Download Chemistry Problems and Solutions: Atomic Mass, Molar Mass, and Balanced Equations and more Lecture notes Chemistry in PDF only on Docsity!
Chapter 5
Solutions to Supplementary Check for Understanding Problems
Moles and Molar Mass
- Indicate the appropriate quantity for each of the following. a) A mole of N atoms contains _________________ atoms. b) A mole of N 2 molecules contains _________________ molecules. c) A mole of N 2 molecules contains _________________ atoms. d) A mole of N atoms has a mass of _________________ grams. e) A mole of N 2 molecules has a mass of _________________ grams. Answers: a) 6.022 x 10 23 atoms b) 6.022 x 10 23 molecules c) 1.2044 x 10 24 atoms d) 14.01 g e) 28.02 g Solutions a) A mole of anything always contains 6.022 x 10 23 items of that material. b) A mole of anything always contains 6.022 x 10 23 items of that material. c) Since a mole of Nof N per molecule of N 2 molecules contains 6.022 x 10 23 molecules of N 2 and there are 2 atoms 2 , the total number of N atoms is given by: 6.022 x10 23 N molecules 2 2 2 x 2 atoms N mol N 1 molecule N
24 2 = 1.2044 10^ atoms N mol N
d) The mass in grams of a mole of atoms of any element (its molar mass) is numerically equally to the weighted average atomic mass in atomic mass units. Since the atomic weight of nitrogen is 14.01, its weighted atomic mass is 14.01 u and a mole of nitrogen atoms weighs 14.01 g.
S.5.
e) The mass in grams of a mole of N 2 molecules (its molar mass) is obtained by summing the molar masses of the atoms in the chemical formula. Since the molar mass of N is 14.01 g/mol (see part d), the molar mass of N 2 is: 14.01g mol N x2 mol N 2 2 = 28.01g 1mol N mol N
- a) What is the mass of 725 sodium atoms in atomic mass units? Answer: 1.67 x 10^3 u Solution The numerical value of the atomic weight of sodium (22.99) refers to the mass of a single sodium atom in atomic mass units. Applying this yields:
725 Na atoms x Na atom22.99 u = 1.67 x10 u^3 b) What is the mass of 725 sodium atoms in grams? Answer: 2.77 x 10-20^ g Solution What we know: number of Na atoms Desired answer: g Na The solution map for this calculation is: atom Na ö mol Na ö g Na The conversion factor needed in the first step is the Avogadro constant expressed in the form
6.022 x101mol Na 23 Na atoms.
The conversion factor needed in the second step is the molar mass of sodium. The numerical value for the molar mass is obtained from the atomic weight of sodium (22.99) and is expressed in the form 22.99 g Na1mol Na.
The solution map for this calculation is: g Ag ö mol Ag ö mol Zn ö g Zn
The conversion factor needed in the first step is the molar mass of Ag in the form (^) 107.9 g Ag1mol Ag.
The problem indicates that the number of atoms of Zn is the same as the number of atoms of Ag. Therefore, mol Zn = mol Ag. This relationship can be applied as the conversion factor in the second step in the form 1mol Zn1mol Ag.
The conversion factor needed in the last step is the molar mass of Zn expressed in the form 65.39 g Znmol Zn.
Putting these together yields:
16.1 g Ag x 107.9 g Ag1 mol Ag x 1 mol Zn1 mol Ag x 65.39 g Zn1 mol Zn = 9.76 g Zn
- One atom of an element is found to weigh 2.107 x 10-22^ g. What is the atomic weight of this element? Answer: 126. Solution What we know: g/atom Desired answer: atomic weight of element We know that the atomic weight of this element is numerically equal to the molar mass of this element. The solution map for calculating the molar mass is:
atom^ g^ ö molg The conversion factor needed is the Avogadro constant expressed in the form 6.022 x10mol 23 atoms.
Applying this yields:
2.107 x10-22 g atom x 6.022 x10^23 atoms =126.9 g mol mol Since the molar mass of this element is 126.9 g/mol, then the atomic weight of the element is 126.9.
