# The Mole. Chapter 2. Solutions for Practice Problems

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1 Chapter 2 The Mole Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the full set of steps: Problem, What Is Required, What Is Given,, Act on Your Strategy, and. Where a shorter solution is appropriate, the Sample Problem contains only two steps: Problem and Solution. Where relevant, a Check Your Solution step is also included in the shorter Sample Problems. Solutions for Practice Problems Student Textbook page Problem The two stable isotopes of boron exist in the following proportions: 19.78% 11 (10.01 u) and 80.22% B (1.01 u). Calculate the average atomic mass of 2 boron. You need to find the average atomic mass of boron. You are given the relative abundance (proportions) and the atomic mass of each stable isotope. Express each isotopic proportion as a decimal and multiply this by its atomic mass. The average mass of boron is the sum of these products. Average atomic mass of B = u (0.1978) u (0.8022) = u Boron contains far more boron-11 than boron-10. Therefore, the atomic mass of the naturally occurring boron should be closer to u than to u. The calculated atomic mass matches this reasoning. A check in the periodic table confirms the value. 2. Problem In nature, silicon is composed of three isotopes. These isotopes (with their isotopic 28 abundances and atomic masses) are Si 29 (92.5, u), Si (4.67%, u), 30 and Si (3.10%, u). 14 Calculate the average mass of silicon. You need to find the atomic mass of silicon. You are given the relative abundance and the atomic mass of each isotope B 19

2 Express each isotopic proportion as a decimal and multiply this by its atomic mass. The average mass of silicon is the sum of these products. Average atomic mass of Si = (27.98 u)(0.92) + (28.97 u) (0.0467) + (29.97 u)(0.0310) = u Silicon contains far more silicon-28, than silicon-29 or silicon-30. Therefore, the atomic mass of the naturally occurring silicon should be closer to u than to u or u. The calculated atomic mass matches this reasoning and a check in the periodic table confirms the value. 3. Problem Copper is a corrosion-resistant metal that is used extensively in plumbing and wiring. 63 Copper exists as two naturally occurring isotopes: Cu 65 (62.93 u) and Cu (64.93 u). These isotopes have isotopic abundances of 69.1% and 30.9%, respectively. Calculate the average atomic mass of copper. You need to find the atomic mass of copper. You are given the relative abundance and the atomic mass of each isotope. Express each isotopic proportion as a decimal and multiply this by its atomic mass. The average mass of copper is the sum of these products. Average atomic mass of Cu = (62.93 u)(0.691) + (64.93 u)(0.309) = u Copper contains more copper-63 than copper-65. Therefore, the atomic mass of the naturally occurring copper should be closer to u than to u. The mean of the two isotopic masses is u. So, the calculated atomic mass of copper, which is lower than this mean value, is a reasonable answer. A check in the periodic table confirms the atomic mass calculated. 4. Problem Rubidium has as two isotopes: rubidum-85 (84.91 u) and rubidium-87 (86.91 u). If the average atomic mass of rubidium is u, determine the percentage abundance of each isotope. You have to find the percentage abundance of rubidium-85 and rubidium-87. The atomic mass of each isotope is given, as well as the average atomic mass of rubidium. First express the abundance of each isotope as a decimal instead of a percentage, giving a total abundance of both isotopes as 1. Let the abundance of be x and Rb 20

3 87 Rb 37 be 1 - x. Set up the equation and solve for x. Then multiply x by 100% to express it as percentage abundance. Average atomic mass = x( mass rubidium-85 ) + ( 1 - x) (mass rubidium-87) u = x(84.91) + ( 1 - x) (86.91) = x x 2x = x = 072. Therefore, the percentage abundance of rubidium is % = 72%. The percentage abundance of rubidium-87, then, is 28%. The mean of the atomic masses of the two isotopes is u. The average atomic mass of rubidium (85.47 u) is slightly less than this mean, and closer to the value of rubidium-85 (84.91 u). So you would expect rubidium to comprise of a greater abundance of rubidium-85, which matches the result. Solutions for Practice Problems Student Textbook pages Problem A small pin contains mol of iron, Fe. How many atoms of iron are in the pin? You need to find the number of atoms of iron in the pin. The pin contains mol of Fe. N A = atoms/mol. N = N A n The number of atoms would be the Avogadro constant multiplied by the number of moles of Fe in the pin. The number of atoms of Fe = mol ( atoms/mol) = atoms Therefore, there are iron atoms in the pin. Work backwards. Dividing the number of atoms by the Avogadro constant should give you the number of moles of Fe. 6. Problem A sample contains mol of gold, Au. How many atoms of gold are there in the sample? You need to find the number of atoms of gold in the sample. The sample contains mol of Au. N A = atoms/mol 21

