## Chemistry: Molecular Approach (4th Edition)

1. Calculate the mass of one mole of electrons in grams as follows: \begin{align} & \text{m(electrons})=\left( \text{m(electron}) \right)\times \left( Avogadros\text{ number} \right) \\ & \text{=}\left( \text{9}\text{.10938}\times \text{1}{{\text{0}}^{-\text{28}}}\text{ g/electron} \right)\left( 6.022\times \text{1}{{\text{0}}^{\text{23}}}\text{ electrons/mol} \right) \\ & =\text{5}\text{.48567}\times \text{1}{{\text{0}}^{-\text{4}}}\text{ }g/mol \end{align} 2. Calculate the mass of one mole of protons in grams as follows: \begin{align} & \text{m(}proton\text{s})=\left( \text{m(}proton) \right)\times \left( Avogadros\text{ number} \right) \\ & \text{=}\left( \text{1}\text{.67262}\times \text{1}{{\text{0}}^{-\text{24}}}\text{ g/}proton \right)\left( 6.022\times \text{1}{{\text{0}}^{\text{23}}}\text{ }proton\text{s/mol} \right) \\ & =1.00725\text{ }g/mol \end{align} 3. Calculate the mass of one mole of neutrons in grams as follows: \begin{align} & \text{m(}neutron\text{s})=\left( \text{m(}neutron) \right)\times \left( Avogadros\text{ number} \right) \\ & \text{=}\left( \text{1}\text{.67493}\times \text{1}{{\text{0}}^{-\text{24}}}\text{ g/}neutron \right)\left( 6.022\times \text{1}{{\text{0}}^{\text{23}}}\text{ }neutron\text{s/mol} \right) \\ & =1.00864\text{ }g/mol \end{align} 4. Calculate the mass of one mole of $\text{C-12}$ atoms in grams as follows: \begin{align} & \text{m(C-12 atom}s)=\left( \text{m(C-12 atom}) \right)\times \left( Avogadros\text{ number} \right) \\ & \text{=}\left( \text{1}\text{.992646}\times \text{1}{{\text{0}}^{-\text{23}}}\text{ g/C-12 atom} \right)\left( 6.022\times \text{1}{{\text{0}}^{\text{23}}}\text{ C-12 atoms/mol} \right) \\ & =11.9997\text{ }g/mol \end{align} 5. Calculate the mass of one mole of doughnuts in grams as follows: \begin{align} & \text{m(}doughnuts)=\left( \text{m(}doughnut) \right)\times \left( Avogadros\text{ number} \right) \\ & \text{=}\left( \text{74 g/}doughnut \right)\left( 6.022\times \text{1}{{\text{0}}^{\text{23}}}\text{ C-12 }doughnut\text{s/mol} \right) \\ & =\text{4}\text{.45628}\times \text{1}{{\text{0}}^{\text{25}}}\text{ }g/mol \end{align} 6. One mole of atoms of carbon contains six moles of each of electrons, protons, and neutrons. Compare the mass of $1\text{ mol(C-atoms)}$ to the sum of the masses of its constituent particles as follows: \begin{align} & \frac{\text{m(1 mole of C-12 atom}s)}{sum\text{ }of\text{ }the\text{ }masses\text{ }of\text{ }particles}=\frac{\text{m(1 mole of C-12 atom}s)}{\text{6 mol}\times \left\{ \text{m(electrons}s)+\text{m(proton}s)+\text{m(neutron}s) \right\}} \\ & =\frac{11.9997\text{ }g/mol}{6\text{ mol}\times \left\{ \left( \text{5}\text{.48567}\times \text{1}{{\text{0}}^{-\text{4}}}\text{ }g/mol \right)+\left( 1.00725\text{ }g/mol \right)+\left( 1.00864\text{ }g/mol \right) \right\}} \\ & =0.99 \\ & \sim 1 \end{align}7. Calculate the number of atoms of carbon in a doughnut as follows: \begin{align} & \text{n(C-atoms)}=\frac{mass\text{ }of\text{ }1\text{ doughnut}}{mass\text{ }of\text{ 1 C-atom}} \\ & =\left( \frac{\text{74 }g/doughnut}{\text{1}\text{.992646}\times \text{1}{{\text{0}}^{-\text{23}}}\text{ g/C-atom}} \right) \\ & =37.1\times {{10}^{23}}\ \text{C-atoms}/doughnut \end{align} The mass of one mole of each item in grams is: \begin{align} & \text{m(electrons})\text{=5}\text{.48567}\times \text{1}{{\text{0}}^{-\text{4}}}\text{ }g/mol \\ & \text{m(protons})\text{=}1.00725\text{ }g/mol \\ & \text{m(neutron})\text{=}1.00864\text{ }g/mol \\ & \text{m(C-12 atoms})\text{=}11.9997\text{ }g/mol \\ & \text{m(doughnuts})\text{=4}\text{.45628}\times \text{1}{{\text{0}}^{\text{25}}}\text{ }g/mol \\ \end{align} Mass of $1\text{ mol(C-atoms)}$is similar to the sum of the masses of the constituent particles and the number of atoms of carbon in a doughnut is $\underline{\text{37}\text{.1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\ \text{C-atoms/doughnut}}$.