How many atoms of oxygen are there in 36.5 grams of potassium nitrate, KNO
? [FW(KNO
) = 101.10 amu; N
= 6.02×10
] 5.56×10
atoms 6.28×10
atoms 6.52×10
atoms 5.28×10
atoms 6.85×10
atoms
The Correct Answer and Explanation is:
To determine how many oxygen atoms are in 36.5 grams of potassium nitrate (KNO₃), follow this step-by-step method used in textbooks:
Step 1: Understand the compound KNO₃
The chemical formula of potassium nitrate (KNO₃) shows that each molecule contains 3 oxygen atoms.
Step 2: Calculate the number of moles of KNO₃
We are given:
- Mass of KNO₃ = 36.5 g
- Formula weight (FW) of KNO₃ = 101.10 g/mol
Moles of KNO₃=massformula weight=36.5101.10≈0.3611 mol\text{Moles of KNO₃} = \frac{\text{mass}}{\text{formula weight}} = \frac{36.5}{101.10} \approx 0.3611\ \text{mol}Moles of KNO₃=formula weightmass=101.1036.5≈0.3611 mol
Step 3: Use Avogadro’s number to find number of molecules
Avogadro’s number is: 6.022×1023 molecules/mol6.022 \times 10^{23}\ \text{molecules/mol}6.022×1023 molecules/mol Number of molecules=0.3611 mol×6.022×1023=2.175×1023 molecules\text{Number of molecules} = 0.3611\ \text{mol} \times 6.022 \times 10^{23} = 2.175 \times 10^{23}\ \text{molecules}Number of molecules=0.3611 mol×6.022×1023=2.175×1023 molecules
Step 4: Multiply by 3 (oxygen atoms per molecule)
Oxygen atoms=2.175×1023×3=6.525×1023 atoms\text{Oxygen atoms} = 2.175 \times 10^{23} \times 3 = 6.525 \times 10^{23}\ \text{atoms}Oxygen atoms=2.175×1023×3=6.525×1023 atoms
✅ Final Answer:
6.52×1023 atoms of oxygen\boxed{6.52 \times 10^{23}\ \text{atoms of oxygen}}6.52×1023 atoms of oxygen
📘 Explanation
To calculate the number of oxygen atoms in a given mass of a compound, you must consider both the chemical formula and the molar relationships involved. Potassium nitrate (KNO₃) is composed of one potassium atom (K), one nitrogen atom (N), and three oxygen atoms (O). Thus, for every 1 mole of KNO₃, there are 3 moles of oxygen atoms.
We start by converting the mass of potassium nitrate into moles, using the molar mass (101.10 g/mol). For 36.5 grams, this gives about 0.3611 moles of KNO₃.
Using Avogadro’s number (6.022 × 10²³), we then convert moles into molecules:
0.3611 moles × 6.022 × 10²³ molecules/mole ≈ 2.175 × 10²³ molecules of KNO₃.
Since each molecule has 3 oxygen atoms, we multiply by 3 to find the total number of oxygen atoms:
2.175 × 10²³ × 3 = 6.525 × 10²³ atoms.
Therefore, 36.5 grams of potassium nitrate contains approximately 6.52 × 10²³ oxygen atoms, which matches the correct choice in the options.
