Question

If the observed absorbance is 0.200 and the blank absorbance is 0.049,

what is the concentration of Fe (mg/mL) in the serum?

Short answer

The concentration of Fe (mg/mL) in the serum is$2.23\mathrm{\mu g}/\mathrm{mL}$

Step-by-step solution

Calculate concentration of Fe:

$\mathrm{Concentration}\mathrm{of}\mathrm{Fe}\mathrm{in}\mathrm{serum}(\mathrm{\mu g}/\mathrm{mL})=\frac{\mathrm{Mass}\mathrm{of}\mathrm{Fe}\mathrm{in}\mathrm{serum}}{\mathrm{Volume}(\mathrm{mL})\mathrm{of}\mathrm{serum}}$ →(1)

Equation for Mass of Fe is

$\mathrm{Mass}\mathrm{of}\mathrm{Fe}\mathrm{in}\mathrm{serum}\left(\mathrm{\mu g}\right)=\frac{\mathrm{Corrected}\mathrm{absorbance}-0.0015}{0.0670}$

Formula for corrected absorbance is

Corrected absorbance = Observed absorbance - Blank absorbance

Given,

observed absorbance=0.200

the blank absorbance=0.049

Substitute in corrected absorbance,

localid="1663667659354" $\mathrm{Corrected}\mathrm{absorbance}=0.200-0.049$

Corrected absorbance$=0.151$

Substitute above value in mass,

$\mathrm{Mass}\mathrm{of}\mathrm{Fe}\mathrm{in}\mathrm{serum}\left(\mathrm{\mu g}\right)=\frac{\mathrm{Corrected}\mathrm{absorbance}-0.0015}{0.0670}$

$\mathrm{Mass}\mathrm{of}\mathrm{Fe}\mathrm{in}\mathrm{serum}\left(\mathrm{\mu g}\right)=\frac{0.151-0.0015}{0.0670}$

localid="1663667676297" $\mathrm{Mass}\mathrm{of}\mathrm{Fe}\mathrm{in}\mathrm{serum}\left(\mathrm{\mu g}\right)\approx 2.23\mathrm{\mu g}$

Substitute in (1),

$\mathrm{Concentration}\mathrm{of}\mathrm{Fe}\mathrm{in}\mathrm{serum}(\mathrm{\mu g}/\mathrm{mL})=\frac{\mathrm{Mass}\mathrm{of}\mathrm{Fem}\mathrm{serum}}{\mathrm{Volume}(\mathrm{mL})\mathrm{of}\mathrm{serum}}$

$\mathrm{Concentration}\mathrm{of}\mathrm{Fe}\mathrm{in}\mathrm{serum}(\mathrm{\mu g}/\mathrm{mL})=\frac{2.23\mathrm{\mu g}}{1\mathrm{mL}}$

$\mathrm{Concentration}\mathrm{of}\mathrm{Fe}\mathrm{in}\mathrm{serum}(\mathrm{\mu g}/\mathrm{mL})=2.23\mathrm{\mu g}/\mathrm{mL}$