Mole Conversions - Answer Key

Section A Calculating Molar Mass - Answers

Problem 1

Calculate the mass in grams of each:

a. \(3.44 \times 10^{-22}\) g
b. \(1.79 \times 10^{-22}\) g
c. \(2.99 \times 10^{-23}\) g
d. \(3.88 \times 10^{-22}\) g

Strategy: Mass of 1 atom = \(\frac{\text{molar mass}}{6.02 \times 10^{23}}\)

a. Pb: molar mass = 207.2 g/mol
\[\frac{207.2 \text{ g/mol}}{6.02 \times 10^{23} \text{ atoms/mol}} = 3.44 \times 10^{-22} \text{ g/atom}\]
b. Ag: 107.9 g/mol → \(1.79 \times 10^{-22}\) g
c. H₂O: (2×1.01 + 16.00) = 18.02 g/mol → \(2.99 \times 10^{-23}\) g
d. C₃H₈(NO₃)₃: (3×12.01 + 8×1.01 + 3×14.01 + 9×16.00) = 233.11 g/mol → \(3.88 \times 10^{-22}\) g

Section B Atoms, Molecules & Avogadro's Number - Answers

Problem 2

How many atoms are contained in each?

a. 99 atoms
b. 25.0 atoms
c. \(4.52 \times 10^{20}\) atoms
d. \(2.76 \times 10^{24}\) atoms

a. C₂H₅OH has 9 atoms per molecule: 11 molecules × 9 = 99 atoms

b. Given directly: 25.0 atoms

c. 0.0986 g Xe:
\[\frac{0.0986 \text{ g}}{131.3 \text{ g/mol}} = 7.51 \times 10^{-4} \text{ mol}\] \[7.51 \times 10^{-4} \text{ mol} \times 6.02 \times 10^{23} = 4.52 \times 10^{20} \text{ atoms}\]
d. 72.5 g CHCl₃ (molar mass = 119.37 g/mol, 5 atoms/molecule):
\[\frac{72.5}{119.37} = 0.607 \text{ mol} \times 6.02 \times 10^{23} = 3.66 \times 10^{23} \text{ molecules}\] \[3.66 \times 10^{23} \times 5 \text{ atoms/molecule} = 1.83 \times 10^{24} \text{ atoms}\]

Problem 3

1 mol of CS₂ contains:

a. \(6.02 \times 10^{23}\) CS₂ molecules
b. \(6.02 \times 10^{23}\) C atoms
c. \(1.20 \times 10^{24}\) S atoms
d. \(1.81 \times 10^{24}\) total atoms

a. 1 mol = \(6.02 \times 10^{23}\) molecules (by definition)
b. 1 C per CS₂ → \(6.02 \times 10^{23}\) C atoms
c. 2 S per CS₂ → \(2 \times 6.02 \times 10^{23} = 1.20 \times 10^{24}\) S atoms
d. 3 atoms total per CS₂ → \(3 \times 6.02 \times 10^{23} = 1.81 \times 10^{24}\) atoms

Problem 4

Oxygen atoms in each:

a. \(6.02 \times 10^{23}\) O atoms
b. \(3.74 \times 10^{23}\) O atoms
c. \(2.41 \times 10^{23}\) O atoms

a. 16.0 g O₂ (molar mass 32.00 g/mol):
\[\frac{16.0}{32.00} = 0.500 \text{ mol O}_2\] 2 O atoms per O₂: \(0.500 \times 2 \times 6.02 \times 10^{23} = 6.02 \times 10^{23}\)

b. 0.622 mol MgO (1 O per MgO):
\[0.622 \times 6.02 \times 10^{23} = 3.74 \times 10^{23}\)

c. \(6.00 \times 10^{22}\) molecules C₉H₈O₄ (4 O per molecule):
\[6.00 \times 10^{22} \times 4 = 2.40 \times 10^{23}\]

Problem 5

Avogadro's number of paper sheets:

\(5.54 \times 10^{19}\) m (about 587 light-years!)