4 N = N A n The number of atoms would be the Avogadro constant multiplied by the number of moles of gold in the sample. The number of atoms of Au = mol ( atoms/mol) = atoms Therefore, there are gold atoms in the sample. Work backwards. Dividing the number of atoms by the Avogadro constant should give you the number of moles of Au. 7. Problem How many formula units are contained in 0.21 mol of magnesium nitrate, Mg(NO 3 ) 2? You need to find the number of formula units in the magnesium nitrate. The sample has 0.21 mol. N A = formula units/mol N = N A n The number of formula units would be the Avogadro constant multiplied by the number of moles of Mg(NO 3 ) 2. The number of formula units of Mg(NO 3 ) 2 = 0.21 mol ( formula units/mol) = formula units Therefore, there are formula units of Mg(NO 3 ) 2 in the sample. Work backwards. Dividing the number of formula units by Avogadro constant should give you the number of moles of Mg(NO 3 ) Problem A litre of water contains 55.6 mol of water. How many molecules of water are in this sample? You need to find the number of molecules of water in the 1 L volume. There is 55.6 mol of water in the 1 L volume. N A = molecules/mol. N = N A n The number of molecules would be the Avogadro constant multiplied by the number of moles of water. The number of molecules of H 2 O = 55.6 mol ( molecules/mol) = molecules Therefore, there are water molecules in one litre of water. 22

5 Work backwards. Dividing the number of molecules by Avogadro s number should give you the number of moles of water. 9. Problem Ethyl acetate, C 4 H 8 O 2, is frequently used in nail polish remover. A typical bottle of nail polish remover contains about 2.5 mol of ethyl acetate. (a) How many molecules are in the bottle of nail polish remover? (b) How many atoms are in the bottle? (c) How many carbon atoms are in the bottle? (a) You need to find the number of molecules in the sample. (b) You need to find the number of atoms in the bottle. (c) You need to find the number of carbon atoms in the bottle. The sample consists of 2.5 mol. The molecular formula of ethyl acetate is C 4 H 8 O 2. The Avogadro constant N A = molecules/mol. (a) N = N A n The number of molecules would be the Avogadro constant multiplied by the number of moles of ethyl acetate. (b) The number of atoms per molecule, from the molecular formula of ethyl acetate, is 14. The total number of atoms in the bottle would be 14 multiplied by the number of molecules. (c) The number of carbon atoms in the molecular formula of ethyl acetate is 4. The total number of atoms in the bottle would be 4 multiplied by the number of molecules. (a) Number of molecules of C 4 H 8 O 2 = 2.5 mol ( molecules/mol) = molecules Therefore, there are molecules in the bottle. (b) The number of atoms = 14 atoms/molecule ( molecules) = atoms Therefore, there are atoms in the bottle. (c) The number of C atoms = 4 atoms/molecule ( molecules) = atoms Therefore, there are atoms of carbon in the bottle. (a) Work backwards. Dividing the number of molecules by Avogadro s number should give you the number of moles of ethyl acetate. (b) Work backwards. Dividing the number of atoms by the number of molecules in the bottle should give you the number of atoms in each molecule. (c) Work backwards. Dividing the number of carbon atoms in the bottle by the number of molecules in the bottle should give you the number of carbon atoms in each molecule. 10. Problem Consider a mol sample of sodium sulfate, Na 2 SO 4. (a) How many formula units are in the sample? (b) How many sodium ions, Na +, are in the sample?

6 (a) You need to find the number of formula units in the sample. (b) You need to find the number of sodium ions in the bottle. (c) How many sulfate ions, SO 4 2 (aq) are in the sample? (d) How many oxygen atoms are in the sample? The sample consists of mol. The molecular formula of sodium sulfate is Na 2 SO 4. N A = formula units/mol (a) The number of formula units would be the Avogadro constant multiplied by the number of moles of sodium sulfate. (b) There are 2 sodium ions per formula unit of sodium sulfate. The total number of sodium ions would be 2 multiplied by the number of formula units in the sample. 2 2 (c) There is one SO 4 (aq) ion per formula unit. The number of SO 4 (aq) ions will be the same as the number of formula units. (d) There are 4 sodium ions per formula unit. The total number of oxygen atoms will be 4 multiplied by the number of formula units. (a) Number of formula units = mol ( formula units/mol) = formula units Therefore, there are formula units in the sample. (b) The number of sodium ions = 2 ions/formula unit ( formula units) = sodium ions. Therefore, there are sodium ions in the sample. 2 (c) number of SO 4 (aq) ions = (d) number of oxygen atoms = = (a) Work backwards. Dividing the number of formula units by Avogadro s number should give you the number of mol of sodium sulfate. (b) Work backwards. Dividing the number of sodium ions by the number of formula units in the sample should give you the number of ions per formula unit. (c) Work backwards. Dividing the number of SO 4 2 (aq) ions by the number of formula units in the sample should give the number of ions per formula unit. (d) Work backwards. Dividing the number of oxygen atoms by the number of formula units in the sample should give the number of atoms per formula unit. 11. Problem A certain coil of pure copper wire has a mass of 5.00 g. This mass of copper corresponds to mol of copper. (a) How many atoms of copper are in the coil? (b) How many atoms of copper would there be in a coil of pure copper wire with a mass of g? (c) How many atoms of copper would there be in a coil of pure copper wire with a mass of 1.00 g? What Is Required? (a) You need to find the number of copper atoms in the coil. (b) You need to find the number of copper atoms in a coil with a mass of 10.0 g. (c) You need to find the number of copper atoms in a coil with a mass of 1.00 g. 24

