Conversion Factors Between Moles of Each Element & Moles Compound Assigment

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Chapter 6 Conversion Factors 1. Write out conversion factors between the moles of each element and the moles of the compound C5H12 2. Write out conversion factors between the moles of each element and the moles of the compound C27H46O 3. Write out conversion factors between the moles of each element and the moles of the compound C6H12O6 4. Write out the conversion factor of the given element in the given compound using the mass percent composition a) Carbon dioxide is 27.19% carbon by mass b) Butane is 17.34% hydrogen by mass c) C6H8O7 is 58.29% oxygen by mass 5. Write out the conversion factor of the given element in the given compound using the mass percent composition a) Ethanol is 13.13% hydrogen by mass b) Aluminum nitrate is 19.73% nitrogen by mass c) Methyl butanoate is 58.8% carbon by mass 6. Write out the conversion factor of the given element in the given compound using the mass percent composition a) Propane is 81.68% carbon by mass b) Ethylene is 14.14% hydrogen by mass c) Aluminum oxide is 52.9% aluminum by mass The Mole Concept 1. How many moles of atoms are in the following? a) 8.22 x 1024 As atoms b) 4.41 x 1022 Ne atoms c) 6.77 x 1023 Li atoms d) 2.10 x 1021 S atoms 2. How many atoms are in the following given numbers of moles of an element? a) 0.994 mol F b) 4.39 mol Sr c) 9.1 x 10-3 mol B d) 13.8 mol Cl 3. How many moles of atoms are in the following? a) 2.13 x 1025 He atoms b) 3.65 x 1022 Ga atoms c) 6.022 x 1023 Mg atoms d) 1.04 x 1024 Cu atoms 4. Complete the table: Element Se Moles 0.0112 K Ti Br Number of Atoms 1.55 x 1023 2.34 7.12 x 1022 5. Complete the table: Element Moles P 13.2 Au 0.222 Number of Atoms F 5.02 x 1024 Be 1.98 x 1023 6. Complete the table: Element Moles O Number of Atoms 9.54 x 1021 Zr 1.30 x 10-6 Bk 25.3 Ba 7.71 x 1024 7. Complete the table: Element Moles Mass N 75.4g W 13.9g Ge 0.00784 Rb 4.09 8. Complete the table: Element Moles Co 62.4g Hg 2.23 x 10-4 Te 18.6 Al Mass 4.97g 9. Complete the table: Element Moles Li 244 Y 0.653 Mass B 20.7g He 0.00422g 10. Calculate the mass in grams of the following: a) 4.45 x 1019 Ag atoms b) 5.50 x 1024 Bi atoms c) 2.08 x 1021 Se atoms d) 7.31 x 1024 Ar atoms 11. Calculate the number of atoms in each of the following: a) 14.8g Al b) 3.98g Ca c) 1.90g Co d) 26.7g S 12. Calculate the mass of the following: a) 3.72 x 1020 Cd atoms b) 6.01 x 1023 Sb atoms c) 9.43 x 1026 F atoms d) 2.84 x 1022 Re atoms 13. Complete the table: Element Mass Moles Sn 3.65 Xe 0.0983 H Number of Atoms 65.5mg Si 4.35 x 1024 14. Complete the table: Element Mass Moles Ca Number of Atoms 6.18 x 1021 Br 0.138kg Ga 9.29 x 1023 K 10.4 15. Complete the table: Element Mass Be Moles 226mg S 7.36 x 1020 Nb N 0.104 0.00391kg 16. Calculate the mass of the following in grams: a) 6.10 mol carbon monoxide b) 0.364 mol sulfur hexafluoride c) 4.44 mol potassium hydroxide d) 1.71 mol lithium nitrate Number of Atoms 17. Calculate the mass of the following in grams: a) 13.4 mol sodium chloride b) 0.223 kmol nitrogen dioxide c) 0.00322 mol boron tribromide d) 144 mmol xenon tetroxide 18. Calculate the number of moles of molecules for the following: a) 29.8g ammonium iodide b) 8.34g potassium cyanide c) 3.50mg sodium sulfate d) 0.0888kg hydrochloric acid 19. Complete the table: Compound H2SO4 Mass Number of Molecules 188kg NaBr KClO3 Moles 0.318 4.05mg NH3 7.10 20. Complete the table: Compound Mass NO 0.929 CCl4 0.133kg HF 16.9g CS2 Moles 0.00268 Number of Molecules 21. Complete the table: Compound Mass CH4 NaOH Moles Number of Molecules 3.45 42.2mg LiCl AgNO3 0.00215 311g 22. Calculate the mass of the following in grams: a) 3.19 x 1024 SrCl2 molecules b) 2.81 x 1021 Na3PO4 molecules c) 7.70 x 1022 KBrO3 molecules d) 4.51 x 1019 HNO3 molecules 23. Calculate the mass of the following in grams: a) 4.30 x 1023 KCN molecules b) 6.25 x 1025 FeCl3 molecules c) 9.64 x 1022 HF molecules d) 7.38 x 1018 CO2 molecules 24. Calculate the number of molecules of the following: a) 13.2g SO2 b) 9.03g CH4 c) 0.00888g H2 d) 157g LiBr 25. A sample of CuS has a mass of 0.67mg. Calculate the number of formula units. 26. A sample of VF4 has a mass of 0.0076kg. Calculate the number of formula units. 