Mole Conversion Practice Quiz

Moles = Mass / Molar mass = 48 g / 24 g/mol = 2 moles

Mass = Moles × Molar mass = 0.5 mol × 18 g/mol = 9 g

One mole of any substance contains Avogadro's number (6.022 x 10²³) of particles (atoms, molecules, ions, etc.).

Moles = Mass / Molar mass = 11 g / 44 g/mol = 0.25 moles

Molecules = Moles × Avogadro's number = 2 mol × 6.022 x 10²³ molecules/mol = 1.2044 x 10²⁴ molecules

At STP (0°C and 1 atm), 1 mole of any ideal gas occupies 22.4 L.

Moles = Volume / 22.4 L/mol = 44.8 L / 22.4 L/mol = 2 moles

Moles = 11.2 L / 22.4 L/mol = 0.5 moles.
Mass = 0.5 mol × 28 g/mol = 14 g

Mass = Moles × Molar mass = 0.25 mol × 98 g/mol = 24.5 g

Explanation: Empirical Formula: C: 85.7/12=7.14, H: 14.3/1=14.3. Ratio is approximately 1:2. So empirical formula is CH2. Molar mass of gas = 1.25g/ (1/22.4) = 28g/mol. Empirical formula mass = 14g/mol. n= 28/14 =2. Molecular formula is (CH2)2 = C2H4.

Moles = Number of molecules / Avogadro’s number = 3.011 x 10²³ / 6.022 x 10²³ = 0.5 moles

0.5x22.4 = 11.2

1. Find moles of oxygen: Divide the number of oxygen atoms by Avogadro's number: (1.204 x 10²⁴) / (6.022 x 10²³) = 2 moles of O atoms.

2. Relate moles of O to moles of Fe₂O₃: One mole of Fe₂O₃ contains three moles of O atoms (from the formula Fe₂O₃). Therefore, if you have 2 moles of O atoms, you have less than 1 mole of Fe₂O₃.

3. Calculate moles of Fe₂O₃: Divide the moles of oxygen by 3 (because there are 3 oxygens in each Fe₂O₃): 2 moles O / 3 = 0.667 moles of Fe₂O₃.

Moles = 5.6 L / 22.4 L/mol = 0.25 moles.

Molar mass = 11 g / 0.25 mol = 44 g/mol.

The balanced equation is 2H₂ + O₂ → 2H₂O.

Moles of H₂ = 4 g / 2 g/mol = 2 moles.

From the equation, 2 moles of H₂ produce 2 moles of H₂O.

Mass of H₂O = 2 mol × 18 g/mol = 36 g.

Explanation: CH₄ + 2O₂ → CO₂ + 2H₂O.

Moles of CH₄ = 4 g / 16 g/mol = 0.25 mol.

This requires 0.5 mol of O₂. (from equation)

Volume of O₂ = 0.5 mol × 22.4 L/mol = 11.2 L.