7 What Is Given? m = 5.00 g n = mol (a) Use the formula N = n N A to determine the number of copper atoms in the coil. (b) Multiply your answer for (a) by 2. (c) Divide your answer for (b) by 10. atoms (a) N = mol mol = atoms (b) atoms = atoms 22 (c) atoms = atoms 10 (a) Work backwards. Dividing the number of atoms by Avogadro s number will give you the number of moles of copper. Solutions for Practice Problems Student Textbook page Problem A sample of bauxite ore contains molecules of aluminum oxide, Al 2 O 3. How many moles of aluminum oxide are in the sample? You need to find the number of moles in molecules of aluminum oxide. The number of molecules is N A = molecules/mol n = N N A Divide the number of molecules by the Avogadro constant to get the number of moles of aluminum oxide molecules Number of moles = molecules/mol = mol Therefore, there are 12.8 moles of aluminum oxide in the sample. Work backwards. Dividing the number of molecules by the number of moles should give you Avogadro s number. 13. Problem A vat of cleaning solution contains molecules of ammonia, NH 3. How many moles of ammonia are in the vat? You need to find the number of moles in molecules of ammonia. The number of molecules is N A = molecules/mol 25

8 Divide the number of molecules by Avogadro s number to get the number of moles of ammonia. 26 Number of moles molecules = ( molecules/mol) 3 = mol Therefore, there are mol of ammonia in the vat. Work backwards. Dividing the number of molecules by the number of moles should give you Avogadro s number. 14. Problem A sample of cyanic acid, HCN, contains atoms. How many moles of cyanic acid are in the sample? Hint: Find the number of molecules of HCN first. You need to find the number of moles of HCN in atoms of HCN. The number of atoms is The formula of cyanic acid is given. N A = molecules/mol From the formula, you know there are 3 atoms in every molecule of cyanic acid. The number of molecules of cyanic acid is the number of atoms divided by 3. Then, divide the number of molecules by the Avogadro constant to get the number of mol of HCN atoms Number of molecules of HCN = 3 atoms/molecule 22 = mol molecules Number of moles of HCN = ( molecules/mol) -2 = mol Therefore, there are mol or mol of cyanic acid in the sample. Work backwards. Dividing the number of molecules by the number of moles should give you the Avogadro constant. 15. Problem A sample of pure acetic acid, CH 3 COOH, contains 1.40 many moles of acetic acid are in the sample? 10 carbon atoms. How You need to find the number of moles of acetic acid that carries carbon atoms. The number of carbon atoms is The molecular formula of acetic acid is CH 3 COOH. N A = molecules/mol From the formula, you know there are 2 carbon atoms in every molecule of acetic acid. The number of molecules of acetic acid is the number of carbon atoms divided 26

9 by 2. Then, divide the number of molecules by the Avogadro constant to get the number of mol of acetic acid C atoms Number of molecules of acetic acid = 2 C - atoms/molecule Number of moles of acetic acid = molecules ( / molecules/mol) = mol Therefore, there are mol or mol of acetic acid in the sample. Work backwards. Dividing the number of molecules by the number of moles should give you the Avogadro constant. Solutions for Practice Problems Student Textbook page Problem State the molar mass of each element. (a) xenon, Xe (b) osmium, Os (c) barium, Ba (d) tellurium, Te You have to find the molar mass of the elements listed. The chemical symbols and chemical names are given. With no other information given, the only source of the answer is a periodic table. The average atomic mass of the element is numerically equivalent to its molar mass. From the average atomic mass in the periodic table, the numerically equivalent molar masses are: (a) Xe: g/mol (b) Os: 190. g/mol (c) Ba: g/mol (d) Te: g/mol Periodic table values can vary in some publications, depending on the number of significant digits recorded. Check with other books or online sources to verify the most commonly accepted atomic mass units (and hence the molar mass) of the elements. 17. Problem Find the molar mass of each compound. (a) ammonia, NH 3 (b) glucose, C 6 H 12 O 6 (c) potassium dichromate, K 2 Cr 2 O 7 (d) iron(iii) sulfate, Fe 2 (SO 4 ) 3 = You have to find the molar masses of the compounds listed. 22 molecules 27