27. A sample of C6H6 has a mass of 15ng. Calculate the number of formula units. Chemical Formulas as Conversion Factors 1. Calculate the number of moles of O in the following: a) 11.4 mol CO2 b) 0.761 mol H2O c) 4.08 mol N2O5 d) 0.302 mol HNO3 2. Calculate the number of moles of H in the following: a) 1.44 mol H2SO4 b) 0.00373 mol C3H8 c) 22.4 mol NO2 d) 0.0200 mol C3H9N 3. How many moles of H are in 1.85 mol [Cr(OH2)5Cl] Cl•H2O? 4. Calculate the mass in kg of the following which each contain 3.5 x 102 kg of lithium. a) Li2SO4 b) LiOH c) LiBr 5. Calculate the mass in mg of the following which each contain 2.2 x 104 mg of silver. a) Ag2O b) AgI c) Ag3N 6. Calculate the mass in kg of the following which each contain 7.1 x 102kg of sodium. a) NaCl b) Na2CO3 c) Na3PO4 7. Which of the following contains the greatest number of moles of oxygen? Show all work. a) 3.03 mol C2H6O2 b) 16.9 mol N2O3 c) 8.27 mol Sr(NO3)2 d) 0.0180 mol BaSO4 8. Which of the following contains the greatest number of moles of hydrogen? a) 0.712 mol C3H6O3 b) 2.40 mol H2O c) 3.49 mol C4H10 d) 7.48 mol NH4OH 9. Of the following compounds containing the ammonium ion, which contains the greatest number of moles of nitrogen? Show all work. a) 11.4 mol NH4NO3 b) 9.65 mol (NH4)2O c) 83.4 mol NH4I d) 0.997 mol (NH4)3PO4 Mass Percent Composition 1. A 8.91g sample of an organic compound is found to contain 3.34g C, 0.98g H, 2.61g O, and 1.98g N. Determine the mass percent of each element. 2. BaSO4 is approximately 58.84% barium by mass. Calculate the mass of barium in grams in a 7.22g sample of BaSO4. 3. Calculate the mass percent composition of oxygen in the following compounds. a) CrO3 b) SO2 c) NaOH d) Al2(SO4)3 4. Calculate the mass percent composition of all the individual elements in each of the following compounds. a) SrCl2 b) Fe2O3 c) CsF d) NaCN 5. Calculate the mass percent composition of all the individual elements in each of the following compounds. a) HClO3 b) AgBr c) PbSO4 d) CH4 6. Calculate the mass percent composition of all the individual elements in each of the following compounds. a) Be3N2 b) IF7 c) ZnF2 d) Ca(OH)2 Calculating Empirical Formulas 1. An organic compound is found to have 36.03g of carbon, 6.048g hydrogen, and 16.00g oxygen. Determine the empirical formula of the organic compound. 2. An organic compound is found to have 32.07g of carbon, 3.36g of hydrogen, 18.63g of nitrogen, and 10.72g of nitrogen. Determine the empirical formula of the organic compound. 3. An organic compound is found to have 15.97g of carbon, 2.016g of hydrogen, and 21.28g of oxygen. Determine the empirical formula of the organic compound. 4. The nonsteroidal anti-inflammatory drug Ibuprofen consists of 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen. Determine the empirical formula of ibuprofen. 5. The chemical compound cinnamaldehyde is what gives cinnamon its flavor and odor. When tested in a lab, it is found to consist of 81.79% carbon, 6.10% hydrogen, and 12.11% oxygen. Determine the empirical formula of cinnamaldehyde. 6. The antiviral drug zanamivir is used in influenza vaccinations. It consists of 43.37% carbon, 6.07% hydrogen, 16.86% nitrogen, and 33.70% oxygen. Determine the empirical formula of zanamivir. Calculating Molecular Formulas 1. Determine the molecular formula of each of the following using the given empirical formulas and total molar masses. a) CO2H, 225.09g/mol b) CClN, 184.41g/mol c) C2H3O, 86.09g/mol d) C4H8O, 144.21g/mol 2. Determine the molecular formula of each of the following using the given empirical formulas and total molar masses. a) C5H8, 136.23 g/mol b) CH3O4P, 660.02 g/mol c) C2H5N, 86.14g/mol d) CH, 78.11g/mol 3. Determine the molecular formula of each of the following using the given empirical formulas and total molar masses. a) C6H8O, 288.37g/mol b) CH2N2O2, 296.18g/mol c) C2H7NS, 77.16g/mol d) C11H22O, 340.57g/mol Challenge Problems 1. How many ethanol molecules are in a 613mL container of pure ethanol? Ethanol has a density of 0.7893 g/cm3. 2. Complete the table: Substance Mass Moles CdF2 Number of Particles 0.0389 N 17.3mg Ag 1.02kg HNO3 4.55 x 1021 3. Complete the table: Formula Name Mass %Na %Cl %O NaClO2 NaClO3 NaClO4 4. Aldrin is an insecticide that was commonly used in soil fields until it was banned in many counties. It is found to consist of 39.50% carbon, 2.21% hydrogen, and 58.29% chlorine. Its molar mass is 364.884g/mol. Using this information, determine the molecular formula of Aldrin. 5. A solution contains approximately 45.4% KBr by mass within it. The solution has a density of 2.21g/mL. What volume of the solution contains 2.0g of potassium? 6. A sample of an acid is analyzed in a laboratory and is found to consist of 40.00% C, 6.71% H, and 53.29% O. Its molar mass is 90.078g/mol. Determine the acid’s molecular compound. Search the formula in your text or an internet source and write its name.
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1. Write out conversion factors between the moles of each element and the
moles of the compound C5H12
Answer:
1 mole of 𝐶5 𝐻12 consist 5 moles of 𝐶 atom and 12 moles of 𝐻 atom.
Conversion factor for Carbon and Hydrogen element are as follows:
5 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶
Number of Carbon atom (C) =
× 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶5 𝐻12
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶5 𝐻12
12 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻

Number of Hydrogen atom (H) =

1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶5 𝐻12

×

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶5 𝐻12
2. Write out conversion factors between the moles of each element and the
moles of the compound C27H46O
Answer:
1 mole of 𝐶27 𝐻46 𝑂 consist 27 moles of 𝐶 atom, 46 moles of 𝐻 atom and 1
mole of O atom.
Conversion factor for Carbon, Hydrogen and Oxygen element are as follows:
27 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶
Number of Carbon atom (C) =
×
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶27 𝐻46 𝑂

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶27 𝐻46 𝑂
Number of Hydrogen atom (H) =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶27 𝐻46 𝑂
Number of Oxygen atom (O) =

46 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶27 𝐻46 𝑂
1 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂

1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶27 𝐻46 𝑂

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶27 𝐻46 𝑂

×

×

3. Write out conversion factors between the moles of each element and the
moles of the compound C6H12O6
Answer:
1 mole of 𝐶6 𝐻12 𝑂6 consist 6 moles of 𝐶 atom, 12 moles of 𝐻 atom and 6 mole
of O atom.
Conversion factor for Carbon, Hydrogen and Oxygen element are as follows:
6 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶
Number of Carbon atom (C) =
×
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶27 𝐻46 𝑂

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶6 𝐻12 𝑂6
Number of Hydrogen atom (H) =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶6 𝐻12 𝑂6
Number of Oxygen atom (O) =

12 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻
1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶27 𝐻46 𝑂
6 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂

1 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶27 𝐻46 𝑂

×

×

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶6 𝐻12 𝑂6
4. Write out the conversion factor of the given element in the given compound
using the mass percent composition
a) Carbon dioxide is 27.19% carbon by mass
Answer:
Molar mass of 𝐶𝑂2 = 44 g/mol
27.19
44 g of 𝐶𝑂2 contain =
× 44 ≈ 11.96 g of carbon
Conversion factor =