- Which has the larger mass, 1.0 mmol of calcium or 1.5 mmol of sulfur? Justify your choice. Answer: 1.5 mmol sulfur Solution The most direct way to make this determination is to calculate the mass of each and compare. The general solution map for this calculation is: mmol element ö mol element ö g element The conversion factor needed in the first step is that between mmol and mol in the form (^10) 1mmol-3 mol.
The conversion factor needed in the second step is the molar mass of the element expressed in the form (^) molg.
Applying these yields:
1.0 mmol Ca x^10 1 mmol Ca^ -3mol Ca x 40.08g Camol Ca = 0.040 g Ca
1.5 mmolSx^10 1 mmolS-3^ molS x 32.07 g SmolS = 0.048g S ³ larger mass
- Which has the larger number of atoms, 0.045 μg of nickel or 0.032 μg of potassium? Justify your choice. Answer: 0.032 μg potassium
The chemical formula for potassium hydrogen phosphate is K 2 HPO 4. The formula indicates that in 1 mole of potassium hydrogen phosphate there are 2 moles of potassium, 1 mole of hydrogen, 1 mole of phosphorus and 4 moles of oxygen. From the periodic table we see that 1 mole of potassium atoms weighs 39.10 g, 1 mole of hydrogen atoms weighs 1.008 g, 1 mole of phosphorus atoms weighs 30.97 g and 1 mole of oxygen atoms weighs 16.00 g. Thus, one mole of K 2 HPO 4 will weigh:
2 K 2 mol K x 39.10 gmol K = 78.20 g
1 H 1 mol H x 1.008gmol H = 1.008g
1 P 1 mol P x 30.97 gmol P = 30.97 g
4 O 4 mol O x 1 mol O16.00 g = 64.00 g 174.18 g The molar mass of potassium hydrogen phosphate is 174.18 g/mol. b) What we know: Pb(C 2 H 3 O 2 ) 2 ; atomic weights of elements Desired answer: g Pb(C 2 H 3 O 2 ) 2 /mol
The chemical formula indicates that in 1 mole of lead(II) acetate there are 1 mole of lead, 4 moles of carbon, 6 moles of hydrogen and 4 moles of oxygen. From the periodic table we see that 1 mole of lead atoms weighs 207.2 g, 1 mole of carbon atoms weighs 12.01 g, 1 mole of hydrogen atoms weighs 1.008 g and 1 mole of oxygen atoms weighs 16.00 g. Thus, one mole of K 2 HPO 4 will weigh:
1 Pb 1 mol Pb x 1 mol Pb207.2 g = 207.2 g
4 C 4 mol C x 1 mol C12.01^ g^ = 48.04 g
6 H 6 mol H^ x 1 mol H1.008g =^ 6.048g
4 O 4 mol O x 1 mol O16.00 g = 64.00 g 325.29 g The molar mass of lead(II) acetate is 325.29 g/mol.
- Calculate the number of moles of compound in each of the following samples. a) 2.239 g C 2 H 5 OH b) 63.1 ng sulfur trioxide c) 1.48 x 10 2 kg potassium permanganate Answers: a) 0.04860 mol b) 7.88 x 10 -10^ mol c) 9.36 x 10 2 mol Solutions a) What we know: g C 2 H 5 OH Desired answer: mol C 2 H 5 OH The solution map for this calculation is g C 2 H 5 OH ö mol C 2 H 5 OH
The formula for potassium permanganate is KMnO 4. The conversion factor needed in the first step is that between kg and g in the form 10 g1kg^3. The conversion factor needed in the second step is the molar mass of KMnO 4. From the periodic table we can get the molar masses of potassium, manganese and oxygen and add them as follows. 39.10 g K + 54.94 g Mn + 4(16.00 g) O = 158.04 g. Thus the molar mass of KMnO 4 is 158.04 g/mol. This is used in the form (^) 158.04 g KMnO1mol KMnO^44 to convert units properly.