10 The chemical formulas of the compounds are given. The molar mass of a compound is the sum of all the molar masses of its component elements. Find the atomic mass (which is numerically equivalent to the molar mass) of the elements from the periodic table and add them together. The molar masses of the compounds are: (a) NH 3 : g/mol + 3(1.01 g/mol) = g/mol (b) C 6 H 12 O 6 : 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = g/mol (c) K 2 Cr 2 O 7 : 2(39.10 g/mol) + 2(52.00 g/mol) + 7(16.00 g/mol) = g/mol (d) Fe 2 (SO 4 ) 3 : 2(55.85 g/mol) + 3[32.07 g/mol + 4(16.00 g/mol)] = g/mol Using round numbers for a quick check, for example in the case of (a), = This estimate is close to the answer of g/mol. 18. Problem Strontium may be found in nature as celestite, SrSO 4. Find the molar mass of celestite. You have to find the molar mass of celestite. The chemical formula of celestite is SrSO 4. The average atomic mass can be obtained from any Periodic Table. The molar mass of celestite is the sum of all the molar masses of its component elements. Find the atomic mass of 1 mol of Sr, 1 mol of S, and 4 mol of O from the Periodic Table and add them together. Molar mass of SrSO 4 = g/mol g/mol + 4(16.00 g/mol) = g/mol Using round numbers for a quick check, you get (4 16) = 183. This estimate is close to the answer of g/mol. 19. Problem What is the molar mass of the ion [Cu(NH 3 ) 4 ] 2+? You have to find the molar mass of the [Cu(NH 3 ) 4 ] 2+ ion. The chemical formula of the ion is [Cu(NH 3 ) 4 ] 2+. The average atomic mass can be obtained from any Periodic Table. The molar mass of the ion is the sum of all the molar masses of its component elements. Find the atomic mass of 1 mol of Cu, 4 mol of N, and 12 mol of H from the Periodic Table and add them together. The mass of the two missing electrons from each ion may be neglected since the mass of an electron is extremely small compared to the mass of an atom. 28

11 Molar mass of [Cu(NH 3 ) 4 ] 2+ = g/mol + 4(14.01 g/mol) + 12(1.01 g/mol) = g/mol g/mol g/mol = g/mol Using round numbers for a quick check, you get = 131. This estimate is close to the answer of g/mol. Solutions for Practice Problems Student Textbook page Problem Calculate the mass of each molar quantity. (a) 3.90 mol of carbon, C (b) 2.50 mol of ozone, O 3 (c) mol of propanol, C 3 H 8 O (d) mol of ammonium dichromate, (NH 4 ) 2 Cr 2 O 7 You need to find the mass of the substances listed. The number of moles of the substances are given as well as the chemical formulas. The molar mass of the elements that make up the substances can be found in a Periodic Table. In order to convert moles to grams, you need to determine the molar mass (M) of the substance. The molar mass of a compound is the sum of all the molar masses of its component elements. Find the atomic mass (which is numerically equivalent to the molar mass) of the elements from the periodic table and add them together. Then multiply the molar mass by the given number of moles to obtain the mass (m) of the substance. (a) For C: M = g/mol m = 3.90 mol g/mol = 46.8 g (b) For O 3 : M = 3(16.00 g/mol) = g/mol m = 2.50 mol g/mol = 120 g (c) For C 3 H 8 O:M = 3(12.01 g/mol) + 8(1.01 g/mol) g/mol = g/mol m = mol g/mol = g (d) For (NH 4 ) 2 Cr 2 O 7 : M = 2[14.01 g/mol + 4(1.01 g/mol) + 2(52.00 g/mol) + 7(16.00 g/mol) = g/mol m = mol g/mol = g 29

12 Work backwards. Divide the calculated mass by the number of moles given. It should g give the molar mass of the substance. For example, in (a), = 12.0 g/mol. This matches the molar mass of carbon mol 21. Problem For each group, which sample has the largest mass? (a) 5.00 mol of C, 1.50 mol of Cl 2, 0.50 mol of C 6 H 12 O 6 (b) 7.31 mol of O 2, 5.64 mol of CH 3 OH, 12.1 mol of H 2 O You need to find the substance with the largest mass among the substances listed. The number of moles of the substances are given as well as the chemical formulas. The molar mass of the elements that make up the substances can be found in a Periodic Table. In order to convert moles to grams, you need to determine the molar mass (M) of the substance. The molar mass of a compound is the sum of all the molar masses of its component elements. Find the atomic mass (which is numerically equivalent to the molar mass) of the elements from the periodic table and add them together. Then multiply the molar mass by the given number of moles to obtain the mass (m) of the substance. (a) For C: M = g/mol m = 5.00 mol g/mol = g For Cl 2 : M = g/mol m = 1.50 mol g/mol = g For C 6 H 12 O 6 : M = g/mol m = 0.50 mol g/mol = g Therefore, the sample of Cl 2 has the largest mass. (b) For O 2 : M = g/mol; m = 7.31 mol g/mol = 4 g For CH 3 OH: M = g/mol; m = 5.64 mol g/mol = 181 g For H 2 O: M = g/mol; m = 12.1 mol g/mol = 218 g Therefore, the sample of O 2 has the largest mass. Work backward to check the mass. Divide the calculated mass by the number of moles given. It should give the molar mass of the substance. For example, for C in (a), 60.6 g =12.0 g/mol. This matches the molar mass of carbon mol 22. Problem A litre, 1000 ml, of water contains 55.6 mol. What is the mass of a litre of water. You need to find the mass of 1000 ml of water. The number of moles of water is given as well as the volume. The molar mass of the elements that make up water can be found in a Periodic Table. 30