100
11.96𝑔 𝐶
44𝑔 𝐶𝑂2

b) Butane is 17.34% hydrogen by mass
Answer:
Molar mass of Butane = 58.12 g/mol
17.34
58.12 g of butane contain = 58.12 ×
≈ 10.08 g H
Conversion factor =

10.08𝑔 𝐻
58.12𝑔 𝐵𝑢𝑡𝑎𝑛𝑒

100

c) C6H8O7 is 58.29% oxygen by mass
Answer:
Molar mass of 𝐶6 𝐻8 𝑂7 = 192.12 g/mol
58.29
192.12 g of 𝐶6 𝐻8 𝑂7 = 192.12 ×
≈ 111.98 g O
Conversion factor =

100

111.98𝑔 𝑂

192.12𝑔 𝐶6 𝐻8 𝑂7

5. Write out the conversion factor of the given element in the given compound
using the mass percent composition
a) Ethanol is 13.13% hydrogen by mass
Answer:
Molar mass of 𝐸𝑡ℎ𝑎𝑛𝑜𝑙 𝐶2 𝐻5 𝑂𝐻 = 46.07 g/mol
13.13
46.07 g of 𝐶2 𝐻5 𝑂𝐻 = 46.07 ×
≈ 6.05 g H
Conversion factor =

6.05𝑔 𝐻

100

46.07𝑔 𝐶2 𝐻5 𝑂𝐻

b) Aluminum nitrate is 19.73% nitrogen by mass
Answer:
Molar mass of 𝐴𝑙𝑢𝑚𝑖𝑛𝑢𝑚 𝑁𝑖𝑡𝑟𝑎𝑡𝑒 = 212.996 g/mol
19.73
212.996 g of 𝐴𝑙𝑢𝑚𝑖𝑛𝑢𝑚 𝑁𝑖𝑡𝑟𝑎𝑡𝑒 = 212.996 ×
≈ 42.02 g N
Conversion factor =

100

42.02𝑔 𝑁
46.07𝑔 𝐴𝑙𝑢𝑚𝑖𝑛𝑢𝑚 𝑁𝑖𝑡𝑟𝑎𝑡𝑒

c) Methyl butanoate is 58.8% carbon by mass
Answer:
Molar mass of 𝑀𝑒𝑡ℎ𝑦𝑙 𝑏𝑢𝑡𝑎𝑛𝑜𝑎𝑡𝑒 = 102.13 g/mol
58.8
102.13 g of 𝑀𝑒𝑡ℎ𝑦𝑙 𝑏𝑢𝑡𝑎𝑛𝑜𝑎𝑡𝑒 = 102.13 ×
≈ 60.05 g C
Conversion factor =

60.05𝑔 𝐶
102.13𝑔 𝑀𝑒𝑡ℎ𝑦𝑙 𝑏𝑢𝑡𝑎𝑛𝑜𝑎𝑡𝑒

100

6. Write out the conversion factor of the given element in the given compound
using the mass percent composition
a) Propane is 81.68% carbon by mass
Answer:
Molar mass of 𝑃𝑟𝑜𝑝𝑎𝑛𝑒 = 44.1 g/mol
81.68
44.1 g of 𝑃𝑟𝑜𝑝𝑎𝑛𝑒 = 44.1 ×
≈ 36.02 g C
Conversion factor =

100
36.02𝑔 𝐶

44.1𝑔 𝑃𝑟𝑜𝑝𝑎𝑛𝑒

b) Ethylene is 14.14% hydrogen by mass
Answer:
Molar mass of 𝐸𝑡ℎ𝑦𝑙𝑒𝑛𝑒 = 28.05 g/mol
14.14
28.05g of 𝐸𝑡ℎ𝑦𝑙𝑒𝑛𝑒 = 28.05 ×
≈ 3.96 g H
Conversion factor =

3.96𝑔 𝐻

100

28.05𝑔 𝐸𝑡ℎ𝑦𝑙𝑒𝑛𝑒

c) Aluminum oxide is 52.9% aluminum by mass
Answer:
Molar mass of 𝐴𝑙𝑢𝑚𝑖𝑛𝑢𝑚 𝑂𝑥𝑖𝑑𝑒 = 101.96 g/mol
52.9
101.96g of 𝐴𝑙𝑢𝑚𝑖𝑛𝑢𝑚 𝑂𝑥𝑖𝑑𝑒 = 101.96 ×
≈ 53.94 g Al
Conversion factor =

