Question

One molecule of haemoglobin combines with four molecules of oxygen. If \(({\rm{1}}{\rm{.0 g}}\) of haemoglobin combines with \(({\rm{1}}{\rm{.53 mL}}\) of oxygen at body temperature (\(({\rm{3}}{{\rm{7}}^{\rm{o}}}{\rm{C}}\)) and a pressure of \(({\rm{743 torr}}\), what is the molar mass of haemoglobin?

Short answer

The molar mass of hemoglobin is 6.8×10⁴g/mol.

Step-by-step solution

Calculation of moles of oxygen

One molecule of haemoglobin will combine with four molecules of oxygen. So, one mole of haemoglobin will combine with four moles of \({{\rm{O}}_2}\).

First, we calculate the number of moles of oxygen react with\(({\rm{1 g}}\)haemoglobin using the ideal gas equation:

\(\begin{aligned}{}{\rm{P = 743 torr \times }}\frac{{{\rm{1\;atm}}}}{{{\rm{760 torr }}}}\\{\rm{ = 0}}{\rm{.978\;atm}}\end{aligned}\)

\({\rm{T = 3}}{{\rm{7}}^{\rm{^\circ }}}{\rm{C + 273 = 310\;K, V = 1}}{\rm{.53\;mL = 1}}{\rm{.53 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;L}}\)

The number of moles of oxygen combined is

\(\begin{aligned}{}{\rm{n = }}\frac{{{\rm{PV}}}}{{{\rm{RT}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.978\;atm \times 1}}{\rm{.53 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;L}}}}{{\left( {{\rm{0}}{\rm{.08206\;L\;atm mo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{\;}}{{\rm{K}}^{{\rm{ - 1}}}}} \right){\rm{ \times 310\;K}}}}\\{\rm{ = 5}}{\rm{.88 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{\;mol}}\end{aligned}\)

Calculating molecular mass

One mole of oxygen combines with \(\frac{{\rm{1}}}{{\rm{4}}}{\rm{ moles}}\) of haemoglobin. So, the number of moles of haemoglobin combined with \(({\rm{5}}{\rm{.88 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{\;mol}}\)es of oxygen are

\(\frac{{\rm{1}}}{{\rm{4}}}{\rm{ \times 5}}{\rm{.88 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{\;mol = 1}}{\rm{.47 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{\;mol}}\)

Moles of haemoglobin\({\rm{ = }}\frac{{{\rm{ Mass of hemoglobin }}}}{{{\rm{ Molar mass of hemoglobin }}}}\)

Molar mass of haemoglobin\({\rm{ = }}\frac{{{\rm{ Mass of hemoglobin }}}}{{{\rm{ Moles of hemoglobin }}}}\)

\(\begin{aligned}{}{\rm{ = }}\frac{{{\rm{1\;g}}}}{{{\rm{1}}{\rm{.47 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{\;mol}}}}\\{\rm{ = 6}}{\rm{.8 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{\;g/mol}}\end{aligned}\)

Therefore, the required value ofmolar mass is \(({\rm{6}}{\rm{.8 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{\;g/mol}}\).