Answer :
Certainly! Let's go through each part of the question step-by-step.
### 5.1 Write down the NAME of the compound above.
The name of the compound Al2(SO4)3 is Aluminum sulfate.
### 5.2 Define the term molar mass.
Molar mass is the mass of one mole of a given substance (chemical element or chemical compound) expressed in grams per mole.
### 5.3 Calculate the following for Al2(SO4)3:
#### 5.3.1 Its molar mass
To find the molar mass of Aluminum sulfate, Al2(SO4)3, we need to sum the masses of all the atoms in the formula.
- The atomic mass of Aluminum (Al) = 26.98 g/mol
- The atomic mass of Sulfur (S) = 32.07 g/mol
- The atomic mass of Oxygen (O) = 16.00 g/mol
The formula Al2(SO4)3 indicates:
- 2 aluminum atoms
- 3 sulfate ions (each with 1 sulfur atom and 4 oxygen atoms)
So, the total molar mass can be calculated as:
[tex]\[ \text{Molar mass} = 2(Al) + 3(SO4) \][/tex]
[tex]\[ \text{Molar mass} = 2 \times 26.98 + 3 \times (32.07 + 4 \times 16.00) \][/tex]
[tex]\[ \text{Molar mass} = 2 \times 26.98 + 3 \times (32.07 + 64.00) \][/tex]
[tex]\[ \text{Molar mass} = 2 \times 26.98 + 3 \times 96.07 \][/tex]
[tex]\[ \text{Molar mass} = 2 \times 26.98 + 288.21 \][/tex]
[tex]\[ \text{Molar mass} = 53.96 + 288.21 \][/tex]
[tex]\[ \text{Molar mass} = 342.17 \text{ g/mol}\][/tex]
#### 5.3.2 Its percentage composition
To find the percentage composition of each element in Al2(SO4)3, we use the molar masses of the individual elements and the total molar mass calculated above.
- Mass of Aluminum:
[tex]\[ \text{mass of Al} = 2 \times 26.98 = 53.96 \][/tex]
- Mass of Sulfur:
[tex]\[ \text{mass of S} = 3 \times 32.07 = 96.21 \][/tex]
- Mass of Oxygen:
[tex]\[ \text{mass of O} = 12 \times 16.00 = 192.00 \][/tex]
Using the total molar mass of 342.17 g/mol, we calculate the percentage composition as follows:
- Percentage of Aluminum:
[tex]\[ \% \text{Al} = \left(\frac{53.96}{342.17}\right) \times 100 \approx 15.77\% \][/tex]
- Percentage of Sulfur:
[tex]\[ \% \text{S} = \left(\frac{96.21}{342.17}\right) \times 100 \approx 28.12\% \][/tex]
- Percentage of Oxygen:
[tex]\[ \% \text{O} = \left(\frac{192.00}{342.17}\right) \times 100 \approx 56.11\% \][/tex]
So, the percentage composition of Al2(SO4)3 is approximately:
- Aluminum: 15.77%
- Sulfur: 28.12%
- Oxygen: 56.11%
#### 5.3.3 The number of moles present in 85.5 g
To calculate the number of moles in 85.5 grams of Al2(SO4)3, we use the formula:
[tex]\[ \text{number of moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
Given mass = 85.5 g and molar mass = 342.17 g/mol:
[tex]\[ \text{number of moles} = \frac{85.5}{342.17} \approx 0.2499 \text{ moles} \][/tex]
#### 5.3.4 The number of aluminum atoms present in 85.5 g
To find the number of Aluminum (Al) atoms in 85.5 grams of Al2(SO4)3, we first find the number of moles (as above) and then use Avogadro's number.
- Number of moles of Al2(SO4)3 = 0.2499 moles
- Each molecule of Al2(SO4)3 contains 2 atoms of Aluminum.
- Avogadro's number (number of atoms per mole) = [tex]\(6.022 \times 10^{23} \text{ atoms/mole}\)[/tex]
So, the number of aluminum atoms is:
[tex]\[ \text{number of Al atoms} = 0.2499 \times 2 \times 6.022 \times 10^{23} \][/tex]
[tex]\[ \text{number of Al atoms} \approx 3.01 \times 10^{23} \text{ atoms} \][/tex]
Therefore, there are approximately [tex]\(3.01 \times 10^{23}\)[/tex] aluminum atoms in 85.5 grams of Al2(SO4)3.
