Question

A 49.5 g sample of a substance which contains C, H, and O is burned in excess oxygen, producing 110.0 g of CO₂ and 22.5 g of H₂O. A mass spectrum confirms that the molar mass of the sample is 162.0 g/mol.

a.) how many moles and grams of carbon were in the original sample?

b.) How many moles and grams of hydrogen were in the original sample?

c.) How many moles and grams of oxygen were in the original sample?

d.) What is the empiracle formula for the sample?

e.) What is the molecular formula for the sample?

Answer 1

Empirical formula is the least coefficient formula, that is

CxHyOz

we must find x,y,z via gravimetry.

Typically we do this relating to moles of C,H,O

mol of CO2 = mass of CO2/MW of CO2 = (110)/(44) = 2.5 mol of CO2

1 mol of CO2 = 1 mol of C ---> 2.5 mol of CO2 = 2.5 mol of C

mol of H2O = mass of H2O/MW of H2O= (22.5)/(18) = 1.25 mol of H2O

1 mol of H2O= 2 mol of H ---> 1.25mol of H2O= 1.25*2 = 2.50 mol of H

Now... mol of Oxygen:

Mass of O = Mass of sample - Mass of C - Mass of H

Mass of C = mol of C * MW of C = 2.5 *12 = 30.0 g of C

Mass of H = mol of H * MW of H = 2.5 *1= 2.50 g of H

Mass of O = 49.5 - 30-2.50= 17.0 g of O

mol of O = mass of O / MW of O = (17)/(16) = 1.0625 mol of O

Ratios:

H:C =2.5/2.5= 1

O:C =  2.5/1.6 = 1.56 = 3/2

H:O = 2.5/1.6 = 1.56 = 3/2

then:

for 3 mol of C --> 3 mol of H and 2 mol of

C3H3O

finally

molar mass --> ratio ?

MW of empirical = 12*3 + 3*1 + 16 = 55 g/mol

ratio = MW of formula / MW of empirical = 162/55 = 2.95 = 3

then

3xC3H3O

C9H9O3