Putting these together yields:
1.48 x10^2 kg KMnO 4 x^103 g KMnO^4 1 kg KMnO 4 4 4 x 1mol KMnO 158.04 g KMnO = 9.36 x10 mol KMnO^24
- How many CO molecules are present in 18.4 metric tons of carbon monoxide? One metric tons equals 1000 kg. Answer: 3.96 x 10^29 molecules Solution What we know: metric tons CO Desired answer: number of molecules CO The solution map for this calculation is metric tons CO ö kg CO ö g CO ö mol CO ö molecules CO The conversion factor needed in the first step is that between metric tons and kg in the form
1metric ton1000 kg. The conversion factor needed in the second step is that between^ kg^ and^ g^ in the form 10 g1kg^3. The conversion factor needed in the third step is the molar mass of CO. From the periodic table we can get the molar masses of carbon and oxygen and add them as follows. 12.01 g C + 16.00 g O = 28.01 g. Thus the molar mass of CO is 28.01 g/mol. This is used in the form (^) 28.01g CO1mol CO to convert units properly.
The conversion factor needed in the last step is the Avogadro constant in the form 6.022 x10mol CO 23 molecules CO.
Putting these together yields:
18.4 metric tons CO x 1 metric ton CO1000 kg CO x^10 1 kg CO^3 g CO x 28.01 g CO1 mol CO x 6.022 x101 mol CO^23 molecules CO = 3.96 10^29 molecules CO
- Calculate the mass in grams of each of the following samples. a) 9.44 mol copper(II) sulfate b) 7.11 mmol Li 2 CO 3 Answers: a) 1.51 x 10 3 g b) 0.427 g Solutions a) What we know: mol copper(II) sulfate Desired answer: g copper(II) sulfate The solution map for this calculation is mol copper(II) sulfate ö g copper(II) sulfate The formula for copper(II) sulfate is CuSO 4. The conversion factor needed is the molar mass of CuSO 4. From the periodic table we can get the molar masses of copper, sulfur and oxygen and add them as follows. 63.55 g Cu + 32.07 g S + 4(16.00 g) O = 159.62 g. Thus the molar mass of CuSO 4 is 159.62 g/mol. This is used in the form 159.62 g CuSO1mol CuSO 44 to convert units properly. Applying this yields: 9.44 mol CuSO 4 x 159.62 g CuSO1 mol CuSO 44 = 1.51x10 g CuSO^34
The formula for sodium thiosulfate is Na 2 S 2 O 3. The conversion factor needed in the first step is the molar mass of Na 2 S 2 O 3. From the periodic table we can get the molar masses of sodium, sulfur and oxygen and add them as follows. 2(22.99 g) Na + 2(32.07 g) S + 3(16.00 g) O = 158.12 g. Thus the molar mass of Na 2 S 2 O 3 is 158.12 g/mol. This is needed in the form
158.12 g Na S O1mol Na S O^2 22 2 . The formula of Na 2 S 2 O 3 indicates that in 1 mole of Na 2 S 2 O 3 there are 2 moles of sulfur atoms. This relationship can be applied as the conversion factor in the second step in the form
1mol Na S O2 molS 2 2 3. Putting these together yields:
4.63 g Na S O 2 2 3 x158.12 g Na S O1 mol Na S O^2 22 2 33 x 1 mol Na S O2 molS 2 2 3 = 0.0586 molS
b) What we know: μg Na 2 S Desired answer: mol S atoms The solution map for this calculation is: μg Na 2 S ö g Na 2 S ö mol Na 2 S ö mol S
The conversion factor needed in the first step is that between μg and g in the form (^101) μ-6^ gg. The conversion factor needed in the second step is the molar mass of Na 2 S. From the periodic table we can get the molar masses of sodium and sulfur and add them as follows. 2(22.99 g) Na + 32.07 g S = 78.05 g. Thus the molar mass of Na 2 S is 78.05 g/mol. This is needed in the form
78.05g Na S1mol Na S^22. The formula of Na 2 S indicates that in 1 mole of Na 2 S there is 1 mole of sulfur atoms. This relationship can be applied as the conversion factor in the last step in the form
1mol Na S1molS 2. Putting these together yields:
5.81 μg Na S 2 x^10 -6^ g Na S^2 1 μg Na S 2 x 1 mol Na S^2 78.05 g Na S 2 2 x 1molS 1 mol Na S = 7.44 x10-8 molS
- Calculate the number of carbon atoms in a 3.92-g sample of C 6 H 4 Cl 2. Answer: 9.64 x 10^22 atoms Solution What we know: g C 6 H 4 Cl (^2) Desired answer: number of C atoms The solution map for this calculation is: g C 6 H 4 Cl 2 ö mol C 6 H 4 Cl 2 ö mol C ö atoms C The conversion factor needed in the first step is the molar mass of Ctable we can get the molar masses of carbon, hydrogen and chlorine and add them as follows. 6 H 4 Cl 2. From the periodic 6(12.01 g) C + 4(1.008 g) H + 2(35.45 g) Cl = 146.99 g. Thus the molar mass of C 6 H 4 Cl 2 is 146.99 g/mol. This is needed in the form (^) 146.99 g C H Cl1mol C H Cl^6 6 4 4 .
