An analysis shows that 5.85g of chromium metal combines with oxygen to form 8.62g of chromium oxide. What is the empirical formula of the compound? Please help!!
find the moles of chromium ... (sample mass) / (molar mass)
find the mass of oxygen ... (chromium oxide mass) - (chromium metal mass)
find the moles of oxygen ... (mass of oxygen) / (molar mass of oxygen)
find the LCM of the moles
... empirical formula is the ratio of the moles
To find the empirical formula of the compound, we first need to determine the moles of chromium and oxygen in the given amounts.
Step 1: Find the moles of chromium.
The molar mass of chromium (Cr) is 52 g/mol.
moles of chromium = mass of chromium / molar mass of chromium
moles of chromium = 5.85 g / 52 g/mol
moles of chromium = 0.1125 mol (rounded to 4 decimal places)
Step 2: Find the moles of oxygen.
The molar mass of oxygen (O) is 16 g/mol.
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = (8.62 g - 5.85 g) / 16 g/mol
moles of oxygen = 0.1725 mol (rounded to 4 decimal places)
Step 3: Find the mole ratio.
Divide the moles of each element by the smaller value of moles.
moles of chromium (Cr) = 0.1125 mol
moles of oxygen (O) = 0.1725 mol
Therefore, the mole ratio of chromium to oxygen is approximately 1:1.
Step 4: Write the empirical formula.
The empirical formula represents the simplest whole number ratio of atoms in a compound. Since the mole ratio is approximately 1:1, the empirical formula is CrO.
Therefore, the empirical formula of the compound is CrO.
To determine the empirical formula of a compound, you need to calculate the ratio of elements present in the compound using their masses. Here's how you can do it step by step:
Step 1: Convert the given masses to moles.
To do this, divide the mass of each element by its molar mass. The molar mass of chromium (Cr) is 52 g/mol, and the molar mass of oxygen (O) is 16 g/mol.
Mass of chromium (Cr) = 5.85 g
Moles of chromium (Cr) = Mass / Molar mass = 5.85 g / 52 g/mol ≈ 0.1125 mol
Mass of oxygen (O) = 8.62 g
Moles of oxygen (O) = Mass / Molar mass = 8.62 g / 16 g/mol ≈ 0.53875 mol
Step 2: Find the simplest whole number ratio.
Divide both moles of chromium and oxygen by the smallest value among them. Since the moles of chromium is smaller, divide both values by 0.1125.
Moles of Cr (Chromium) = 0.1125 mol / 0.1125 mol = 1
Moles of O (Oxygen) = 0.53875 mol / 0.1125 mol ≈ 4.78
Step 3: Convert the subscripts to whole numbers.
Multiply the number of moles of each element by a factor that will give you the simplest whole number ratio. In this case, multiply both subscripts by 4 to get a whole number. Remember, you can only round to a whole number and not a fraction.
Empirical formula: Cr₁O₄
Therefore, the empirical formula of the compound is CrO₄.
I do this a slightly different way than R_Scott as follows:
% Cr = (5.85/8.62)*100 = about 67.86
% O = 100% - 67.86 = about 32.14
g Cr = 67.86
g O = 32.14
mols Cr = 67.86/52 = about 1.3
mols O = 32.14/16 = about 2.0
Divide both numbers by the smaller; i.e., by 1.3 like this
1.3/1.3 = 1
2.0/1.3 = 1.54 and that 4 is not significant.
You want whole numbers so multiply by an integer to obtain whole numbers. Multiplying by 2 gives Cr = 2 and O = 3.08 and this rounds to 2:3 so the empirical formula is Cr2O3.