A compound containing only C, H, and O, was extracted from the bark of the sassa

Question

A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 27.7 mg produced 75.2 mg of CO2 and 15.4 mg of H20. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas 1st attempt Part 1 (0.5 point) 111 See Periodic Table Note that formulas of organic compounds should first list the carbon and hydrogen with the rest of the atoms listed in alphabetical order. For this problem use the format: CHO wherex, y, and z are subscripts of 2 or greater. Empirical Formula Part 2 (0.5 point) See Hint Molecular Formula:

Explanation / Answer

1)

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 0.0752/44

= 0.0017091

Number of moles of H2O = mass of H2O / molar mass H2O

= 0.0154/18

= 0.0008556

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.0017091

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.0008556 = 0.0017111

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 0.0277 - 0.0017091*12 - 0.0017111*1

= 0.0054798

number of mol of O = mass of O / molar mass of O

= 0.0054798/16.0

= 0.0003425

Divide by smallest to get simplest whole number ratio:

C: 0.0017091/0.0003425 = 5

H: 0.0017111/0.0003425 = 5

O: 0.0003425/0.0003425 = 1

So empirical formula is:CHO

2)

Molar mass of C5H5O = 5*MM(C) + 5*MM(H) + 1*MM(O)

= 5*12.01 + 5*1.008 + 1*16.0

= 81.09 g/mol

Now we have:

Molar mass = 162.0 g/mol

Empirical formula mass = 81.09 g/mol

Multiplying factor = molar mass / empirical formula mass

= 162.0/81.09

= 2

So molecular formula is:CHO

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