Question

a. Write an equality and two conversion factors for Greg's serum iron level.

b. How many micrograms of iron were in the 8.0-ml Sample of Greg's blood?

Short answer

a. $42\mathrm{mcg}\mathrm{of}\mathrm{iron}=1\mathrm{dL}\mathrm{of}\mathrm{blood}$

b. $3.4\mathrm{mcg}\mathrm{of}\mathrm{iron}$

Step-by-step solution

Part (a) Given Information

To write two conversion factors.

Part (a) Explanation

Micro and milli are used to measure smaller units.

Greg's iron test showed a blood serum iron level of $42\mathrm{mcg}/\mathrm{dL}$.

The unit conversion method that is used for Greg's serum iron level is:

$$\begin{gathered} 42\mathrm{mcg}\mathrm{of}\mathrm{iron}=1\mathrm{dL}\mathrm{of}\mathrm{blood} \\ \frac{42\mathrm{mcg}\mathrm{of}\mathrm{iron}}{1\mathrm{dL}\mathrm{of}\mathrm{blood}}\mathrm{and}\frac{1\mathrm{dL}\mathrm{of}\mathrm{blood}}{42\mathrm{mcg}\mathrm{of}\mathrm{iron}} \end{gathered}$$

Part (b) Given Information

The volume of Greg's blood sample=8.0 mL

The density of the sample is 42 mcg/dL.

part (b) Explanation

Unit conversion for mL into dL:

$$\begin{gathered} 1\mathrm{dL}=100\mathrm{mL} \\ \mathrm{or} \\ 1\mathrm{mL}=\frac{1}{10}\mathrm{dL}=0.1\mathrm{dL} \end{gathered}$$

The calculation for mass:

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{iron}=\mathrm{Volume}\mathrm{of}\mathrm{sample}\times \mathrm{Density} \\ \mathrm{Mass}\mathrm{of}\mathrm{iron}=0.08\mathrm{dL}\times 42\mathrm{mcg}/\mathrm{dL} \\ \mathrm{Mass}\mathrm{of}\mathrm{iron}=3.4\mathrm{mcg} \end{gathered}$$

Thus, the mass of iron in the sample is 3.4 mcg.