13 In order to convert moles to grams, you need to determine the molar mass (M) of water. Find the atomic mass (which is numerically equivalent to the molar mass) for 2 mol H and 1 mol O from the periodic table and add them together. Then multiply the molar mass by the given number of moles to obtain the mass (m). Molar mass = 2(1.01 g/mol) g/mol = g/mol m = 55.6 mol g/mol = g = g Since the density of water is 1.0 g/ml, the mass of 1000 ml is 1000 g or 1 kg. This agrees with the result calculated by the mole method.. Problem To carry out a particular reaction, a chemical engineer needs 255 mol of styrene, C 8 H 8. How many kilograms of styrene does the engineer need? You need to find the mass of styrene needed. The number of moles of the styrene is given as well as the chemical formula. The molar mass of the elements that make up styrene can be found in a Periodic Table. In order to convert moles to grams, you need to determine the molar mass (M) of the styrene. The molar mass of a compound is the sum of all the molar masses of its component elements. Find the atomic mass (which is numerically equivalent to the molar mass) of the elements from the periodic table and add them together. Then multiply the molar mass by the given number of moles to obtain the mass (m) of the substance. Divide the mass in gram by 10 3 g/kg for a kilogram conversion. M = 8(12.01 g/mol) + 8(1.01 g/mol) = g/mol m = 255 mol g/mol = g Mass in kilograms = g/10 3 g/kg = kg Therefore, the engineer needs 26.6 kg of styrene for the reaction. Work backwards. Divide the calculated mass by the number of moles given. It should g give the molar mass of the substance. = 104 g mol, which agrees with 255 mol the calculations. Solutions for Practice Problems Student Textbook page Problem Calculate the number of moles in each sample. (a) 103 g of Mo (b) g of Pd (c) kg of Cr (d) 56.3 mg of Ge You need to find the number of moles in the given mass of elements listed. 31

14 The masses of the elements are given. You have access to a periodic table. In each case, determine the molar mass (M) from a periodic table and divide it into the given sample mass (m) to obtain the number of moles (n). Use n = m. In the M case of (c) and (d), first convert the given amount to grams before proceeding with the calculation. Act On Your Strategy (a) For Mo: M = g/mol; m = 103 g 103 g n = = 107. mol g/mol (b) For Pd: M = g/mol; m = g g n = = 124 mol g/mol (c) For Cr: M = g/mol; m = kg 10 3 g/kg = 736 kg 736 g n = = mol g/mol (d) For Ge: M = g/mol; m = 56.3 mg 103 mg/g = g g - n = = mol g/mol In each case, work backward to reestablish the given mass of the element. For example, in (a), 1.07mol g/mol = 103 g. This value matches the amount given in the question. 25. Problem How many moles of compound are in each sample? (a) 39.2 g of silicon dioxide, SiO 2 (b) 7.34 g of nitrous acid, HNO 2 (c) kg of carbon tetrafluoride, CF 4 (d) mg of 1-iodo-2,3,-dimethylbenzene, C 8 H 9 I You need to find the number of moles in the given mass of substances listed. The masses of the substances are given as well as the chemical formulas. You have access to a periodic table. In each case, determine the molar mass (M) from a periodic table and divide it into the given sample mass (m) to obtain the number of moles (n). Use n = m. In the M case of (c) and (d), first convert the given amount to grams before proceeding with the calculation. Act On Your Strategy (a) For SiO 2 : M = g/mol; m = 39.2 g 39.2 g n = = mol g/mol (b) For HNO 2 : M = g/mol; m = 7.34 g 7.34 g n = = mol g/mol 32

15 3 8 (c) For CF 4 : M = g/mol; m = ( g 10 kg) 10 g/kg = g g n = 10 3 g/kg = mol g/mol (d) For C 8 H 9 I: M = 2.07 g/mol; m = mg/10 3 mg/g = g g -8 n = = mol 2.07 g/mol In each case, work backward to reestablish the given mass of the element. For example, in (a), mol g/mol = 39.2 g. This value matches the amount given in the question. 26. Problem Sodium chloride, NaCl, can be used to melt snow. How many moles of sodium chloride are in a 10 kg bag? You need to find the number of moles of NaCl in the bag. The mass of the salt is 10 kg. You have access to a Periodic Table. Determine the molar mass (M) from a Periodic Table and divide it into the given sample mass (m) to obtain the number of moles (n). Use n = m. Remember to convert the given amount to grams before proceeding with the calculation. M Act On Your Strategy M = g/mol; m = 10 kg 10 3 g/kg = g 10,000 g -2 n = = mol g/mol There is mol of salt in 10 kg. Work backward to reestablish the given mass of the salt. ( mol) g/mol = g. This value closely matches the amount given in the question. 27. Problem Octane, C 8 H 18, is a principal ingredient of gasoline. Calculate the number of moles in a 20.0 kg sample of octane. You need to find the number of moles of octane in the sample of gasoline. The mass of the octane is 20.0 kg and its chemical formula is given. You have access to a Periodic Table. Determine the molar mass (M) of octane using the atomic mass units from a periodic table. Divide M into the given sample mass (m) to obtain the number of moles (n). Use n = m. Remember to convert the given amount to grams before proceeding M with the calculation. Act On Your Strategy M = g/mol; m = 20.0 kg 10 3 g/kg = g; 20,000 g n = = 175 mol g/mol 5 33

