A mass of aluminium metal is heated in a such in a such a way that is was exposed to air.(1.751g Al) A compound containing aluminium and oxygen is formed. It has a mass of 3.311 g. What is the empirical formula of aluminum oxide according to this experimental data?
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Since the aluminum oxide has a mass of 3.311g and the aluminum originally has a mass of 1.751g the mass of oxygen in the compound is:
3.311-1.751 = 1.56g O
Now in order to find the empirical formula you have to find the moles of aluminum and the moles of oxygen in the aluminum oxide. The molar mass of aluminum is about 26.98g/mol and the molar mass of oxygen is 16g/mol.
Therefore, the moles of aluminum and oxygen are as follows:
1.751g Al / (26.98g/mol Al) = 0.0649mol Al
1.56g O / (16g/mol O) = 0.0975mol O
Finally find the molar ratios of aluminum and oxygen in the aluminum oxide:
Al: 0.0649/0.0649 = 1 * 2 = 2
O: 0.0975/0.0649 = 1.5 * 2 = 3
The empirical formula of aluminum oxide is: Al2O3
To find the empirical formula of aluminum oxide using the experimental data given, we need to determine the ratio of aluminum to oxygen in the compound.
Step 1: Calculate the moles of aluminum and oxygen in the compound.
- Moles of aluminum = mass of aluminum / molar mass of aluminum
The molar mass of aluminum is 26.98 g/mol.
Moles of aluminum = 1.751 g / 26.98 g/mol
- Moles of oxygen = mass of compound - mass of aluminum
Moles of oxygen = 3.311 g - 1.751 g = 1.56 g
Step 2: Divide the moles of each element by the smallest number of moles.
Dividing both moles of aluminum and oxygen by the smallest number of moles, which is 1.56 g (moles of oxygen), we get:
- Moles of aluminum = 1.751 g / 1.56 g = 1.12
- Moles of oxygen = 1.56 g / 1.56 g = 1.00
Step 3: Use the whole number ratio to determine the empirical formula.
The mole ratio indicates that the empirical formula is Al₁.₁₂O₁.
However, we need to convert these decimals to whole numbers by multiplying them by a common factor. In this case, the common factor is 8.
- Moles of aluminum = 1.12 × 8 = 8.96 ≈ 9
- Moles of oxygen = 1.00 × 8 = 8.00 ≈ 8
Therefore, the empirical formula of aluminum oxide according to the experimental data is Al₉O₈.
To find the empirical formula of aluminum oxide, we need to determine the ratio of aluminum (Al) to oxygen (O) in the compound using the given experimental data.
First, let's find the number of moles of aluminum and oxygen in the compound.
1. Calculate the number of moles of aluminum:
- Mass of aluminum = 1.751 g
- Molar mass of aluminum (Al) = 27 g/mol
- Moles of aluminum = Mass of aluminum / Molar mass of aluminum
Moles of aluminum = 1.751 g / 27 g/mol ≈ 0.0648 mol
2. Calculate the number of moles of oxygen:
- Mass of compound = Mass of aluminum oxide = 3.311 g
- Moles of oxygen = Mass of compound - Mass of aluminum
Moles of oxygen = 3.311 g - 1.751 g ≈ 1.56 g
Now, let's find the simplest ratio between aluminum and oxygen by dividing the number of moles of each element by the smaller number of moles.
Dividing both moles of aluminum and oxygen by 0.0648 (the smaller value), we get:
Moles of aluminum = 0.0648 mol / 0.0648 mol ≈ 1 mol
Moles of oxygen = 1.56 g / 0.0648 mol ≈ 24 mol
Rounded to the nearest whole number, the ratio between aluminum and oxygen is approximately 1:24.
Therefore, the empirical formula of aluminum oxide according to this experimental data is Al2O3, indicating that there are 2 moles (or atoms) of aluminum and 3 moles (or atoms) of oxygen in the compound.