## Chemistry: The Central Science (13th Edition)

$4.4 \times 10^{-3}$ grams of carbon monoxide (CO).
Step 1: Find the area $A$ = $(length)$ $\times$ $(width)$ $\times$ $(height)$ of the apartment using the given dimensions. Keep exact value, but acknowledge significant figures in the value for later calculation. Step 2: $(10.6 ft \times 14.8 ft \times 20.5 ft)$ = 3216.04 $ft^{3}$ (three significant figures) - a.) We are given the conversion factor 48 micrograms (CO) per cubic meter: $\frac{(48)micrograms (CO)}{(1)m^{3}}$ or [48 micrograms of carbon monoxide (CO)] per/for every [1 cubic meter $(m^{3})$] - b.) We also know the apartment/room's volume is: 3216.04 $ft^{3}$ -(limited to three sig. figs., but if you happen to use the rounded value of 3220 $ft^{3}$ in dimensional analysis the answer should still be the same). Step 3: Note: There are a few other possible conversion factors suitable for this problem. The conversion must have the same or a similar working order as the examples below, all possibilities should begin with the exact calculated area of the apartment/room, and applied using appropriate dimensional analysis to find grams of carbon monoxide (CO): $(ft^{3})$ to $(m^{3})$ to $(micrograms)$ to $(grams)$ $(ft^{3})$ to $(cm^{3})$ to $(m^{3})$ to $(micrograms)$ to $(grams)$ $(ft^{3})$ to $(in^{3})$ to $(m^{3})$ to $(micrograms)$ to $(grams)$ Step 4: 3216.04 $ft^{3}$ $\times$ $\frac{(0.305)^{3}m^{3}}{(1)ft^{3}}$ $\times$ $\frac{(48)micrograms}{(1)m^{3}}$ $\times$ $\frac{(10^{-6})grams}{(1)microgram}$ = $4.4 \times 10^{-3}$ grams (CO) The end result should be to two significant digits, unless however, the given value of (48 micrograms) is for some reason considered to be an exact value. Then the answer would be to three significant digits, because the calculated cubic feet of the apartment/room was noted as having only three sig figs. However, it is unlikely that (48 micrograms of carbon monoxide gas) for every (1 cubic meter) is an exact value. Human error, equipment error, and experimental error all might prove the conversion as nothing more than an approximate value. To add, the apartment/room was not described as being a sealed or closed environment, therefore other gaseous compounds/elements are undoubtedly shifting the equilibrium and therefore altering the amount of carbon monoxide present at any given moment. Lastly, we have assumed the apartment/room to be vacant, but if decorative items and furnishings are inside the space it would reduce the volume available for carbon monoxide to roam about which again alters the approximate conversion.