16 Work backward to reestablish the given mass of octane. 175 mol g/mol = g or 20.0 kg. This value closely matches the 20.0 kg amount given in the question. Solutions for Practice Problems Student Textbook page Problem Determine the mass of each sample. (a) formula units of ZnCl 2 (b) formula units of Pb 3 (PO 4 ) 2 (c) molecules of C 15 H 21 N 3 O 15 (d) molecules of N 2 O 5 You need to find the masses of the substances listed. You are given the number of formula units or molecules of the substances. N A = formula units/mol Convert the number of formula units or the number of molecules into moles (n) by dividing by the Avogadro constant. Then multiply the number of moles by the molar mass (M) to get the mass in grams (m) of the substance. 24 ( formula units) (a) n = = mol ( ) formula units / mol M of ZnCl 2 = g / mol; 3 Therefore, m = mol g / mol = g or g 21 (b) n = formula units = mol ( ) formula units / mol M of Pb 3 (PO 4 ) 2 = g/mol; Therefore, m = 0.01 mol g/mol = 9.98 g (c) ( molecules) n = = 151. mol ( ) molecules / mol M of C 15 H 21 N 3 O 15 = g/mol; Therefore, m = 1.51 mol g/mol = g 29 (d) ( molecules) 5 n = = mol ( ) molecules / mol M of N 2 O 5 = g/mol; Therefore, m = 1.99 mol 10 5 g/mol g/mol = 9.98 g or kg. Work backwards. Dividing the mass by the number of moles should give the molar mass of the substance. For example, in (a), 10.0 mol = 136. The results match. 29. Problem What is the mass of lithium in 254 formula units of lithium chloride, LiCl? You need to find the mass of lithium in the LiCl. 34

17 You are given the number of formula units of lithium chloride as well as the chemical formula. N A = formula units/mol. From the chemical formula, you know there is 1 unit of lithium for every formula unit of lithium chloride, so you can assume that there are 254 units of lithium in the sample. Convert the number of units of the lithium into moles (n) by dividing by the Avogadro constant. Then multiply the number of moles by the molar mass (M) of lithium to get the mass in grams (m) of the lithium. n = 254 formula units -22 = mol ( ) formula units / mol M of Li = 6.94 g/mol; Therefore, m = ( ) mol 6.94 g/mol = g. Work backwards. Dividing the mass by the number of moles should give the molar mass of lithium. = 6.94 g/mol. The results match g mol 30. Problem A sample of potassium oxide, K 2 O, contains K + ions. What is the mass of the sample? You must find the mass of potassium oxide in that contains K + ions. You know the number of K + ions and the Avogadro constant is ions/mol. Convert the number of ions to moles (n) by dividing by the Avogadro constant. From the formula of potassium oxide, K 2 O, one mol of K 2 O contains 2 potassium ions. Therefore the moles of K 2 O must be half the number of moles of K +. Divide the moles of K + by 2. Change this number of moles of K 2 O to grams by multiplying by the molar mass of K 2 O. M of K 2 O is 94.2 g/mol ion n = mol K = ions ions/mol mol of K 2 O = 3.31 mol Therefore m = 3.31 mol K 2 O 94.2 g/mol = 312 g Work backwards. 312 g K 2O 94.2 g/mol + 2 mol K ion/ mol = ions 1 mol K O This is the number of ions given in the problem Problem A sample of ammonium chloride, NH 4 Cl, contains hydrogen atoms. What is the mass of the sample? 35

18 You must find the mass of ammonium chloride in that contains hydrogen atoms. You know the number of hydrogen atoms and the Avogadro constant is ions/mol. Convert the number of atoms to moles (n) by dividing the Avogadro constant. From the formula of ammonium chloride, NH 4 Cl, one mol of NH 4 Cl contains 4 hydrogen 1 atoms. Therefore, the moles of NH 4 Cl must be the number of moles of hydrogen 4 atoms. Divide the number of moles of hydrogen atoms by 4. Change this number of moles of NH 4 Cl to grams by multiplying by the molar mass of NH 4 Cl. M of NH 4 Cl is 53.5 g/mol atom n = =1.50 mol hydrogen atoms atoms/mol mol of NH 4 Cl = mol Therefore m = mol NH 4 Cl 53.5 g/mol = 20.1 g NH 4 Cl Work backwards g NH4Cl 53.5 g/mol 4 mol H atom atom/mol 1 mol NH Cl 4 = atoms This closely matches the number of hydrogen atoms given in the problem. 32. Problem Express the mass of a single atom of titanium, Ti, in grams. You have to find the mass of a single Ti atom. The chemical symbols and chemical names are given. You have access to a periodic table, and the Avogadro constant, N A = atoms/mol. According to the Avogadro constant, there are atoms of Ti in every mole of titanium. Therefore, the number of moles (n) of 1 atom of Ti is 1 mol divided by atoms. Multiply this value by the molar mass (M) of Ti to obtain the mass (m) of 1 atom of Ti. Number of moles of 1 atom = 1 mol/ atoms = mol M of Ti = g/mol Therefore, m = mol g/mol = g. Work backwards. Dividing the mass by the number of moles should give the molar mass of titanium. g. The results match = g/mol.. - mol 36

