How many atoms of O there in 86.2 grams of cadmium (II) permanganate?
The correct answer and explanation is:
Let’s solve the problem step-by-step:
Question:
How many atoms of oxygen (O) are in 86.2 grams of cadmium(II) permanganate?
Step 1: Write the formula of cadmium(II) permanganate
- Cadmium(II) means Cd has a +2 charge → Cd²⁺
- Permanganate ion is MnO₄⁻ (charge = -1)
- To balance the charges, 1 Cd²⁺ will combine with 2 MnO₄⁻ ions to form Cd(MnO₄)₂
So the formula is Cd(MnO₄)₂.
Step 2: Calculate the molar mass of Cd(MnO₄)₂
- Atomic masses (approximate):
Cd = 112.41 g/mol
Mn = 54.94 g/mol
O = 16.00 g/mol
Molar mass of Cd(MnO₄)₂ =
= (1 × 112.41) + (2 × 54.94) + (2 × 4 × 16.00)
= 112.41 + 109.88 + (8 × 16.00)
= 112.41 + 109.88 + 128.00
= 350.29 g/mol
Step 3: Calculate the number of moles in 86.2 grams
moles=massmolar mass=86.2 g350.29 g/mol≈0.246 mol\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{86.2 \text{ g}}{350.29 \text{ g/mol}} \approx 0.246 \text{ mol}
Step 4: Calculate the number of oxygen atoms in the sample
- Each mole of Cd(MnO₄)₂ contains 8 moles of oxygen atoms (since MnO₄ has 4 O, and there are 2 MnO₄ groups, total 8 O atoms per formula unit).
- Moles of oxygen atoms = 0.246 mol Cd(MnO₄)2×8=1.968 mol O atoms0.246 \text{ mol Cd(MnO₄)}_2 \times 8 = 1.968 \text{ mol O atoms}
- Number of atoms = moles × Avogadro’s number
=1.968 mol×6.022×1023 atoms/mol≈1.185×1024 oxygen atoms= 1.968 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 1.185 \times 10^{24} \text{ oxygen atoms}
Final answer:
There are approximately 1.19 × 10²⁴ atoms of oxygen in 86.2 grams of cadmium(II) permanganate.
Explanation (about 300 words):
This problem involves converting a given mass of a compound into the number of atoms of a specific element within it. To do this, you first need the correct chemical formula for the compound—in this case, cadmium(II) permanganate, which is Cd(MnO₄)₂. The Roman numeral II indicates the oxidation state of cadmium is +2, which balances with two permanganate ions (MnO₄⁻).
Once the formula is confirmed, calculate the molar mass by summing the atomic masses of all atoms in one mole of the compound: one Cd atom, two Mn atoms, and eight oxygen atoms (from two MnO₄ groups). Using atomic masses (Cd = 112.41 g/mol, Mn = 54.94 g/mol, O = 16.00 g/mol), the total molar mass is 350.29 g/mol.
Next, find the number of moles in the given mass (86.2 g) by dividing by the molar mass, yielding about 0.246 moles of the compound.
Since the compound contains 8 oxygen atoms per formula unit, multiply the moles of compound by 8 to get the moles of oxygen atoms. This gives approximately 1.968 moles of oxygen atoms.
Finally, multiply the moles of oxygen atoms by Avogadro’s number (6.022 × 10²³ atoms/mol) to convert moles into atoms. The result is roughly 1.19 × 10²⁴ oxygen atoms in 86.2 grams of cadmium(II) permanganate.
This stepwise approach—finding the molar mass, moles, and then converting to atoms—is standard in stoichiometric problems involving counting atoms in a given mass of a compound.