# Practice Homework 16: Mass ⇔ Moles ⇔ Particles

Download Practice Homework 16: Mass ⇔ Moles ⇔ Particles

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Practice Homework 16: Mass Moles Particles

This assignment is for practice only. It is not due. Remember: write your conversion factors for both MM (molar mass) and Avogadro’s Number when

doing these problems. Go to Questions 7, 8, 9, 10, and 11 for “twist” problems.

1) Calculate the number of carbon atoms in a 20.00-gram sample of carbon. The molar mass of carbon is 12.01 g/mol.

2) A sample of lead contains 4.750 × 1022 atoms of lead. Determine the mass (in grams) of the lead in the sample. The molar mass of lead is 207.2 g/mol.

3) A sample of zinc(II) chloride, ZnCl2, contains 3.204 × 1025 form of ZnCl2. Determine the mass (in grams) of zinc(II) chloride in the sample. The molar mass of zinc(II) chloride is 136.29 g/mol.

4) A sample of phosphorus triiodide contains 5.28 × 1022 molecules of phosphorus triiodide. Calculate the mass (in grams) of phosphorus triiodide in the sample. The molar mass of PI3 is 411.7 g/mol.

5) A 0.400-gram sample of acetylsalicylic acid (also called “aspirin”, C9H8O4, 180.16 g/mol) is dissolved in 2.00 L of water. How many molecules of C9H8O4 are present in the liquid? (Note: The volume of water, 2.00 L, is not needed in this calculation. But, in two weeks, it will be!

6) A sample of magnesium sulfate (espom salts, MgSO4, 120.37 g/mol) is found to have a mass of 0.24 grams. Calculate the number of magnesium sulfate formula units in the sample.

7) Calculate the volume (in mL) of bromine liquid, Br2(l), that contains 2.394 × 1023 molecules of Br2(l). The molar mass of bromine liquid is 159.8 g/mol. The density of bromine liquid is 3.1 g/mL. Treat molar mass, density, and Avogadro’s Number like exact numbers in this question.

Hint: Write all three conversion factors: the MM conversion factor, the Av# conversion factor, and the density conversion factor.

Note: The solution for this problem is included in this answer key.

8) Octane liquid, C8H18, is a component of gasoline. Calculate the number of octane molecules in a 212.4-liter sample of octane. The molar mass of octane is 114.23 g/mol. The density of octane is 0.703 g/mL. Again, you must use the original value of density in this calculation.

9) A sample of sucrose (table sugar), C12H22O11, has a mass of 12.00 grams. The molar mass of sucrose is 342.30 g/mol. Calculate the number of carbon atoms in the sample. The solution for this problem is included in this answer key.

10) Seratonin is a biological chemical that is believed to have a role in regulating sleeping patterns. Its chemical formula is C10H12N2O, and its molar mass is 176.22 g/mol. A sample of seratonin contains 2.99 × 1019 nitrogen atoms. Calculate the mass (in mg) of seratonin in the sample.

11)* Challenge Problem: A sample of carbon tetrachloride, an organic liquid, has a molar mass of 153.82 g/mol. A sample of carbon tetrachloride contains 4.02 × 1017 atoms of chlorine. Calculate the volume of carbon tetrachloride in nanoliters. Don’t worry: this would be an extra credit type question. The density of carbon tetrachloride is 1.59 g/mL.

Page 1 of 3

Answers

Remember: You can always write answers in scientific notation form with no penalty (as long as the values are written correctly in scientific notation form.)

1) 1.003 × 1024 C atoms

2) 163.4 g Pb (atoms)

OR 1.634 × 102 g Pb (atoms)

3) 7.251 × 103 g ZnCl2 (form)

OR 7.251 × 103 g ZnCl2 (form)

4) 35.9 g PI3 (molec)

OR 3.59 × 101 g PI3 (molec)

5) 1.34 × 1021 C9H8O4 molec

6) 1.2 × 1021 MgSO4 form

Question 7 Solution 159.8 g Br2 = 1 mol Br2

1 mol Br2 = 6.022 × 1023 Br2 molec

3.1 g Br2 = 1 mL Br2

2.394 × 1023 Br2 molec

1 mol Br2

159.8 g Br2 1 mL Br2

( 1 ) (6.022 × 1023 Br2 molec) ( 1 mol Br2 ) (3.1 g Br2) =

20.49 mL Br2

8) 7.872 × 1026 C8H18 molec

Page 2 of 3

Question 9 Solution 342.30 g C12H22O11 = 1 mol C12H22O11

1 C12H22O11 molec = 12 C atoms

12.00 g C12H22O11

1 mol C12H22O11

6.022 × 1023 C12H22O11 molec

12 C atoms

23

9) (

1

) (342.30 g C H O ) (

1 mol C H O

) (1 C H O molec) = 2.553 × 10 C atoms

12 22 11

12 22 11

12 22 11

10) 4.37 mg C10H12N2O 11) 16.2 mL CCl4

Page 3 of 3

This assignment is for practice only. It is not due. Remember: write your conversion factors for both MM (molar mass) and Avogadro’s Number when

doing these problems. Go to Questions 7, 8, 9, 10, and 11 for “twist” problems.