53.94𝑔 𝐴𝑙
101.96𝑔 𝐴𝑙𝑢𝑚𝑖𝑛𝑢𝑚 𝑂𝑥𝑖𝑑𝑒

100

Answers:
1 mole = 6.02𝑥 1023 𝑎𝑡𝑜𝑚𝑠
8.22 𝑥 1024 𝐴𝑠 𝑎𝑡𝑜𝑚𝑠

a)
b)
c)

6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
6.77 𝑥 1023 𝐿𝑖 𝑎𝑡𝑜𝑚𝑠

6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
2.10 𝑥 1021 𝑆 𝑎𝑡𝑜𝑚𝑠

d)

≈ 13.65 As atoms

6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
4.41 𝑥 1022 𝑁𝑒 𝑎𝑡𝑜𝑚𝑠

≈ 0.07 Ne atoms

≈ 1.12 Li atoms

6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙

≈ 3.48 x 10³ S atoms

Answers:
1 mole = 6.02𝑥 1023 𝑎𝑡𝑜𝑚𝑠
a) 0.994 𝑚𝑜𝑙 𝐹 𝑥 6.02 𝑥 1023

𝑎𝑡𝑜𝑚𝑠

𝑚𝑜𝑙𝑒
23 𝑎𝑡𝑜𝑚𝑠

b) 4.39 𝑚𝑜𝑙 𝑆𝑟 𝑥 6.02 𝑥 10
c) 9.1 𝑥 10−3 𝑚𝑜𝑙 𝐵 𝑥

≈ 5.98 𝑥 1023 𝐹 𝑎𝑡𝑜𝑚𝑠
≈ 2.642 𝑥 1024 𝑆𝑟 𝑎𝑡𝑜𝑚𝑠

𝑚𝑜𝑙𝑒
𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023
𝑚𝑜𝑙𝑒
23 𝑎𝑡𝑜𝑚𝑠

d) 13.8 𝑚𝑜𝑙 𝐶𝑙 𝑥 6.02 𝑥 10

𝑚𝑜𝑙𝑒

≈ 5.478 𝑥 1021 𝐵 𝑎𝑡𝑜𝑚𝑠

≈ 8.307 𝑥 1024 𝐶𝑙 𝑎𝑡𝑜𝑚𝑠

Answers:
1 mole = 6.02𝑥 1023 𝑎𝑡𝑜𝑚𝑠
a)
b)
c)
d)

2.13 𝑥 1025 𝐻𝑒 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
3.65 𝑥 1022 𝐺𝑎 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
6.022 𝑥 1023 𝑀𝑔 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
1.04 𝑥 1024 𝐶𝑢 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙

≈ 35.38 He atoms
≈ 0.06 Ga atoms
≈ 1 Mg atoms
≈ 1.7 Cu atoms

Answers:
1 mole = 6.02𝑥 1023 𝑎𝑡𝑜𝑚𝑠
𝑎) 0.0112 𝑚𝑜𝑙 𝑆𝑒 𝑥 6.02 𝑥 1023
b)

1.55 𝑥 1023 𝐾 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
7.12 𝑥 1022 𝐵𝑟 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙

𝑚𝑜𝑙𝑒

≈ 6.74 𝑥 1021 𝑆𝑒 𝑎𝑡𝑜𝑚𝑠

≈ 0.25 K atoms

𝑐) 2.34 𝑚𝑜𝑙 𝑇𝑖 𝑥 6.02 𝑥 1023
d)

𝑎𝑡𝑜𝑚𝑠

𝑎𝑡𝑜𝑚𝑠
𝑚𝑜𝑙𝑒

≈ 1.408 𝑥 1024 𝑇𝑖 𝑎𝑡𝑜𝑚𝑠

≈ 0.118 Br atoms

Answers:
𝑎) 13.2 𝑚𝑜𝑙 𝑃 𝑥 6.02 𝑥 1023
𝑏) 0.222 𝑚𝑜𝑙 𝐴𝑢 𝑥 6.02 𝑥
c)
d)

5.02 𝑥 1024 𝐹 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
1.98 𝑥 1023 𝐵𝑒 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙

𝑎𝑡𝑜𝑚𝑠

≈ 7.946 𝑥 1024 𝑃 𝑎𝑡𝑜𝑚𝑠

𝑚𝑜𝑙𝑒
23 𝑎𝑡𝑜𝑚𝑠
10
𝑚𝑜𝑙𝑒

≈ 1.336 𝑥 1023 𝐴𝑢 𝑎𝑡𝑜𝑚𝑠

≈8.34 F atoms
≈ 0.328 Be atoms

Answers:
a)