19 33. Problem Vitamin B 2, C 17 H 20 N 4 O 6, is also called riboflavin. What is the mass, in grams, of a single molecule of riboflavin? You have to find the mass of a single riboflavin molecule. The chemical formula of the vitamin is given. You have access to a periodic table, and the Avogadro constant, N A = molecules/mol. According to the Avogadro constant, there are molecules of riboflavin in every mole of the vitamin. Therefore, the number of moles (n) of 1 molecule of riboflavin is 1 mol divided by molecules. Multiply this value by the molar mass (M) of C 17 H 20 N 4 O 6 to obtain the mass (m) of a single molecule. 1mol Number of moles of 1 molecule = = mol molecules M of C 17 H 20 N 4 O 6 = g/m; Therefore, m = mol g/mol = g. Work backwards. Dividing the mass by the number of moles should give the molar mass of titanium. 22 g g/mol. The results match. Solutions for Practice Problems Student Textbook page Problem Determine the number of molecules or formula units in each sample. (a) 10.0 g of water, H 2 O (b) 52.4 g of methanol, CH 3 OH (c).5 g of disulfur dichloride, S 2 Cl 2 (d) g of lead(ii) phosphate, Pb 3 (PO 4 ) 2 You need to find the number of formula units or molecules of the substances listed. You are given the masses of the substances as well as their molecular formulas. N A = formula units/mol First convert the mass (m) to moles (n), using the molar mass (M) of the substance and the formula n = m/m. Multiplying this value by the Avogadro constant will yield the number of formula units (or the number of molecules) g g/mol (a) n = = mol Therefore, the number of molecules of water = mol ( ) molecules/mol = molecules. (b) n = 52.4 g g/mol = mol = 1.63 mol Therefore, the number of molecules of CH 3 OH. = 1.63 mol ( ) molecules/mol = molecules. 37

20 .5 g g/mol (c) n = = mol Therefore, the number of molecules of S 2 Cl 2 = mol ( ) molecules/mol = molecules. (d) n = g g/mol = mol Therefore, the number of formula units of Pb 3 (PO 4 ) 2 = mol ( ) formula units/mol = formula units. In each case, work backwards to reestablish the mass given in the question. For example, in (a), each mole of water has a mass of g. Therefore, molecules of water has a mass of: molecules ( 1 mol ) ( g ) molecules 1 mol = 9.99 g The answer is close to the 10.0 g given in the question (a). Your answer is reasonable. 35. Problem How many atoms of hydrogen are in molecules of sodium glutamate, NaC 5 H 8 NO 4? You need to find the number of H atoms in the given number of molecules. You are given the number of molecules of sodium glutamate as well as its molecular formula. From the formula, you can tell that there are 8 hydrogen atoms for every formula unit of sodium glutamate. The total number of hydrogen atoms is therefore 8 multiplied by the given number of molecules. Total number of H atoms = 8 atoms/molecule ( ) molecules = atoms Using round numbers for a quick check, = or This estimate is close to the answer. 36. Problem How many molecules are in a 64.3 mg sample of tetraphosphorus decoxide, P 4 O 10? You need to find the number of molecules in the P 4 O 10 sample. You are given the mass of the tetraphosphorus decoxide as well as its molecular formula. You know the Avogadro constant, N A = molecules/mol. First convert the mass (m) to moles (n), using the molar mass (M) of the tetraphosphorus decoxide and the formula n = m/m. Multiplying this value by Avogadro s number will yield the number of molecules. Remember to convert the mass to grams before proceeding with the calculation. m = mg = g; M = g/mol 3 10 mg/g 38

21 Therefore, n = g = mol g/mol Hence, the number of molecules of P 4 O 10 = mol ( ) molecules/mol = molecules. Work backwards to reestablish the mass given in the question. Therefore, molecules of P 4 O 10 has a mass of: molecules ( 1 mol molecules ) ( g 1 mol = g or 64.1 mg. The answer is close to the 64.3 mg given in the question. 37. Problem (a) How many formula units are in a g sample of potassium chlorate, KClO 3? (b) How many ions (chlorate and potassium) are in this sample? (a) You need to find the number of formula units in the given sample. (b) You need to find the number of chlorate and potassium ions in this sample. (a) You are given the mass of the potassium chlorate. (b) You are given its molecular formula. You know Avogadro s number, N A = formula units/mol. (a) First convert the mass (m) to moles (n), using the molar mass (M) of the potassium chlorate and the formula n = m. Multiplying this value by the Avogadro M constant will yield the number of formula units. (b) From the chemical formula of potassium chlorate, you know there is 1 ion of potassium and 1 ion of chlorate for every formula unit of potassium chlorate. The total number of ions will therefore be the number of formula units calculated in (a) multiplied by 2. (a) m = g; M = g/mol; Therefore, n = g = mol g/mol Hence, the number of formula units of KClO 3 = mol ( ) molecules/mol = formula units (b) Each formula unit of KClO 3 consists of two ions, so there are 2 ions/formula unit ( ) formula units = ions in the sample. (a) Work backwards to reestablish the mass given in the question. Therefore, formula units of KClO 3 has a mass of: formula units ( 1 mol molecules ) ) ( g 1 mol = g. The answer is close to the g given in the question. (b) Using round numbers for a quick check, 2 ( ) = This estimate is close to the answer. ) 39