1) Calculate the number of carbon atoms in a 20.00-gram sample of carbon. The molar mass of carbon is 12.01 g/mol.

2) A sample of lead contains 4.750 × 1022 atoms of lead. Determine the mass (in grams) of the lead in the sample. The molar mass of lead is 207.2 g/mol.

3) A sample of zinc(II) chloride, ZnCl2, contains 3.204 × 1025 form of ZnCl2. Determine the mass (in grams) of zinc(II) chloride in the sample. The molar mass of zinc(II) chloride is 136.29 g/mol.

4) A sample of phosphorus triiodide contains 5.28 × 1022 molecules of phosphorus triiodide. Calculate the mass (in grams) of phosphorus triiodide in the sample. The molar mass of PI3 is 411.7 g/mol.

5) A 0.400-gram sample of acetylsalicylic acid (also called “aspirin”, C9H8O4, 180.16 g/mol) is dissolved in 2.00 L of water. How many molecules of C9H8O4 are present in the liquid? (Note: The volume of water, 2.00 L, is not needed in this calculation. But, in two weeks, it will be!

6) A sample of magnesium sulfate (espom salts, MgSO4, 120.37 g/mol) is found to have a mass of 0.24 grams. Calculate the number of magnesium sulfate formula units in the sample.

7) Calculate the volume (in mL) of bromine liquid, Br2(l), that contains 2.394 × 1023 molecules of Br2(l). The molar mass of bromine liquid is 159.8 g/mol. The density of bromine liquid is 3.1 g/mL. Treat molar mass, density, and Avogadro’s Number like exact numbers in this question.

Hint: Write all three conversion factors: the MM conversion factor, the Av# conversion factor, and the density conversion factor.

Note: The solution for this problem is included in this answer key.

8) Octane liquid, C8H18, is a component of gasoline. Calculate the number of octane molecules in a 212.4-liter sample of octane. The molar mass of octane is 114.23 g/mol. The density of octane is 0.703 g/mL. Again, you must use the original value of density in this calculation.

9) A sample of sucrose (table sugar), C12H22O11, has a mass of 12.00 grams. The molar mass of sucrose is 342.30 g/mol. Calculate the number of carbon atoms in the sample. The solution for this problem is included in this answer key.

10) Seratonin is a biological chemical that is believed to have a role in regulating sleeping patterns. Its chemical formula is C10H12N2O, and its molar mass is 176.22 g/mol. A sample of seratonin contains 2.99 × 1019 nitrogen atoms. Calculate the mass (in mg) of seratonin in the sample.

11)* Challenge Problem: A sample of carbon tetrachloride, an organic liquid, has a molar mass of 153.82 g/mol. A sample of carbon tetrachloride contains 4.02 × 1017 atoms of chlorine. Calculate the volume of carbon tetrachloride in nanoliters. Don’t worry: this would be an extra credit type question. The density of carbon tetrachloride is 1.59 g/mL.

Page 1 of 3

Answers

Remember: You can always write answers in scientific notation form with no penalty (as long as the values are written correctly in scientific notation form.)

1) 1.003 × 1024 C atoms

2) 163.4 g Pb (atoms)

OR 1.634 × 102 g Pb (atoms)

3) 7.251 × 103 g ZnCl2 (form)

OR 7.251 × 103 g ZnCl2 (form)

4) 35.9 g PI3 (molec)

OR 3.59 × 101 g PI3 (molec)

5) 1.34 × 1021 C9H8O4 molec

6) 1.2 × 1021 MgSO4 form

Question 7 Solution 159.8 g Br2 = 1 mol Br2

1 mol Br2 = 6.022 × 1023 Br2 molec

3.1 g Br2 = 1 mL Br2

2.394 × 1023 Br2 molec

1 mol Br2

159.8 g Br2 1 mL Br2

( 1 ) (6.022 × 1023 Br2 molec) ( 1 mol Br2 ) (3.1 g Br2) =

20.49 mL Br2

8) 7.872 × 1026 C8H18 molec

Page 2 of 3

Question 9 Solution 342.30 g C12H22O11 = 1 mol C12H22O11

1 C12H22O11 molec = 12 C atoms

12.00 g C12H22O11

1 mol C12H22O11

6.022 × 1023 C12H22O11 molec

12 C atoms

23

9) (

1

) (342.30 g C H O ) (

1 mol C H O

) (1 C H O molec) = 2.553 × 10 C atoms

12 22 11

12 22 11

12 22 11

10) 4.37 mg C10H12N2O 11) 16.2 mL CCl4

Page 3 of 3

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