Equal number of protein mixture containing, Haemoglobin $15.5 K g / m o l$ ,
Ribonuclease $13.7 K g / m o l$ , Myoglobin $17.2 K g / m o l$ . Calculate the mass average molecular weight (approximate) of the protein solution.

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Solution

Given,
Mass of Haemoglobin $= M_{1} = 15.5 K g / m o l$
Mass of Ribonuclease $M_{2} = 13.7 K g / m o l$
Mass of Myoglobin $= M_{3} = 17.2 K g / m o l$

Also,
it is given that we have an equal number of all these protein
$∴ n_{1} = n_{2} = n_{3} = n$

Now, The mass average molecular weight is given by formula:
${- M}_{w} = \frac{\sum n_{i} M_{i}^{2}}{\sum n_{i} M_{i}}$

Thus by putting all the values,

${- M}_{w} = \frac{n_{1} M_{1}^{2} + n_{2} M_{2}^{2} + n_{3} M_{3}^{2}}{n_{1} M_{1} + n_{2} M_{2} + n_{3} M_{3}} = \frac{n (15.5)^{2} + n (13.7)^{2} + n (17.2)^{2}}{n \times 15.5 + n \times 13.7 + n \times 17.2} = 15.59 \approx 16 K g / m o l$

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