9.54 𝑥 1021 𝑂 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
−6

𝑏) 1.30 𝑥 10

≈ 0.01584 O atoms

𝑚𝑜𝑙 𝑍𝑟 𝑥 6.02 𝑥 1023

𝑐) 25.3 𝑚𝑜𝑙 𝐵𝑘 𝑥 6.02 𝑥 1023
d)

7.71 𝑥 1024 𝐵𝑎 𝑎𝑡𝑜𝑚𝑠
6.02 𝑥 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙

𝑎𝑡𝑜𝑚𝑠
𝑚𝑜𝑙𝑒

𝑎𝑡𝑜𝑚𝑠
𝑚𝑜𝑙𝑒

≈ 7.826 𝑥 1017 𝑍𝑟 𝑎𝑡𝑜𝑚𝑠

≈ 1.523 𝑥 1025 𝐴𝑢 𝑎𝑡𝑜𝑚𝑠

≈ 12.807 Ba atoms

Answers:
a)
b)

75.4 𝑔 𝑁
14

𝑔
𝑚𝑜𝑙 𝑁

≈ 5.3857 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁

13.9 𝑔 𝑁
183.84

𝑔
𝑚𝑜𝑙 𝑁

≈ 0.0756 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑊

c) 0.00784 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐺𝑒 × 72.64
d) 4.09 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑅𝑏 × 85.46

𝑔

𝑔
𝑚𝑜𝑙

𝑚𝑜𝑙

≈ 0.5694 g of Ge

≈ 349.53 g of Rb

Answers:
62.4 𝑔 𝐶𝑜
a)
≈ 1.0588 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶𝑜
𝑔
58.93

𝑚𝑜𝑙 𝐶𝑜

b) 2.23 × 10−4 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻𝑔 × 200.59
c) 18.6 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑇𝑒 × 127.6
d)

4.97 𝑔 𝐴𝑙
26.98

𝑔
𝑚𝑜𝑙 𝐴𝑙

𝑔
𝑚𝑜𝑙

𝑔
𝑚𝑜𝑙

≈ 0.447 g of Hg

≈ 2,373.36 g of Te

≈ 0.1842 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐴𝑙

Answers:
a) 244 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐿𝑖 × 6.941

𝑔
𝑚𝑜𝑙
𝑔

b) 0.653 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑌 × 88.90
c)
d)

20.7 𝑔 𝐵

𝑚𝑜𝑙

≈ 58.05 g of Y

≈ 1.9147 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐵

𝑔

10.811
𝑚𝑜𝑙 𝐵
0.00422 𝑔 𝐻𝑒
4.002

≈ 1,693.604 g of Li

𝑔
𝑚𝑜𝑙 𝐻𝑒

≈ 0.001 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻𝑒

Mass of atoms = moles of atoms × Molar mass (g/mol) of element.
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠
⇒ Mass of atoms =
× 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡.

𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟

Answers:
a)
b)
c)
d)

4.45 𝑥 1019
6.02 𝑥 1023
5.50×1024
6.022×1023
2.08×1021
6.022×1023
7.31×1024
6.022×1023

× 107.87 ≈ 0.008 𝑔 𝑜𝑓 𝐴𝑔
× 208.98 ≈ 1908.652 g of Bi
× 78.96 ≈ 0.273 g of Se
× 39.95 ≈ 484.946 g of Ar

Answers:
Number of atoms = moles of element × Avogadro’s number
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
Moles of element =
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡

⇒ Number of atoms =
a)
b)
c)
d)

14.8
26.98
3.98
40.08
1.90
58.93
26.7
32.07

𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡

× 6.022 × 1023

× 6.022 × 1023 ≈ 3.303 × 1023 atoms of Al
× 6.022 × 1023 ≈ 4.067 × 1022 atoms of Ca
× 6.022 × 1023 ≈ 1.942 × 1022 atoms of Co
× 6.022 × 1023 ≈ 5.014 × 1023 atoms of S

Answers:
Mass = moles of element x mass of Element
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠
Moles of element =

