Understanding the Mole Concept
The mole is the fundamental unit in chemistry that bridges the gap between the microscopic world of atoms and the macroscopic world of laboratory measurements. To solve these problems, we first need the atomic masses of Hydrogen (H) and Sulfur (S):
- Atomic mass of H: 1.008 g/mol (often rounded to 1.01)
- Atomic mass of S: 32.06 g/mol
Molar Mass of H₂S
To find the molar mass of $H_2S$, we sum the masses of its constituent atoms: $$M(H_2S) = (2 \times 1.008) + 32.06 = 34.076 \text{ g/mol}$$
Step-by-Step Solutions
a. Grams of H₂S in 0.400 mole
Use the formula: $\text{mass} = \text{moles} \times \text{molar mass}$ $$\text{Mass} = 0.400 \text{ mol} \times 34.076 \text{ g/mol} = 13.6304 \text{ g} \approx 13.63 \text{ g}$$
b. Gram-atoms of H and S
One mole of $H_2S$ contains 2 moles of H atoms and 1 mole of S atoms.
- H: $0.400 \times 2 = 0.800 \text{ gram-atoms of H}$
- S: $0.400 \times 1 = 0.400 \text{ gram-atoms of S}$
c. Grams of H and S
Multiply the moles of atoms by their respective atomic masses.
- Mass of H: $0.800 \text{ mol} \times 1.008 \text{ g/mol} = 0.8064 \text{ g}$
- Mass of S: $0.400 \text{ mol} \times 32.06 \text{ g/mol} = 12.824 \text{ g}$
d. Molecules of H₂S
Use Avogadro's Number ($N_A = 6.022 \times 10^{23} \text{ molecules/mol}$): $$\text{Molecules} = 0.400 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 2.4088 \times 10^{23} \text{ molecules}$$
e. Atoms of H and S
Multiply the number of molecules by the number of specific atoms per molecule.
- Atoms of H: $(2.4088 \times 10^{23} \text{ molecules}) \times 2 = 4.8176 \times 10^{23} \text{ atoms of H}$
- Atoms of S: $(2.4088 \times 10^{23} \text{ molecules}) \times 1 = 2.4088 \times 10^{23} \text{ atoms of S}$