Height per sheet:
\[\frac{4.60 \text{ cm}}{500 \text{ sheets}} = 0.0092 \text{ cm/sheet}\]
Total height:
\[6.02 \times 10^{23} \times 0.0092 \text{ cm} = 5.54 \times 10^{21} \text{ cm}\] \[5.54 \times 10^{21} \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = 5.54 \times 10^{19} \text{ m}\]

Section C Grams ↔ Moles - Answers

Problem 6

Grams of sodium with same atoms as 10.0 g K:

5.88 g Na

Moles of K:
\[\frac{10.0 \text{ g}}{39.10 \text{ g/mol}} = 0.256 \text{ mol K}\]
Same moles of Na:
\[0.256 \text{ mol} \times 22.99 \text{ g/mol} = 5.88 \text{ g Na}\]

Problem 7

Molar mass of unknown element:

10.8 g/mol (likely Boron)

\[\text{Molar mass} = \frac{\text{mass of 1 atom} \times N_A}{1 \text{ mol}}\] \[= 1.79 \times 10^{-23} \text{ g} \times 6.02 \times 10^{23} = 10.8 \text{ g/mol}\]

Problem 8

Sugar molecules in 5-lb bag:

\(4.0 \times 10^{24}\) molecules

Convert to grams:
\[5 \text{ lb} \times \frac{454 \text{ g}}{1 \text{ lb}} = 2270 \text{ g}\]
Molar mass C₁₂H₂₂O₁₁ = 342.34 g/mol
\[\frac{2270 \text{ g}}{342.34 \text{ g/mol}} = 6.63 \text{ mol}\] \[6.63 \text{ mol} \times 6.02 \times 10^{23} = 3.99 \times 10^{24} \text{ molecules}\]

Section D Real-World Scale - Answers

Problem 9

Water drops calculations:

a. \(8.1 \times 10^{16}\) drops
b. \(7.2 \times 10^5\) cubic miles

a. 1 cubic mile in drops:
\[1 \text{ mi}^3 \times \left(\frac{5280 \text{ ft}}{1 \text{ mi}}\right)^3 \times \left(\frac{12 \text{ in}}{1 \text{ ft}}\right)^3 \times \left(\frac{2.54 \text{ cm}}{1 \text{ in}}\right)^3 \times \frac{20 \text{ drops}}{1 \text{ cm}^3}\] \[= 8.1 \times 10^{16} \text{ drops}\]
b. Volume of a mole of drops:
\[\frac{6.02 \times 10^{23} \text{ drops}}{8.1 \times 10^{16} \text{ drops/mi}^3} = 7.4 \times 10^6 \text{ mi}^3\]

Problem 10

Copper atoms in penny:

\(2.4 \times 10^{22}\) Cu atoms

\[\frac{2.5 \text{ g}}{63.55 \text{ g/mol}} = 0.0393 \text{ mol}\] \[0.0393 \text{ mol} \times 6.02 \times 10^{23} = 2.37 \times 10^{22} \text{ atoms}\]

Problem 11

Mg needed for twice Al atoms:

16 g Mg

Moles of Al:
\[\frac{18 \text{ g}}{26.98 \text{ g/mol}} = 0.667 \text{ mol Al}\]
Need twice as many atoms = 2 × 0.667 = 1.334 mol Mg:
\[1.334 \text{ mol} \times 24.31 \text{ g/mol} = 32.4 \text{ g Mg}\]

Section E Multi-Step Challenges - Answers

Problem 12

Atoms in 0.350 mol P₄:

\(8.43 \times 10^{23}\) atoms

4 P atoms per P₄ molecule:
\[0.350 \text{ mol} \times 4 \text{ atoms/molecule} \times 6.02 \times 10^{23} = 8.43 \times 10^{23} \text{ atoms}\]

Problem 13

1.00-g samples comparison:

a. H₂O has most molecules
b. H₂O has most atoms

Moles in each sample:
O₂: 1.00/32.00 = 0.03125 mol → 1.88 × 10²² molecules × 2 atoms = 3.76 × 10²² atoms
H₂O: 1.00/18.02 = 0.0555 mol → 3.34 × 10²² molecules × 3 atoms = 1.00 × 10²³ atoms
CH₃OH: 1.00/32.05 = 0.0312 mol → 1.88 × 10²² molecules × 6 atoms = 1.13 × 10²³ atoms

Most molecules: H₂O (0.0555 mol)
Most atoms: CH₃OH (1.13 × 10²³ atoms)

Problem 14

Grams of Fe₂S₃ for Avogadro's number of atoms:

41.5 g

Fe₂S₃ has 5 atoms per formula unit
To get 6.02 × 10²³ total atoms, need 6.02 × 10²³ ÷ 5 = 1.20 × 10²³ formula units = 0.200 mol
Molar mass Fe₂S₃ = 2(55.85) + 3(32.07) = 207.91 g/mol
\[0.200 \text{ mol} \times 207.91 \text{ g/mol} = 41.6 \text{ g}\]

Problem 15

Ca combined with P in Ca₃P₂:

Ratio is 3:2, so answer depends on amount of P given