Using the approximate formula FiO2 = 0.21 + (flow in L/min × 0.04), what is the FiO2 for a 5 L/min nasal cannula?

Using the approximate formula FiO2 = 0.21 + (flow in L/min × 0.04), what is the FiO2 for a 5 L/min nasal cannula?

• A. 0.21
• B. 0.29
• C. 0.41
• D. 0.50

Answer: (C)

The basic idea is that oxygen concentration with a nasal cannula rises roughly 4 percentage points for every extra liter per minute of flow above room air, which is 21% oxygen to begin with. So for 5 L/min, you add 5 × 0.04 = 0.20 to 0.21, giving 0.41, or about 41% FiO2. That’s why 5 L/min corresponds to roughly 41% oxygen. The numbers given in the other choices reflect lower flows (no flow would be 21%, about 2 L/min would be around 29%), and reaching 50% would require a higher flow (around 7–8 L/min). In practice, actual FiO2 can vary with breathing pattern and mouth breathing, but the estimate with the formula points to about 41%.

The basic idea is that oxygen concentration with a nasal cannula rises roughly 4 percentage points for every extra liter per minute of flow above room air, which is 21% oxygen to begin with. So for 5 L/min, you add 5 × 0.04 = 0.20 to 0.21, giving 0.41, or about 41% FiO2. That’s why 5 L/min corresponds to roughly 41% oxygen. The numbers given in the other choices reflect lower flows (no flow would be 21%, about 2 L/min would be around 29%), and reaching 50% would require a higher flow (around 7–8 L/min). In practice, actual FiO2 can vary with breathing pattern and mouth breathing, but the estimate with the formula points to about 41%.