Chapter 7: Problem 72
How many grams of oxygen are contained in \(8.50 \mathrm{~g}\) of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} ?\)
Short Answer
Expert verified
There are 4.76 grams of oxygen in 8.50 g of \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}.
Step by step solution
01
- Determine the molar mass of \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}
Calculate the molar mass of \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} by adding up the atomic masses of all atoms in the formula. - Aluminium (Al): \(2 \times 26.98 \text{ g/mol} = 53.96 \text{ g/mol}\)- Sulfur (S): \(3 \times 32.06 \text{ g/mol} = 96.18 \text{ g/mol}\)- Oxygen (O): \(12 \times 16.00 \text{ g/mol} = 192.00 \text{ g/mol}\)Therefore, the total molar mass is \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} = 53.96 + 96.18 + 192.00 = 342.14 \text{ g/mol}.
02
- Calculate the moles of \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} in 8.50 g
Use the molar mass from Step 1 to find the number of moles: \[ \text{Moles of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} = \frac{8.50 \text{ g}}{342.14 \text{ g/mol}} = 0.0248 \text{ moles} \]
03
- Determine the mass of oxygen in the moles of \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}
Each mole of \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} contains 12 moles of oxygen atoms. Therefore, the moles of oxygen in 0.0248 moles of \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} are:\[ \text{Moles of oxygen} = 0.0248 \text{ moles of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \times 12 \text{ moles of oxygen/mole of } \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} = 0.2976 \text{ moles of oxygen} \]Now, convert the moles of oxygen to grams using the molar mass of oxygen (16.00 g/mol):\[ \text{Mass of oxygen} = 0.2976 \text{ moles} \times 16.00 \text{ g/mol} = 4.7616 \text{ g} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stoichiometry
Stoichiometry is the part of chemistry that deals with the quantitative relationships between the amounts of reactants and products in chemical reactions. It's like a recipe for chemical reactions. Just like you need the right amount of ingredients to bake a cake, you need the right amounts of substances to ensure a chemical reaction proceeds correctly. In this example, stoichiometry helps us figure out how many grams of oxygen are present in a given amount of aluminum sulfate \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\).
- First, you need to determine the molar mass of the compound.
- Next, you calculate the number of moles of the compound you are working with.
- Finally, you use these moles to find the mass of a specific component (like oxygen).
Mole Concept
The mole concept is fundamental to understanding chemical quantities. A mole is like a chemist's dozen and is equivalent to Avogadro's number, approximately \[6.022 \times 10^{23}\] entities (atoms, molecules, ions, etc.).
This concept lets chemists easily count particles by weighing them because counting each particle individually would be impractical due to their tiny sizes.
To use the mole concept:
This concept lets chemists easily count particles by weighing them because counting each particle individually would be impractical due to their tiny sizes.
To use the mole concept:
- Find the molar mass of your substance. This is the mass of one mole of that substance, usually in grams per mole (g/mol).
- Convert a given mass to moles by dividing the mass by the molar mass. In our exercise, this was \( \frac{8.50 \text{ g}}{342.14 \text{ g/mol}} \).
- Use the number of moles to perform further calculations, like finding the mass of a component within that substance.
Chemical Formula Analysis
Chemical formula analysis helps us understand the composition of compounds. A chemical formula provides the types and numbers of atoms in a molecule. Analyzing it lets us determine elemental proportions and relationships within the compound.
Take \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). This formula tells us:
Take \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). This formula tells us:
- The compound contains 2 aluminum (Al) atoms.
- It contains 3 sulfate \( \mathrm{SO}_{4} \) groups.
- Each sulfate group contains 1 sulfur (S) atom and 4 oxygen (O) atoms.
- 2 aluminum atoms.
- 3 sulfur atoms (from 3 \( \mathrm{SO}_{4} \) groups).
- 12 oxygen atoms (since each \( \mathrm{SO}_{4} \) group has 4 oxygens, and we have 3 groups).