Finding vapor pressure / Partial pressure and moles using the Data collection

Convert : 60 grams of O, --> _Moles of O,. (Use 32.00 grams = 1 mole of O, as the molar mass.) * Liters of Gas (at STP) 22 4 Liters 1 mole Divide by Multiply by 22.4 Liters 22.4 Liters 1 mole 22.4 Liters Multiply by Molar Mass Moles Divide by Molar Mass 1 mole (6.02 x 10^23) 6.02 x 10 atoms or molecules 1 mole Multiply by (6.02 x 10^23) Divide by Particles Molar Mass 6.02 x 10 atoms or molecules (Atoms or Molecules) Weight in 1 mole grams 1 mole Molar Mass

Data Set 1 Mass of Mg metal: 0.0462g Temperature of the water in the beaker: 22.0°C Total Volume of Gas Collected: 44.1 mL Barometric Pressure: 753 torr. Height of water column in Eudiometer: 0.0 mm Data Set 2 Mass of Mg metal: 0.0441 g Temperature of the water in the beaker: 22..6°C Total Volume of Gas Collected: 43.6 mL Barometric Pressure: 753 torr. Height of water column in the eudiometer: 19.0 mm

Hi! I am unsure of how to solve this problem. Thank you!