⇒ Mass =

𝐴𝑣𝑜𝑔𝑟𝑎𝑑𝑜 𝑠𝑛𝑢𝑚𝑏𝑒𝑟
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠

𝐴𝑣𝑜𝑔𝑟𝑎𝑑𝑜′ 𝑠𝑛𝑢𝑚𝑏𝑒𝑟
3.72 ×1020

a) Mass of Cd =
b) Mass of Sb =
c) Mass of F =

6.02 ×1023
6.01 ×1023

6.02 ×1023
9.43 ×1026

× 112.41 𝑔 ≈ 0.069 𝑔 𝑜𝑓 𝐶𝑑

× 121.76 𝑔 ≈ 121.55 𝑔 𝑜𝑓 𝑆𝑏

× 19 𝑔 ≈ 29,762.46 𝑔 𝑜𝑓 𝐹

6.02 ×1023
3.72 ×1020

d) Mass of Re =

× 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐸𝑙𝑒𝑚𝑒𝑛𝑡

6.02 ×1023

× 186.20 𝑔 ≈ 115.06 𝑔 𝑜𝑓 𝑅𝑒

Answers:
a) Mass of Sn = 118.71
Atoms of Sn = 6.02 ×

𝑔

× 3.65 𝑚𝑜𝑙𝑒𝑠 ≈ 2.1973 × 1024 𝑎𝑡𝑜𝑚𝑠

𝑚𝑜𝑙
𝑎𝑡𝑜𝑚𝑠
1023
𝑚𝑜𝑙

× 0.0983 𝑚𝑜𝑙𝑒𝑠 ≈ 5.917 × 1022 𝑎𝑡𝑜𝑚𝑠

b) Mass of Xe= 131.293
Atoms of Xe = 6.02 ×
c) Moles of H =

65.5 𝑚𝑔
1

𝑔
𝑚𝑜𝑙

× 0.0983 𝑚𝑜𝑙𝑒𝑠 ≈ 12.9 𝑔

≈ 0.0655 moles

Atoms of H = 6.02 × 1023
d) Moles of Si =

× 3.65 𝑚𝑜𝑙𝑒𝑠 ≈ 433.29 𝑔

𝑚𝑜𝑙
𝑎𝑡𝑜𝑚𝑠
1023
𝑚𝑜𝑙
𝑔

𝑎𝑡𝑜𝑚𝑠

𝑚𝑜𝑙
4.35 ×1024 𝑎𝑡𝑜𝑚𝑠
𝑎𝑡𝑜𝑚𝑠

6.02 ×1023
𝑚𝑜𝑙
𝑔

Mass of Si = 28.08

𝑚𝑜𝑙

× 0.0655 𝑚𝑜𝑙𝑒𝑠 ≈ 3.943 × 1022 𝑎𝑡𝑜𝑚𝑠

≈ 7.226 moles

× 7.226 𝑚𝑜𝑙𝑒𝑠 ≈ 202.90 𝑔

6.18 ×1021 𝑎𝑡𝑜𝑚𝑠

a) Moles of Ca =

𝑎𝑡𝑜𝑚𝑠

6.02 ×1023
𝑚𝑜𝑙
𝑔

Mass of Ca = 40.07
b) Moles of Br =

× 0.010 𝑚𝑜𝑙𝑒𝑠 ≈ 0.40007 𝑔

𝑚𝑜𝑙
0.138 𝑘𝑔

79.90

𝑔
𝑚𝑜𝑙

≈ 1.727 moles

Atoms of Br = 6.02 × 1023
c) Moles of Ga =

𝑎𝑡𝑜𝑚𝑠

𝑚𝑜𝑙
9.29 ×1023 𝑎𝑡𝑜𝑚𝑠

Mass of Ga = 69.7

𝑎𝑡𝑜𝑚𝑠

6.02 ×1023
𝑚𝑜𝑙
𝑔
𝑚𝑜𝑙

d) Mass of K = 39.10
Atoms of K = 6.02 ×

≈ 0.010 moles

× 1.727 𝑚𝑜𝑙𝑒𝑠 ≈ 1.039 × 1024 𝑎𝑡𝑜𝑚𝑠

≈ 1.543 moles

× 1.543 𝑚𝑜𝑙𝑒𝑠 ≈ 107.54 𝑔
𝑔

× 10.4 𝑚𝑜𝑙𝑒𝑠 ≈ 406.64 𝑔

𝑚𝑜𝑙
𝑎𝑡𝑜𝑚𝑠
1023
𝑚𝑜𝑙

× 10.4 𝑚𝑜𝑙𝑒𝑠 ≈ 6.26 × 1024 𝑎𝑡𝑜𝑚𝑠

a) Moles of Be =

226 𝑚𝑔
9

𝑔
𝑚𝑜𝑙

≈ 0.025 moles

Atoms of Be = 6.02 × 1023
b) Moles of S =
Mass of S = 32

× 0.025 𝑚𝑜𝑙𝑒𝑠 ≈ 1.505 × 1022 𝑎𝑡𝑜𝑚𝑠

𝑚𝑜𝑙
7.36 ×1020 𝑎𝑡𝑜𝑚𝑠
𝑎𝑡𝑜𝑚𝑠

6.02 ×1023
𝑚𝑜𝑙
𝑔

𝑚𝑜𝑙

≈ 1.22 × 10−3 moles

× 1.22 × 10−3 𝑚𝑜𝑙 ≈ 0.039 𝑔

c) Mass of Nb = 92.9

𝑔
𝑚𝑜𝑙

× 0.104 𝑚𝑜𝑙𝑒𝑠 ≈ 9.66 𝑔

