An unknown compound contains only c, h, and o. combustion of 8.50 g of this compound produced 20.0 g co2 and 5.46 g h2o. what is the empirical formula of the unknown compound?

The answer to your question is: C₃H₄O

Explanation:

Data

CxHyOz = 8.5 g

CO2 = 20 g

H2O = 5.46 g

Reaction

             CxHyOz + O2     ⇒    CO2   +    H2O

CO2

    MW = 44g

                       44g CO2 12g of C

                        20g CO2   x

                       x = (20 x 12) / 44

                       x = 5.45 g of C

                     # of moles = n = 5.45 / 12 = 0.454 mol of C

H2O

     MW = 18 g

                       18 g H2O 2g of H

                       5.46 g       x

                       x = (5.46 x 2) / 18 = 0.61 g of H

                       n = 0.61 / 1 = 0.61 moles of H2

Mass of O2

              mass CxHyOz = mass CO2 + mass H2 + mass O2

              mass O2 = 8.5 - 5.45 - 0.61

              mass O2 = 2.44g

              n = 2.44 / 16 = 0.153 mol of O2

Now, divide by the lowest number of moles

0.454 mol of C/ 0.153 = 2.97 ≈ 3

0.61 moles of H2/ 0.153 = 3.99 ≈ 4

0.153 mol of O2/ 0.153 =  1

Then, the empirical formula is: C₃H₄O

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.