### 5.1 Write down the NAME of the compound above.
The name of the compound Al2(SO4)3 is Aluminum sulfate.
### 5.2 Define the term molar mass.
Molar mass is the mass of one mole of a given substance (chemical element or chemical compound) expressed in grams per mole.
### 5.3 Calculate the following for Al2(SO4)3:
#### 5.3.1 Its molar mass
To find the molar mass of Aluminum sulfate, Al2(SO4)3, we need to sum the masses of all the atoms in the formula.
- The atomic mass of Aluminum (Al) = 26.98 g/mol
- The atomic mass of Sulfur (S) = 32.07 g/mol
- The atomic mass of Oxygen (O) = 16.00 g/mol
The formula Al2(SO4)3 indicates:
- 2 aluminum atoms
- 3 sulfate ions (each with 1 sulfur atom and 4 oxygen atoms)
So, the total molar mass can be calculated as:
[tex]\[ \text{Molar mass} = 2(Al) + 3(SO4) \][/tex]
[tex]\[ \text{Molar mass} = 2 \times 26.98 + 3 \times (32.07 + 4 \times 16.00) \][/tex]
[tex]\[ \text{Molar mass} = 2 \times 26.98 + 3 \times (32.07 + 64.00) \][/tex]
[tex]\[ \text{Molar mass} = 2 \times 26.98 + 3 \times 96.07 \][/tex]
[tex]\[ \text{Molar mass} = 2 \times 26.98 + 288.21 \][/tex]
[tex]\[ \text{Molar mass} = 53.96 + 288.21 \][/tex]
[tex]\[ \text{Molar mass} = 342.17 \text{ g/mol}\][/tex]
#### 5.3.2 Its percentage composition
To find the percentage composition of each element in Al2(SO4)3, we use the molar masses of the individual elements and the total molar mass calculated above.
- Mass of Aluminum:
[tex]\[ \text{mass of Al} = 2 \times 26.98 = 53.96 \][/tex]
- Mass of Sulfur:
[tex]\[ \text{mass of S} = 3 \times 32.07 = 96.21 \][/tex]
- Mass of Oxygen:
[tex]\[ \text{mass of O} = 12 \times 16.00 = 192.00 \][/tex]
Using the total molar mass of 342.17 g/mol, we calculate the percentage composition as follows:
- Percentage of Aluminum:
[tex]\[ \% \text{Al} = \left(\frac{53.96}{342.17}\right) \times 100 \approx 15.77\% \][/tex]
- Percentage of Sulfur:
[tex]\[ \% \text{S} = \left(\frac{96.21}{342.17}\right) \times 100 \approx 28.12\% \][/tex]
- Percentage of Oxygen:
[tex]\[ \% \text{O} = \left(\frac{192.00}{342.17}\right) \times 100 \approx 56.11\% \][/tex]
So, the percentage composition of Al2(SO4)3 is approximately:
- Aluminum: 15.77%
- Sulfur: 28.12%
- Oxygen: 56.11%
#### 5.3.3 The number of moles present in 85.5 g
To calculate the number of moles in 85.5 grams of Al2(SO4)3, we use the formula:
[tex]\[ \text{number of moles} = \frac{\text{mass}}{\text{molar mass}} \][/tex]
Given mass = 85.5 g and molar mass = 342.17 g/mol:
[tex]\[ \text{number of moles} = \frac{85.5}{342.17} \approx 0.2499 \text{ moles} \][/tex]
#### 5.3.4 The number of aluminum atoms present in 85.5 g
To find the number of Aluminum (Al) atoms in 85.5 grams of Al2(SO4)3, we first find the number of moles (as above) and then use Avogadro's number.
- Number of moles of Al2(SO4)3 = 0.2499 moles
- Each molecule of Al2(SO4)3 contains 2 atoms of Aluminum.
- Avogadro's number (number of atoms per mole) = [tex]\(6.022 \times 10^{23} \text{ atoms/mole}\)[/tex]
So, the number of aluminum atoms is:
[tex]\[ \text{number of Al atoms} = 0.2499 \times 2 \times 6.022 \times 10^{23} \][/tex]
[tex]\[ \text{number of Al atoms} \approx 3.01 \times 10^{23} \text{ atoms} \][/tex]
Therefore, there are approximately [tex]\(3.01 \times 10^{23}\)[/tex] aluminum atoms in 85.5 grams of Al2(SO4)3.