Atoms of Nb = 6.02 × 1023
d) Moles of N =

𝑎𝑡𝑜𝑚𝑠

0.00391 𝑘𝑔
14

𝑔
𝑚𝑜𝑙

𝑎𝑡𝑜𝑚𝑠
𝑚𝑜𝑙

× 0.104 𝑚𝑜𝑙𝑒𝑠 ≈ 6.26 × 1022 𝑎𝑡𝑜𝑚𝑠

≈ 0.279 moles

Atoms of N = 6.02 × 1023

𝑎𝑡𝑜𝑚𝑠
𝑚𝑜𝑙

× 0.279 𝑚𝑜𝑙𝑒𝑠 ≈ 1.68 × 1023 𝑎𝑡𝑜𝑚𝑠

Mass of compound = Moles of compound × Molar mass of compound
Molar mass of carbon monoxide (𝐶𝑂) = 28.01 g/mol
Molar mass of sulphur hexafluoride (𝑆𝐹6 ) = 146.06 g/mol
Molar mass of potassium hydroxide (𝐾𝑂𝐻) = 56.1 g/mol
Molar mass of lithium nitrate (𝐿𝑖𝑁𝑂3 ) = 68.95 g/mol
a) Mass of carbon monoxide = 6.10 × 28.01 ≈ 170.9 g
b) Mass of sulphur hexafluoride = 0.364 × 146.06 ≈ 53.2 g
c) Mass of potassium hydroxide = 4.44 × 56.1 ≈ 249.1 g
d) Mass of lithium nitrate = 1.71 × 68.95 ≈ 117.9 g

Mass of compound = Moles of compound × Molar mass of compound
Molar mass of sodium chloride (𝑁𝑎𝐶𝑙) = 22.99 g/mol
Molar mass of Nitrogen dioxide (𝑁𝑂2 ) = 46 g/mol
Molar mass of boron tribromide (𝐵𝐵𝑟3 ) = 250.52 g/mol
Molar mass of xenon tetroxide (𝑋𝑒𝑂4 ) = 195.29 g/mol
a) Mass of sodium chloride = 13.4 𝑚𝑜𝑙 × 22.99
b) Mass of Nitrogen dioxide = 0.223 𝑘𝑚𝑜𝑙 × 46

𝑔

≈ 308.06 g

𝑚𝑜𝑙
𝑔
𝑚𝑜𝑙

≈ 10,256 g

c) Mass of boron tribromide = 0.00322 𝑚𝑜𝑙 × 250.52
d) Mass of xenon tetroxide = 144 𝑚𝑚𝑜𝑙 × 195.29

Moles of compound =

𝑔
𝑚𝑜𝑙

𝑔
𝑚𝑜𝑙

≈ 0.806 g

≈ 28.12 g

𝑀𝑎𝑠𝑠
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑

Molar mass of ammonium iodide (𝑁𝐻4 𝐼) = 144.94 g/mol
Molar mass of potassium cyanide (𝐾𝐶𝑁) = 65.12 g/mol
Molar mass of sodium sulfate (𝑁𝑎2 𝑆𝑂4 ) = 142.04 g/mol
Molar mass of hydrocholoric acid (𝐻𝐶𝑙) = 36.46 g/mol
a) Moles of ammonium iodide =
b) Moles of potassium cyanide =
c) Moles of sodium sulfate=

29.8 𝑔
144.94 𝑔/𝑚𝑜𝑙
8.34 𝑔

65.12 𝑔/𝑚𝑜𝑙
3.5 𝑚𝑔

≈ 0.128 mol

≈ 2.46 𝑥 10...


Anonymous
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