Question

I have an input stream consisting of glycerol, water, and isopropanol, which are equal parts on a volume basis.

What is the composition of the stream on a molar basis?

Answer 1

The composition of stream on molar basis for glycerol, water and isopropanol is 16.62%, 67.43% and 15.93% respectively.

Step-by-step explanation:

Let suppose the volume is 1
`m^3`. For three equal components volume of each compound is 0.33
`m^3`.

By using the chemical formulas of the compounds, molecular masses are given as

• 
  `mm_g` is the molecular mass of glycerol which is 92 g/ mole
• 
  `mm_w` is the molecular mass of water which is 18 g/mole
• 
  `mm_i` is the molecular mass of isopropanol which is 60 g/mole

• 
  `\rho_g` is the density of glycerol which is 1260
  `kg/m^3`

• 
  `\rho_w` is the density of water which is 1000
  `kg/m^3`

• 
  `\rho_i` is the density of isopropanol which is 786
  `kg/m^3`

Now using the density and volumes of individual compounds, the masses are given as

`V_g=V_w=V_i =0.33 m^3`

Mass of glycerol is

`m_g=\rho_g * V_g\\m_g=1260 * 0.33\\m_g=415.8 kg`

Mass of water is

`m_w=\rho_w * V_w\\m_w=1000 * 0.33\\m_w=330 kg`

Mass of isopropanol is

`m_i=\rho_i * V_i\\m_i=786* 0.33\\m_i=260 kg`

Now number of moles of each compound are given as

Moles of Glycerol are

`n_g=m_g/mm_g\\n_g=415.8 * 1000/92\\n_g=4519.56 \,moles\\`

Moles of water are

`n_w=m_w/mm_w\\n_w=330 * 1000/18\\n_w=18333.33 \,moles\\`

Moles of isopropanol are

`n_i=m_i/mm_i\\n_i=260 * 1000/60\\n_i=4333.33 \,moles\\`

Total number of moles is

`n_t=n_g+n_w+n_i\\n_t=27186.22\, moles`

Now composition on molar basis is given as

Molar composition of glycerol is

`mn_g= (n_g)/(n_t) * 100\\mn_g= (4519.56)/(27186.22) * 100\\mn_g= 16.62 \%`

Molar composition of water is

`mn_w= (n_w)/(n_t) * 100\\mn_w= (18333.33)/(27186.22) * 100\\mn_w= 67.43 \%`

Molar composition of isopropanol is

`mn_i= (n_i)/(n_t) * 100\\mn_i= (4333.33)/(27186.22) * 100\\mn_i= 15.93 \%`

So the composition of stream on molar basis for glycerol, water and isopropanol is 16.62%, 67.43% and 15.93% respectively.