Question

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 1.126 g sample of BHT was combusted in an oxygen rich environment to produce 3.373 g of CO2(g) and 1.105 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT.

Answer 1

Given,

Mass of BHT = 1.126 g

Mass of CO₂ produced after combustion = 3.373 g

Mass of H₂O produced after combustion = 1.105 g

Calculating the number of moles of C and H from the given masses of CO₂ and H₂O,

= 3.373 g of CO₂ x (1 mol /44.01 g) x ( 1 mol C / 1 mol CO₂)

= 0.07664 mol of C

Similarly,

= 1.105 g of H₂O x ( 1 mol / 18.02 g) x (2 mol of H / 1 mol of H₂O)

= 0.1226 mol of H

Now, Converting the number of moles of C and H to grams,

= 0.07664 mol of C x ( 12.01 g /1 mol) = 0.9205 g of C

= 0.1226 mol of H x ( 1.008 g / 1 mol) = 0.1236 g of H

Now, We know, BHT contains C, H, and O,

Thus, Mass of BHT = Mass of C + Mass of H + Mass of O

1.126 g = 0.9205 g +0.1236 g + Mass of O

Mass of O = 0.08191 g of Oxygen

Calculating the number of moles of oxygen,

= 0.08191 mol of O x ( 1 mol / 15.99 g)

= 0.00512 mol of O

Now, WE have,

Moles of C = 0.07664 mol

Moles of H = 0.1226 mol

Moles of O = 0.00512 mol

Dividing each number of moles by the least number of moles,

= 0.07664 mol of C / 0.00512 = 15 mol of C

= 0.1226 mol of H / 0.00512 = 24 mol of H

0.00512 mol of O / 0.00512 = 1 mol of O

Thus, the empirical formula of BHT is C₁₅H₂₄O₁Or C₁₅H₂₄O