How many atoms of aluminum [Al] are present in a sample of aluminum carbonate [AlCar] weighing 3.45E6 amu?
To determine the number of atoms of aluminum in a sample of aluminum carbonate, you can use the concept of moles and Avogadro's number.
1. Find the molar mass of aluminum carbonate [AlCar]:
- The molar mass of aluminum (Al) = 26.98 g/mol
- The molar mass of carbon (C) = 12.01 g/mol
- The molar mass of oxygen (O) = 16.00 g/mol
Aluminum carbonate has the chemical formula Al2(CO3)3, so:
- Molar mass of aluminum carbonate (AlCar) = (2 * Al) + (3 * C) + (9 * O)
- Molar mass of aluminum carbonate (AlCar) = (2 * 26.98 g/mol) + (3 * 12.01 g/mol) + (9 * 16.00 g/mol)
- Molar mass of aluminum carbonate (AlCar) = 233.99 g/mol
2. Convert the sample weight from atomic mass units (amu) to grams (g):
To convert from amu to grams, use the conversion factor:
- 1 g/mol = 1 amu
Therefore, 3.45E6 amu = 3.45E6 g
3. Calculate the number of moles of aluminum carbonate:
- Moles of aluminum carbonate (AlCar) = mass of sample (g) / molar mass of aluminum carbonate (g/mol)
- Moles of aluminum carbonate (AlCar) = 3.45E6 g / 233.99 g/mol
4. Use the stoichiometry from the balanced chemical equation to determine the number of moles of aluminum:
- The balanced chemical equation for aluminum carbonate is: Al2(CO3)3
- The coefficient of aluminum is 2 in this equation, indicating that 2 moles of aluminum are present for every one mole of aluminum carbonate.
Therefore, the number of moles of aluminum = 2 * moles of aluminum carbonate (AlCar)
5. Finally, calculate the number of atoms of aluminum:
- Number of atoms of aluminum = moles of aluminum * Avogadro's number
- Avogadro's number = 6.022E23 mol^(-1)
Number of atoms of aluminum = [2 * (3.45E6 g / 233.99 g/mol)] * 6.022E23 mol^(-1)
By substituting the values and performing the calculations, you can determine the number of atoms of aluminum present in the given sample of aluminum carbonate.
mass Al2(CO3)3 = 3.45E6 amu x (1.66E-24 g/amu) = ? grams.
mols Al2(SO4)3 = grams/molar mass = ?
mols Al = mols Al2(CO3)3 x 2 since there are two mols Al in 1 mol Al2(SO4)3.
#atoms = mols Al x (6.022E23 atoms/mol) =?