The formula of C 6 H 4 Cl 2 indicates that in 1 mole of C 6 H 4 Cl 2 there are 6 moles of carbon atoms. This relationship can be applied as the conversion factor in the second step in the form
1mol C H Cl6 mol C 6 4 2.
The conversion factor needed in the last step is the Avogadro constant in the form 6.022 x101mol C 23 atoms C.
Putting these together yields:
3.92 g C H Cl 6 4 2 x146.99 g C H Cl1mol C H Cl^6 6 4 4 22 x1 mol C H Cl6 mol C 6 4 2 x 6.022 x10^23 atoms C 1 mol C = 9.64 x10 22 atoms C
- How many moles of oxygen atoms are present in 4.40 mmol calcium phosphate? Answer: 0.0352 mol
The formula for barium chloride is BaCl 2. Since the mass percent values apply to any sample of BaCl 2 , it is convenient to consider one mole of this compound. From the periodic table we can get the molar masses of barium and chlorine and add them as follows. 137.3 g Ba + 2(35.45 g) Cl = 208.2 g. Thus the molar mass of BaCl 2 is 208.2 g/mol, so 208.2 g represents the total mass of material. The formula indicates that 1 mole of BaCl 2 contains 1 mole of barium. Therefore the mass of barium (=component mass) present is 137.3 g. These values are used to obtain the mass percent for barium.
2 mass % Ba = 137.3g Ba x 100 = 65.95% 208.2 g BaCl Since chlorine is the only other element present in the compound, the mass % Cl = 100.00 % - 65.95 % = 34.05 %. b) What we know: sodium sulfate; atomic weights of elements Desired answer: mass % of each element present The formula for sodium sulfate is Na 2 SO 4. Since the mass percent values apply to any sample of Na 2 SO 4 , it is convenient to consider one mole of this compound. From the periodic table we can get the molar masses of sodium, sulfur and oxygen and add them as follows. 2(22.99 g) Na + 32.07 g S + 4(16.00 g) O = 142.05 g. Thus the molar mass of Na 2 SO 4 is 142.05 g/mol, so 142.05 g represents the total mass of material. The formula indicates that 1 mole of Na 2 SO 4 contains 2 moles of sodium. Therefore the mass of sodium (=component mass) present is 2(22.99 g) = 45.98 g. These values are used to obtain the mass percent for sodium.
2 4 mass % Na = 45.98g Na x 100 = 32.37 % 142.05g Na SO The formula indicates that 1 mole of Na 2 SO 4 contains 1 mole of sulfur. Therefore the mass of sulfur (=component mass) present is 32.07 g. These values are used to obtain the mass percent for sulfur.
2 4 mass %S = 32.07 g S x 100 = 22.58% 142.05g Na SO
Since oxygen is the only other element present in the compound, the mass % O = 100.00 % - 32.37 % - 22.58% = 45.05 %.
