an unknown compound is known to contain 30.34%M and 60.56%O and has a molar mass of 92.00g. what are the empirical formula and the molecular formula for the compound?( O=16.0g/mol,M=14.1g/mol
To determine the empirical formula and the molecular formula of the compound, we need to follow these steps:
Step 1: Convert the mass percentages to grams.
Given that the compound contains 30.34% M and 60.56% O, we can calculate the mass of M and O in grams using the molar masses of M and O:
Mass of M = 30.34% * 92.00 g = 27.953 g (rounded to three decimal places)
Mass of O = 60.56% * 92.00 g = 55.989 g (rounded to three decimal places)
Step 2: Convert the mass of each element to moles.
To do this, we use the molar mass of each element:
Moles of M = Mass of M / Molar mass of M = 27.953 g / 14.1 g/mol = 1.983 mol (rounded to three decimal places)
Moles of O = Mass of O / Molar mass of O = 55.989 g / 16.0 g/mol = 3.499 mol (rounded to three decimal places)
Step 3: Determine the simplest ratio of moles.
To find the simplest ratio, divide each mole value by the smallest mole value:
Moles of M (rounded) = 1.983 mol / 1.983 mol = 1 mol
Moles of O (rounded) = 3.499 mol / 1.983 mol ≈ 1.765 mol
Step 4: Write the empirical formula.
The empirical formula represents the simplest whole-number ratio of atoms in a compound. Since we determined that the molar ratio is 1:1, the empirical formula is simply MO.
Step 5: Determine the molecular formula.
The molecular formula represents the actual number of atoms of each element in a compound. To find it, we need to compare the molar mass of the empirical formula (MO) with the given molar mass of the compound (92.00 g/mol).
Empirical formula mass = Molar mass of M (14.1 g/mol) + Molar mass of O (16.0 g/mol) = 30.1 g/mol
To find the molecular formula, we divide the molar mass of the compound by the empirical formula mass:
Molecular formula mass = 92.00 g/mol
92.00 g/mol / 30.1 g/mol ≈ 3.054
Since the result is approximately 3, the molecular formula is three times the empirical formula:
Molecular formula = (MO)3 = M3O3 = M3O3
Therefore, the empirical formula for the compound is MO, and the molecular formula is M3O3.