find the empirical formula for a compound that consists of aluminum and chlorine in which the aluminum is 20.2% by mass
To find the empirical formula of a compound, we need to determine the ratio of the elements present in the compound. Given that aluminum (Al) constitutes 20.2% by mass, and chlorine (Cl) makes up the remainder, we can follow these steps:
1. Assume we have 100 grams of the compound.
2. Calculate the mass of aluminum:
Mass of aluminum = 20.2% of 100g = 20.2g
3. Calculate the mass of chlorine:
Mass of chlorine = 100g - 20.2g = 79.8g
Next, we need to determine the moles of each element:
4. Determine the number of moles of aluminum:
Moles of aluminum = Mass of aluminum / molar mass of aluminum
The molar mass of aluminum (Al) is 26.98 g/mol.
Moles of aluminum = 20.2g / 26.98 g/mol = 0.75 mol (rounded to two decimal places)
5. Determine the number of moles of chlorine:
Moles of chlorine = Mass of chlorine / molar mass of chlorine
The molar mass of chlorine (Cl) is 35.45 g/mol.
Moles of chlorine = 79.8g / 35.45 g/mol = 2.25 mol (rounded to two decimal places)
6. Divide the number of moles of each element by the smallest number of moles:
Moles of aluminum (Al) = 0.75 mol / 0.75 mol = 1
Moles of chlorine (Cl) = 2.25 mol / 0.75 mol = 3
7. Based on the mole ratio, we can write the empirical formula:
The empirical formula is AlCl3, indicating that there is 1 atom of aluminum and 3 atoms of chlorine present in the compound.
To find the empirical formula for a compound, you need to determine the ratio of the elements in the compound. In this case, we are given that aluminum is 20.2% by mass.
To start, assume we have a 100g sample of the compound. This means that 20.2g of the compound is made up of aluminum.
To find the mass of chlorine in the compound, subtract the mass of aluminum from the total mass of the compound:
Mass of chlorine = Total mass of compound - Mass of aluminum
Mass of chlorine = 100g - 20.2g = 79.8g
Now that we know the mass of aluminum and chlorine in our assumed 100g sample, we can convert these masses into moles using their respective atomic masses.
The atomic mass of aluminum (Al) is 26.98 g/mol, and the atomic mass of chlorine (Cl) is 35.45 g/mol.
Number of moles of aluminum = Mass of aluminum / Atomic mass of aluminum
Number of moles of aluminum = 20.2g / 26.98g/mol ≈ 0.749 mol
Number of moles of chlorine = Mass of chlorine / Atomic mass of chlorine
Number of moles of chlorine = 79.8g / 35.45g/mol ≈ 2.25 mol
Now, we need to find the simplest whole number ratio between the moles of aluminum and chlorine.
Divide both mole values by the smallest non-zero value (which is 0.749 mol in this case):
Aluminum: 0.749 mol / 0.749 mol ≈ 1 mol
Chlorine: 2.25 mol / 0.749 mol ≈ 3 mol
The ratio between aluminum and chlorine is approximately 1:3.
Thus, the empirical formula for the compound consisting of aluminum and chlorine is AlCl3.