- Which of the following compounds contains the largest mass percent of nitrogen? Justify your choice. a) NH 4 NO 3 b) HNO 3 c) N 2 O 4 d) Al(NO 3 ) (^3) Answer: NH 4 NO 3 Solution The most direct way to answer this is to determine the mass percent N in each compound. This is calculated from the mass of one mole of a compound and the mass of nitrogen present in one mole of that compound. The required data are shown below.
Compound Mass of one mole of compound Mass of N in one mole of compound Mass % N NH 4 NO 3 80.05 g 28.02 g 35.00 % HNO 3 63.02 g 14.01 g 22.23 % N 2 O 4 92.02 28.02 g 30.45 % Al(NO 3 ) 3 213.01 g 42.03 g 19.72 %
Thus, NH 4 NO 3 has the largest mass % N.
- In a particular molecular compound the mass percent sulfur is 50.0% and the mass percent oxygen is 50.0%. What is the ratio of oxygen atoms to sulfur atoms in a molecule of this compound? Answer: 2 O atoms/S atom Solution What we know: mass % of each element present; atomic weights of elements Desired answer: number of O atoms/S atom in compound The ratio of O atoms to S atoms is the same as the ratio of mol O to mol S. In order to obtain themoles of each element, choose an arbitrary mass of the compound (100 g simplifies the math) and use the following solution map: g compound ö g element ö mol element
Putting these together yields:
1.5 kg samplex^10 1 kg sample^3 g sample x 100 g sample18 g Cr x 52.00 g Cr1mol Cr = 5.2 mol Cr
Stoichiometric Calculations (mole-to-mole)
- Balance the following equation and state the meaning of the equation in terms of individual units of reactants and products and in terms of moles of reactants and products. Al(s) + MnO 2 (s) ÷ Mn(s) + Al 2 O 3 (s) Answers: 4Al(s) + 3MnO 2 (s) ÷ 3Mn(s) + 2Al 2 O 3 (s) 4 atoms 3 formula units 3 atoms 2 formula units 4 mol 3 mol 3 mol 2 mol Solution To balance this equation, place a coefficient of 3 in front of MnO 2 and a coefficient of 2 in front of Al 2 O 3 to balance oxygen atoms. This requires a coefficient of 3 in front of Mn to balance manganese atoms. Finally, a coefficient of 4 in front of Al will balance aluminum atoms. The balanced equation is: 4Al(s) + 3MnO 2 (s) ÷ 3Mn(s) + 2Al 2 O 3 (s) At the most fundamental level this equation suggests that 4 atoms of aluminum react with 3 formula units of MnO 2 to produce 3 atoms of Mn and 2 formula units of Al 2 O 3. Multiplying each reactant amount and each product amount by the Avogadro constant yields 4 mol Al reacting with 3 mol MnO 2 to form 3 mol Mn and 2 mol Al 2 O 3.
- How many moles of CO 2 are needed to react completely with 0.675 mol LiOH? LiOH(aq) + CO 2 (g) ÷ Li 2 CO 3 (aq) + H 2 O(l) (unbalanced) Answer: 0.338 mol
Solution What we know: mol LiOH; equation relating CO 2 and LiOH Desired answer: mol CO (^2) The solution map for this calculation is: mol LiOH ö mol CO 2 The conversion factor needed is the mole ratio for these two substances from the balanced equation. To balance this equation place a coefficient of 2 in front of LiOH. The balanced equation is: 2LiOH(aq) + CO 2 (g) ÷ Li 2 CO 3 (aq) + H 2 O(l) Applying the mole ratio yields:
0.675 mol LiOH (^) 2 mol LiOH1mol CO^2 = 0.338 mol CO 2
- Given the reaction 4FeS(s) + 7O 2 (g) ÷ 2Fe 2 O 3 (s) + 4SO 2 (g) how many moles of O 2 are needed to: a) produce 0.693 mol Fe 2 O 3? b) react completely with 9.14 mol FeS? c) form 1.51 mol SO 2? Answers: a) 2.43 mol b) 16.0 mol c) 2.64 mol Solutions a) What we know: mol Fe 2 O 3 ; balanced equation relating O 2 and Fe 2 O 3 Desired answer: mol O 2
