Find The Empirical Formula

Question

Tetrahydrocannabinol (THC), the active agent in marijuana, contains 80.16% carbon, 9.63% hydrogen, and 10.17% oxygen by mass. What is the empirical formula of THC?

Answer 1

The easiest way to answer empirical formula questions is to remember this poem:

"Percent to mass

Mass to mole

Divide by small

Multiply till whole"

Putting this into practice we obtain

80.16 g carbon

9.63 g hydrogen

10.17 g oxygen

Next, convert the mass of each element to moles of each element. We obtain:

6.67 mol carbon

9.55 mol hydrogen

0.64 mol oxygen

We can see that oxygen has the smallest amount of moles, so divide all amounts by 0.64. We obtain:

Carbon: 10.42

Hydrogen: 14.92

Oxygen: 1

We can round the values of carbon and hydrogen to 10.5 and 15 respectively. From here we can multiply by 2 to get whole numbers for all elements.

Carbon: 21

Hydrogen: 30

Oxygen: 2

This gives an empirical formula for THC of C₂₁H₃₀O₂

Answer 2

When given the % composition, as in this problem, we assume we 100 g of the unknown. That way, we can easily just convert % to grams. From grams of each element, we will find the moles. Once we have moles of each element, we will be sure they are in whole number ratios. These will then be the subscripts in the empirical formula.

moles C: 80.16 g C x 1 mol / 12.01 g = 6.674 moles C

moles H: 9.63 g H x 1 mol / 1.01 g = 9.53 moles H

moles O: 10.17 g O x 1 mol / 16.00 g = 0.6356 moles O

In an attempt to get all whole numbers, we will divide all by the lowest value (0.6356)...

moles C = 6.674 mol / 0.6356 = 10.50 moles C

moles H = 9.53 mol / 0.6356 = 14.99 moles H ~ 15.0 moles H

moles O = 0.6356 mol / 0.6356 = 1.00 moles O

We now have whole numbers for both H and O, but not for C, and it isn't close enough to 11 to round.

So, if we simply multiply all values by 2, we will have all whole numbers...

moles C = 10.50 x 2 = 21.0 moles C

moles H = 15.0 x 2 = 30.0 moles H

moles O = 1.00 x 2 = 2.00 moles O

Empirical formula = C₂₁H₃₀O₂

Answer 3

Find the actual mass of each element:

Carbon: 80.16g

Hydrogen: 9.63g

Oxygen: 10.17g

Convert to moles:

C; 80.16g / (12 g/mol) = 6.68 mol

H; 9.63/ (1 g/mol) = 9.63 mol

O; 10.17g/ (16 g/mol) = 0.64 mol

Use the smallest mol (0.64) to divide and find the subscripts:

C; 6.68/0.64 = 10.4

H; 9.63/0.64 = 15.0

O; 0.63/0.64 = 1

Since Hydrogen's mole can not be rounded, we will have to multiply all the numbers by 2 to get numbers that are either whole or can be rounded.

C; 10.4 x 2 = 20.8 (can be rounded to 21)

H; 15.1 x 2 = 30.0

O; 1 x 2 = 2

Therefore the subscripts are 21, 30, and 2 for C, H, and O respectively: C₂₁H₃₀O₂