How many atoms are in 55.3 g of potassium bromite (KBrO2)?
The correct answer and explanation is:
To find how many atoms are in 55.3 g of potassium bromite (KBrO2), the following steps are involved:
Step 1: Determine the molar mass of KBrO2
The molar mass is the sum of the atomic masses of all elements in one mole of the compound.
- Potassium (K) has an atomic mass of approximately 39.1 g/mol.
- Bromine (Br) has an atomic mass of approximately 79.9 g/mol.
- Oxygen (O) has an atomic mass of approximately 16.0 g/mol.
Now, calculate the molar mass of KBrO2: Molar mass of KBrO2=39.1 g/mol (K)+79.9 g/mol (Br)+2×16.0 g/mol (O)=39.1+79.9+32.0=151.0 g/mol\text{Molar mass of KBrO2} = 39.1 \, \text{g/mol (K)} + 79.9 \, \text{g/mol (Br)} + 2 \times 16.0 \, \text{g/mol (O)} = 39.1 + 79.9 + 32.0 = 151.0 \, \text{g/mol}
Step 2: Calculate the number of moles in 55.3 g of KBrO2
The number of moles is calculated by dividing the mass of the substance by its molar mass: Number of moles=55.3 g151.0 g/mol=0.366 mol\text{Number of moles} = \frac{55.3 \, \text{g}}{151.0 \, \text{g/mol}} = 0.366 \, \text{mol}
Step 3: Calculate the total number of formula units in 0.366 moles
Avogadro’s number tells us that one mole of any substance contains 6.022×10236.022 \times 10^{23} formula units (atoms, molecules, or ions). For KBrO2, we have: Number of formula units=0.366 mol×6.022×1023 units/mol=2.2×1023 formula units\text{Number of formula units} = 0.366 \, \text{mol} \times 6.022 \times 10^{23} \, \text{units/mol} = 2.2 \times 10^{23} \, \text{formula units}
Step 4: Calculate the number of atoms
Each formula unit of KBrO2 contains:
- 1 potassium (K) atom
- 1 bromine (Br) atom
- 2 oxygen (O) atoms
So, the total number of atoms in each formula unit is: Total atoms per formula unit=1+1+2=4 atoms\text{Total atoms per formula unit} = 1 + 1 + 2 = 4 \, \text{atoms}
Now, calculate the total number of atoms: Total number of atoms=2.2×1023 formula units×4 atoms=8.8×1023 atoms\text{Total number of atoms} = 2.2 \times 10^{23} \, \text{formula units} \times 4 \, \text{atoms} = 8.8 \times 10^{23} \, \text{atoms}
Final Answer:
The total number of atoms in 55.3 g of potassium bromite (KBrO2) is approximately 8.8×10238.8 \times 10^{23} atoms.