The relative mass of aluminum chloride is 267 and its composition by mass is 20,3% Al and 79,7% chlorine. Determine the empirical and molecular formulas of Aluminum chloride.

Take a 100 g sample.

This will give you 20.3 g Al and 79.7 g Cl. Convert each of these grams into moles.
moles Al = 20.3/atomic mass Al
moles Cl = 79.7/atomic mass Cl.

Now find the ratio, in small whole numbers, to equate the two. The easiest way to do that is to divide the smaller number by itself (which assure you it will be 1.000), then divide the other number by the same small number. You should get AlxCly where x and y are small whole numbers. That will give you the empirical formula.
Then add up the atomic masses to find the mass of the empirical formula, divide into 267 to determine n, round n to a whole number, and multiply by the empirical formula to find the molecular formula. The molecular formula will be (AlxCly)n.

Well, based on the given information, it seems like aluminum chloride has a bit of an identity crisis. It's like it can't decide if it wants to be Al or Cl, so it just settles for having both in its formula.

Let's break it down. We know that aluminum makes up 20.3% of the mass, while chlorine makes up 79.7%. This means that for every 100 grams of aluminum chloride, we have 20.3 grams of aluminum and 79.7 grams of chlorine.

To determine the empirical formula, we need to find the ratio between the two elements. We can start by assuming we have 100 grams of aluminum chloride.

Using the molar masses of aluminum (26.98 g/mol) and chlorine (35.45 g/mol), we can calculate the number of moles for each element:

Aluminum: 20.3 g / 26.98 g/mol = 0.7522 mol
Chlorine: 79.7 g / 35.45 g/mol = 2.2489 mol

Now, let's divide both of these values by the smallest of the two:

Aluminum: 0.7522 mol / 0.7522 mol = 1
Chlorine: 2.2489 mol / 0.7522 mol ≈ 2.99

Rounding the chlorine value to the nearest whole number, we get approximately 3, giving us an empirical formula of AlCl3. This means that for every 1 atom of aluminum, we have 3 atoms of chlorine.

Now, let's talk about the molecular formula. To find this, we need to know the molar mass of the compound. The given relative mass of 267 gives us a clue.

The empirical formula mass of AlCl3 is approximately 27 + (3 x 35.45) = 133.34 g/mol. Now, we can divide the given relative mass by the empirical formula mass:

267 g / 133.34 g/mol ≈ 2

So, the molecular formula of aluminum chloride is Al2Cl6. It's like the empirical formula got multiplied by 2, so now we have 2 atoms of aluminum and 6 atoms of chlorine in every molecule of aluminum chloride.

And there you have it! Aluminum chloride can't decide if it's Al or Cl, so it just chooses to be both with its empirical and molecular formulas. It's like the "two-faced" compound of the chemistry world.

To determine the empirical and molecular formulas of aluminum chloride, we need to find the ratio of aluminum to chlorine atoms present in the compound.

1. Start by assuming we have 100 grams of aluminum chloride. From the given composition, we know that 20.3 grams are aluminum and 79.7 grams are chlorine.

2. Determine the number of moles for each element in the compound using their respective molar masses:
- Aluminum (Al): molar mass = 27 g/mol; moles = mass / molar mass = 20.3 g / 27 g/mol = 0.752 moles
- Chlorine (Cl): molar mass = 35.5 g/mol; moles = mass / molar mass = 79.7 g / 35.5 g/mol = 2.244 moles

3. Divide the number of moles of each element by the smallest number of moles to obtain the empirical formula:
- Al: 0.752 moles / 0.752 moles = 1 mole
- Cl: 2.244 moles / 0.752 moles ≈ 3 moles

The empirical formula is therefore AlCl3, indicating that the ratio of aluminum to chlorine atoms in aluminum chloride is 1:3.

To determine the molecular formula, we need to compare the molar mass of the empirical formula (AlCl3) with the given relative mass (267g/mol).

4. Calculate the molar mass of the empirical formula:
- Al: 27 g/mol × 1 = 27 g/mol
- Cl: 35.5 g/mol × 3 = 106.5 g/mol
Total molar mass of AlCl3 = 27 g/mol + 106.5 g/mol = 133.5 g/mol

5. Divide the given relative mass by the calculated molar mass of the empirical formula:
267 g/mol / 133.5 g/mol = 2

The result indicates that the empirical formula (AlCl3) needs to be multiplied by 2 to obtain the molecular formula. Therefore, the molecular formula of aluminum chloride is Al2Cl6.

To determine the empirical and molecular formulas of aluminum chloride, we need to analyze the information provided about its composition.

1. Empirical Formula:
The empirical formula represents the simplest whole number ratio of atoms in a compound. To find the empirical formula, we can start by assuming we have 100 grams of aluminum chloride, which will allow us to determine the mass of each element in the compound.

Given:
- The relative mass of aluminum chloride is 267.
- The composition by mass is 20.3% Al and 79.7% chlorine.

First, convert the percentages into grams:
- Aluminum (Al): (20.3 / 100) * 100 g = 20.3 g
- Chlorine (Cl): (79.7 / 100) * 100 g = 79.7 g

Next, convert the mass of each element into moles using their atomic masses:
- Aluminum (Al): 20.3 g / 27 g/mol (atomic mass of Al) = 0.752 moles
- Chlorine (Cl): 79.7 g / 35.5 g/mol (atomic mass of Cl) = 2.25 moles

To find the simplest whole number ratio of atoms, divide each mole value by the smallest mole value. In this case, the smallest mole value is 0.752:
- Aluminum (Al): 0.752 moles / 0.752 moles = 1 mole
- Chlorine (Cl): 2.25 moles / 0.752 moles ≈ 2.99 moles

Since we need to express the empirical formula as whole numbers, round the value for chlorine to the nearest whole number:
- Aluminum (Al): 1 mole
- Chlorine (Cl): 3 moles

Thus, the empirical formula for aluminum chloride is AlCl3.

2. Molecular Formula:
The molecular formula represents the actual number of atoms of each element in a molecule.

To determine the molecular formula, we need to know the molar mass of aluminum chloride. Since the given value of 267 is the relative mass, we can assume it represents the molecular mass of the compound.

The molar mass of aluminum chloride is 267 g/mol. To find the molecular formula, we need to compare the empirical formula mass (EFM) with the molar mass (MM) of aluminum chloride.

The empirical formula mass is calculated as follows:
EFM = (Atomic mass of Al) + (Atomic mass of Cl * Number of Cl atoms)
EFM = (27 g/mol) + (35.5 g/mol * 3)

EFM = 27 g/mol + 106.5 g/mol = 133.5 g/mol

Now, find the ratio of the molar mass to the empirical formula mass:
Ratio = MM / EFM
Ratio = 267 g/mol / 133.5 g/mol = 2

Since the ratio is 2, the molecular formula is simply the empirical formula (AlCl3) multiplied by 2:
Molecular formula = (AlCl3)2 = Al2Cl6

Thus, the molecular formula of aluminum chloride is Al2Cl6.