22 Solutions for Practice Problems Student Textbook page Problem A balloon contains 2.0 L of helium gas at STP. How many moles of helium are present? What Is Required? You need to find the number of moles of helium at STP. What Is Given? The gas is at STP, so its molar volume = 22.4 L/mol. That is, V i = 22.4 L, n i = 1 mol, V f = 2.0 L. Use Avogadro s law to solve for n f. n i V i = n f 1mol 2.0 L 22.4 L \ n V f = = mol i Therefore, the number of moles of helium is mol. The molar volume is 22.4 L/mol. Working with the numbers, 2.0 L = 22 L/mol mol The results match. 39. Problem How many moles of gas are present in 11.2 L at STP? How many molecules? What Is Required? (a) You need to find the number of moles of gas at the given volume at STP. (b) You need to find the number of molecules of gas at STP. What Is Given? (a) At STP, V i = 22.4 L; n i = 1 mol; V f = 11.2 L (b) You know the Avogadro constant = molecules/mol. (a) Use Avogadro s law to solve for n f. (b) Multiply n f by the Avogadro constant to obtain the number of molecules at this given volume. n i V i = n f 1mol 11.2 L 22.4 L \ n V f = = mol f Number of molecules of gas = mol ( ) molecules/mol = molecules The molar volume is 22.4 L/mol. Working with the numbers, 11.2 L mol = 22.4 L/mol. The results match. 40. Problem What is the volume, at STP, of 3.45 mol of argon gas? What Is Required? You need to find the volume of the argon gas for the given number of moles. What Is Given? At STP, V i = 22.4 L; n i = 1 mol; n f = 3.45 mol 40

23 Use Avogadro s law to solve for V f. n i V i = n f V f \ V f = 3.45 mol 22.4 L 1.00 mol = 77.3 L The molar volume is 22.4 L/mol. Working with the numbers, 77.3 L The results match mol = 22.4 L/mol. 41. Problem At STP, a container holds g of nitrogen gas, g of oxygen gas, g of carbon dioxide gas, and g of ammonia gas. What is the volume of the container? What Is Required? You need to find the volume of the container. What Is Given? At STP, V i = 22.4 L; n i = 1 mol The mass of each gas is given. You have access to the periodic table for the molar masses of the elements. The volume of the container will comprise the total number of moles of all the gases present. Divide each mass of gas by its own molar mass to obtain its number of moles. Add the number of moles together, and then use this as n f in Avogadro s law to solve for V f g n nitrogen = = mol n oxygen = n carbon dioxide = n ammonia = g/mol g g/mol g g/mol g g/mol = mol = 1.50 mol = 1.00 mol Total number of moles = mol mol mol mol = 3.50 mol Applying Avogadro s law: n i V i = n f 3.50 mol 22.4 L 1.00 mol \ V V f = = 78.4 L f Therefore, the volume of the container is 78.4 L. The molar volume at STP is 22.4 L/mol. Working with the numbers, = 22.4 L/mol. The results match L 3.50 mol 42. Problem (a) What volume do 2.50 mol of oxygen occupy at STP? (b) How many molecules are present in this volume of oxygen? (c) How many oxygen atoms are present in this volume of oxygen? What Is Required? (a) You need to find the volume of oxygen from the given number of moles. (b) You need to find the number of molecules of this oxygen. (c) You need to find the number of atoms present in this volume of oxygen. 41

24 What Is Given? At STP, V i = 22.4 L; n i = 1 mol; n f = 2.50 mol You know Avogadro s number is molecules/mol. You know oxygen is a diatomic gas. (a) Use Avogadro s law to solve for V f. (b) Multiply V f by the Avogadro number to obtain the number of molecules of oxygen gas. (c) Multiply the number of molecules of oxygen gas by 2 atoms to obtain the number of oxygen atoms in the sample. n i 2.50 mol 22.4 L (a) = n f \ V V i V f = = 56.0 L f 1.00 L (b) Number of molecules of oxygen = 250. mol ( ) molecules/ mol = molecules (c) There are two oxygen atoms in one molecule of oxygen gas. Therefore, the number of oxygen atoms = 2( ) molecules = atoms. The molar volume is 22.4 L/mol. Working with the numbers, 56.0 L = 22.4 L/mol mol The results match. 43. Problem What volume do atoms of neon occupy at STP? What Is Required? You need to find the volume of the gas for the given number of moles. What Is Given? At STP, V i = 22.4 L; n i = 1 mol Number of atoms = Avogadro s number is molecules/mol. Neon is a monatomic gas. Neon is a monatomic so the number of atoms will equal the number of molecules of gas. Multiply the number of molecules by Avogadro s number to get the number of moles of gas as n f. Use Avogadro s law and n f to solve for V f Number of moles of Ne = 24 molecules = 3.32 mol n i V i = n f 3.32 mol 22.4 L 1.00 mol molecules/mol \ V V f = = 74.4 L f Therefore, the volume of the atoms of neon at STP is 74.4 L. The molar volume is 22.4 L/mol. Working with the numbers, 74.4 L = 22.4 L/mol mol The results match. 42

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