Issues regarding the conversion between volume ratio and mass ratio of liquefied gases.
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This post was last edited by btmark on 2010-9-1 09:28. The sulfur-containing liquefied gas produced in the workshop contains a high proportion of C5 compounds. I would like to calculate the mass of C5 compounds in the liquefied gas per day, but the laboratory data provided are in volume percentages. The details are as follows: Sample time, Analysis item (volume percentage): H2S (mg/m3), Ethane, Propane, Isobutane, n-Butane, Isopentane, n-Pentane, Total. For sulfur-containing liquefied gas at 8:00, the values are 1202.55, 11.48, 9.58, 31.70, 29.97, 14.72, and 100.00 respectively. Assuming that the daily production volume is 10 tons of liquefied gas, what is the mass percentage of C5 compounds in it, and what is the actual mass? I would appreciate it if experts could share the calculation methods and provide the formulas for future use!
The volume fraction of a gas is equivalent to its molar fraction; converting the molar fraction to a mass fraction will give the desired value
W1=(X1*M1)/(X1*M1+X2*M2+X3*M3+….+Xn*Mn)7. Assuming that the daily production is 10 tons of liquefied gas, with a carbon pentane content of 52% on a mass basis, the amount of carbon pentane is 5.2 tons