33 p (33%) Number of students performing at or above average: 67 (57%) Number of students
More informationCalculation of Molar Masses. Molar Mass. Solutions. Solutions
Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements
More informationCalculations and Chemical Equations. Example: Hydrogen atomic weight = 1.008 amu Carbon atomic weight = 12.001 amu
Calculations and Chemical Equations Atomic mass: Mass of an atom of an element, expressed in atomic mass units Atomic mass unit (amu): 1.661 x 1024 g Atomic weight: Average mass of all isotopes of a given
More informationOxidationReduction Reactions
OxidationReduction Reactions What is an OxidationReduction, or Redox, reaction? Oxidationreduction reactions, or redox reactions, are technically defined as any chemical reaction in which the oxidation
More informationChemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.
1 Introduction to Chemistry Atomic Weights (Definitions) Chemical Calculations: The Mole Concept and Chemical Formulas AW Atomic weight (mass of the atom of an element) was determined by relative weights.
More informationUNIT (4) CALCULATIONS AND CHEMICAL REACTIONS
UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS 4.1 Formula Masses Recall that the decimal number written under the symbol of the element in the periodic table is the atomic mass of the element. 1 7 8 12
More informationChapter Three: STOICHIOMETRY
p70 Chapter Three: STOICHIOMETRY Contents p76 Stoichiometry  The study of quantities of materials consumed and produced in chemical reactions. p70 31 Counting by Weighing 32 Atomic Masses p78 Mass Mass
More informationHOMEWORK 4A. Definitions. OxidationReduction Reactions. Questions
HOMEWORK 4A OxidationReduction Reactions 1. Indicate whether a reaction will occur or not in each of following. Wtiring a balcnced equation is not necessary. (a) Magnesium metal is added to hydrochloric
More informationChemical Composition Review Mole Calculations Percent Composition. Copyright Cengage Learning. All rights reserved. 8 1
Chemical Composition Review Mole Calculations Percent Composition Copyright Cengage Learning. All rights reserved. 8 1 QUESTION Suppose you work in a hardware store and a customer wants to purchase 500
More informationneutrons are present?
AP Chem Summer Assignment Worksheet #1 Atomic Structure 1. a) For the ion 39 K +, state how many electrons, how many protons, and how many 19 neutrons are present? b) Which of these particles has the smallest
More informationChem 115 POGIL Worksheet  Week 4 Moles & Stoichiometry
Chem 115 POGIL Worksheet  Week 4 Moles & Stoichiometry Why? Chemists are concerned with mass relationships in chemical reactions, usually run on a macroscopic scale (grams, kilograms, etc.). To deal with
More informationThe Mole Concept. The Mole. Masses of molecules
The Mole Concept Ron Robertson r2 c:\files\courses\111020\2010 final slides for web\mole concept.docx The Mole The mole is a unit of measurement equal to 6.022 x 10 23 things (to 4 sf) just like there
More informationBalancing Chemical Equations Worksheet
Balancing Chemical Equations Worksheet Student Instructions 1. Identify the reactants and products and write a word equation. 2. Write the correct chemical formula for each of the reactants and the products.
More information