How many electrons are in:

a. One mole of lithium atoms

b. Two moles of carbon atoms

c. One mole of carbon dioxide molecules

d. Half a mole of methane molecules

1 answer

To calculate the number of electrons in each of the provided substances, we can use Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles (atoms or molecules) per mole.

a. One mole of lithium atoms
Lithium (Li) has an atomic number of 3, meaning each lithium atom has 3 electrons.
Number of electrons = \(3 , \text{electrons/atom} \times 6.022 \times 10^{23} , \text{atoms}\) \[ = 1.8066 \times 10^{24} , \text{electrons} \]

b. Two moles of carbon atoms
Carbon (C) has an atomic number of 6, meaning each carbon atom has 6 electrons.
Number of electrons in two moles = \(2 , \text{moles} \times 6 , \text{electrons/atom} \times 6.022 \times 10^{23} , \text{atoms/mole}\) \[ = 2 \times 6 \times 6.022 \times 10^{23} = 7.22784 \times 10^{24} , \text{electrons} \]

c. One mole of carbon dioxide molecules
Each carbon dioxide (CO₂) molecule consists of 1 carbon atom and 2 oxygen atoms. Carbon has 6 electrons, and oxygen (O) has an atomic number of 8, so each oxygen atom has 8 electrons.
Total electrons in one CO₂ molecule: \[ 6 , (\text{from C}) + 2 \times 8 , (\text{from 2 O}) = 6 + 16 = 22 , \text{electrons} \] Number of electrons in one mole of CO₂ = \(22 , \text{electrons/molecule} \times 6.022 \times 10^{23} , \text{molecules}\) \[ = 1.32684 \times 10^{25} , \text{electrons} \]

d. Half a mole of methane molecules
Methane (CH₄) consists of 1 carbon atom and 4 hydrogen atoms.
Total electrons in one CH₄ molecule: \[ 6 , (\text{from C}) + 4 \times 1 , (\text{from 4 H}) = 6 + 4 = 10 , \text{electrons} \] Number of electrons in half a mole of CH₄ = \(0.5 , \text{mole} \times 10 , \text{electrons/molecule} \times 6.022 \times 10^{23} , \text{molecules/mole}\) \[ = 0.5 \times 10 \times 6.022 \times 10^{23} = 3.011 \times 10^{24} , \text{electrons} \]

Summary:

  • a. \(1.8066 \times 10^{24} , \text{electrons}\)
  • b. \(7.22784 \times 10^{24} , \text{electrons}\)
  • c. \(1.32684 \times 10^{25} , \text{electrons}\)
  • d. \(3.011 \times 10^{24} , \text{electrons}\)