Mass of CO in grams

Question

Suppose that polluted air has carbon monoxide levels of 15.0 L CO per 1.0 × 106 L of air. An average human inhales approximately 5.0 × 102 mL of air per breath and takes about 15 breaths per minute. If the density of carbon monoxide gas is 1.2 g/L, what mass of carbon monoxide (in grams) will the average person inhale in 8.0 hours?

Step 1: Identify the FOUR equivalences given to you in the problem. An equivalence consists of two quantities that are equal to one another (Example: 1 minute = 60 seconds). Write the density of CO gas as one of your equivalences.

Step 2: Write your strategy for unit conversion and solving the problem.

Step 3: Follow your strategy and show your work below.

Answer 1

Hello, Naomi,

There are many conversions needed. Be careful in thinking through the units before doing any of the operations. Use units in your calculation to help keep everything straight.

One can quickly spot that the volumes are given in both liters and milliliters. The answer wants to know the mass of CO inhaled over an 8 hour period. Let's start with the total volume of air inhaled by a human in 8 hours, where B is breath.

5E02 ml/B

15 B/min

60 min/hr

Breaths over an 8 hour period is calculated as (15B/min)*(60min/hour)*(8 hours). See how the units cancel to gives Breaths for 8 hours.

Then convert the information to the total volume in ml for the 8 hour period:

(5E+02 ml/B)*(15 B/min)*(60 min/hr)*(8 hours) = 4.5E+05 (ml/8 hr) Let's convert that to Liters by multiplying by (1 Liter/1000 ml). That gives 4.5E+02 Liters in 8 hours.

This is the total volume of air and CO. We can determine the percent CO in air from the initial information. Divide the volume of CO by the total volume of air: 1.50E-05. This is the fraction of the total air volume that is CO. Multiply this times the total breath volume over 8 hours to give the total CO volume over that period:

4.5E+02 (L/8 hr) * 1.50E-05 = 6.75E-03 L CO in 8 hours

The density of CO is given as 1.2 g/L, so multiply the CO volume times this conversion factor to get 8.10E-03 grams of CO, or 0.0081 grams CO.

This seems very small, so double-check my math. The process should be OK, though.

I hope this helps,

Bob