calculate the lowest whole number ratio of
Moles of Mg/moles of O
average of Moles of Mg/Moles of O
Trial 1
mass of empty crucible and lid= 36.560
mass of Mg=0.348
mass of crucible, lid, and product=38.229
Trail 2
mass of empty crucible and lid=32.221
mass of Mg=0.340
mass of crucible, lid, and product=33.757
To calculate the lowest whole number ratio of moles of Mg to moles of O, you need to determine the number of moles of Mg and O individually.
First, calculate the moles of Mg:
1. Determine the change in mass of Mg by subtracting the mass of the empty crucible and lid from the mass of the crucible, lid, and Mg product for each trial.
- Trial 1: Change in mass of Mg = (mass of crucible, lid, and product) - (mass of empty crucible and lid)
= 38.229 g - 36.560 g
= 1.669 g
- Trial 2: Change in mass of Mg = (mass of crucible, lid, and product) - (mass of empty crucible and lid)
= 33.757 g - 32.221 g
= 1.536 g
2. Calculate the moles of Mg using the molar mass of Mg, which is 24.31 g/mol.
- Trial 1: Moles of Mg = (change in mass of Mg) / (molar mass of Mg)
= 1.669 g / 24.31 g/mol
= 0.0686 mol (rounded to 4 decimal places)
- Trial 2: Moles of Mg = (change in mass of Mg) / (molar mass of Mg)
= 1.536 g / 24.31 g/mol
= 0.0631 mol (rounded to 4 decimal places)
Next, calculate the moles of O:
1. Determine the change in mass of O by subtracting the mass of Mg from the mass of the crucible, lid, and product for each trial.
- Trial 1: Change in mass of O = (mass of crucible, lid, and product) - (mass of Mg)
= 38.229 g - 0.348 g
= 37.881 g
- Trial 2: Change in mass of O = (mass of crucible, lid, and product) - (mass of Mg)
= 33.757 g - 0.340 g
= 33.417 g
2. Calculate the moles of O using the molar mass of O, which is 16.00 g/mol.
- Trial 1: Moles of O = (change in mass of O) / (molar mass of O)
= 37.881 g / 16.00 g/mol
= 2.37 mol (rounded to 2 decimal places)
- Trial 2: Moles of O = (change in mass of O) / (molar mass of O)
= 33.417 g / 16.00 g/mol
= 2.09 mol (rounded to 2 decimal places)
Finally, determine the lowest whole number ratio of moles of Mg to moles of O by dividing the number of moles of Mg and O by their smallest common factor. In this case, the smallest common factor is 0.01.
- Trial 1: Ratio of Moles of Mg/Moles of O = (0.0686 mol) / (0.01 mol)
= 6.86 / 0.01
= 686:1 (rounded to the nearest whole number)
- Trial 2: Ratio of Moles of Mg/Moles of O = (0.0631 mol) / (0.01 mol)
= 6.31 / 0.01
= 631:1 (rounded to the nearest whole number)
Therefore, the lowest whole number ratios of moles of Mg to moles of O for Trial 1 and Trial 2 are 686:1 and 631:1, respectively.