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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 6 - Problem 108p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 6 - Problem 108p

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# Complete the table.SubstanceMassMolesNumber of Particles (atoms or molecules)C6H12O615.8

ISBN: 9780321910295 34

## Solution for problem 108P Chapter 6

Introductory Chemistry | 5th Edition

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Problem 108P

Complete the table.

 Substance Mass Moles Number of Particles (atoms or molecules) C6H12O6 15.8 g ______ ______ Pb ______ ______ 9.04 × 1021 CF4 22.5 mg ______ ______ C ______ 0.0388 ______
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Step 1 of 3

Solution  108P

The complete table is given below.

 Substance Mass Moles Number of Particles (atoms or molecules) C6H12O6 15.8 g 0.0877 moles 0.528 x 1023 molecules Pb 3.110 g 0.0150  moles 9.04 × 1021 CF4 22.5 mg 0.000256  moles 1.54 x 1020 molecules. C 0.4659 g 0.0388 0.2337 x 1023 atoms

1. The molar mass of C6H12O6 = 6 x 12.01 g/mol + 12 x 1.008 g/mol + 6 x16 g/mol

= 180.156 g/mol

Thus,

180.156 g = 1 mol

1 mol = 6.023 x 1023 molecules

15.8 g contains = 15.8 g x = 0.0877 mol

15.8 g contains = 15.8 g  x x  = 0.528 x 1023 molecules

(2) The molar mass of Pb  = 207.2  g/mol

207.2 g = 1 mol

1 mol = 6.023 x 1023 atoms  = 207.2 g

1000 mg =1 g

9.04 x 1021 contains = 9.04 x 1021  x = 0.0150  mol

9.04 x 1021 contains  =  9.04 x 1021  atoms x    x  =3.110 g

(3) The molar mass of CF4  =12.01 g/mol + 4 x 18.998 g/mol = 88.002 g/mol

88.002 g  = 1 mol

1 mol = 6.023 x 1023 molecules  = 88.002 g

22.5 mg contains =22.5 mg x   x = 0.000256  mol

22.5 mg  contains  = 22.5 mg x  x   x  = 1.54 x 1020 molecules.

(4) The molar mass of C  =12.01 g/mol

12.01 g  = 1 mol

1 mol = 6.023 x 1023 atoms

0.0388  contains =0.0388 moles  x  = 0.4659 g

0.0 388  mole contains  = 0.0388 moles x  = 0.2337 x 1023 atoms

Step 2 of 3

Step 3 of 3

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