Question

What is the empirical formula of a compound that contains 40 percent carbon, 6.7 percent hydrogen, and 53.3 percent oxygen

Answer 1

40g C 6.7g H 53.3g O -------------- 100g total Divide each element by its molar mass 40g C x 1mol / 12.01g = 3.33mol C 6.7g H x 1mol / 1.01g = 6.09mol H 53.3g O x 1mol/ 16.00g = 3.33mol O Divide each mole number by the smallest # of moles. In this case, 3.33 is the smallest. 3.33mol C / 3.33mol = 1 C 6.09mol H / 3.33mol = 1.83 H 3.33mol O / 3.33mol = 1 O Empirical Formula = CH2O

Answer 2

The empirical formula is CH2O

calculation

step 1: calculate the moles of each element present in the compound.

moles = % composition / molar mass

• from periodic table the molar mass for carbon =12 g/mol ,

for hydrogen = 1 g/mol

for oxygen = 16 g/mol

• moles is therefore

for Carbon (C) = 40 /12 = 3.33 moles

for hydrogen (H) = 6.7 /1= 6.70 moles

for oxygen(O) = 53.3 / 16=3.33 moles

Step 2 : find the mole ratio of C:H:O by diving each mole by smallest mole ( 3.33 moles)

that is carbon (C) = 3.33/3.33 =1

Hydrogen(H) = 6.70 / 3.33=2

Oxygen(O) = 3.33/3.33 =1

Therefore the empirical formula = CH2O