Question

A polymer sample combines five different molecular-weight fractions, each of equal weight. The molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000. Calculate Mn, Mw, and Mz. Based upon these results, explain whether this sample has a broad or narrow molecular-weight distribution compared to typical commercial polymer samples.

Answer 1

Answer:

Mn = 43,783

Mw = 60,000

Mz = 73,333

narrow distribution = 1.37

Explanation:

The molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000. This means their Mi is respectively: (The molar weight (Mi) of the fractions)

Fraction 1 : Mi = 20  *10^3

Fraction 2: Mi = 40 *10^3

Fraction 3 : Mi = 60 *10^3

Fraction 4: Mi = 80 *10^3

Fraction 5 : Mi = 100  *10^3

The ΣMi = 300*10^-3

The Wi (mass of the fractions is for all the fractions the same, let's say 1)

So Wi = 1+1+1+1+1 = 5

Since number of moles = mass / Molar mass

The number of moles is respectively: ni = Wi/Mi (x10^5)

Fraction 1 : ni = Wi/Mi = 1/20000 = 5

Fraction 2: ni = 1/40000 = 2.5

Fraction 3 : ni =1/60000 = 1.67

Fraction 4: ni = 1/80000 = 1.25

Fraction 5 : ni= 1/100000 = 1

The Σni = 11.42

Mn = ΣWi/ni = 5/11.42*10^-5 = 43,783

Mw = (ΣWi * Mi)/ΣWi  = 300,000 /5 = 60,000

Mz = (ΣWi * Mi²)/ΣWi *Mi = (4*10^8 +16*10^8 +36*10^8 +64*10^8 +100*10^8) /300,000  =73,333

Mz/Mn = narrow distribution =60,000/43,783 = 1.37

Answer 2

Answer:

Mn=43783 g/mol

Mw=60000 g/mol

Mz=73333 g/mol

Explanation:

Hello,

In this case, since the molar masses of the fractions are 20x10³, 40x10³, 60x10³, 80x10³ and 100x10³, the total molar mass is:

$M_T=20x10^3+40x10^3 +60x10^3+80x10^3+100x10^3=300x10^3g/mol$

Now, assuming each fraction weights 1 gram, the moles of each fraction turns out:

$n_1=\frac{1g}{20x10^3g/mol}=5x10^{-5} mol\\n_2=\frac{1g}{40x10^3g/mol}=2.5x10^{-5} mol\\n_3=\frac{1g}{60x10^3g/mol}=1.7x10^{-5} mol\\n_4=\frac{1g}{80x10^3g/mol}=1.25x10^{-5} mol\\n_5=\frac{1g}{100x10^3g/mol}=1x10^{-5} mol$

And the total moles:

$n_T=11.42x10^{-5} mol$

Therefore,

$Mn=\frac{\Sigma W_i}{n_i} =\frac{5g}{11.42x10^{-5} mol}=43783g/mol$

$Mw=\frac{\Sigma W_iM_i }{\Sigma W_i}=\frac{300x10^{3}g/mol}{5} =60000g/mol$

$Mz=\frac{\Sigma W_iM_i^{2}}{\Sigma W_iM_i}=\frac{(4x10^8+1.6x10^9+3.6x10^9+6.4x10^9+10x10^9)g^2/mol^2}{300x10^{3}g/mol}=73333g/mol$

Best regards.