Question

What is the empirical formula of a compound with a percent composition of 22.5% Phosphorous and 77.5% Chlorine?

Answer 1

`\boxed {\boxed {\sf PCl_3}}`

Step-by-step explanation:

We are given the percent composition: 22.5% phosphorus and 77.5% chlorine.

We can assume there are 100 grams of this compound. We choose 100 because we can simply use the percentages as the masses.

• 22.5 g P
• 77.5 g Cl

Next, convert these masses to moles, using the molar masses found on the Periodic Table.

• P: 30.974 g/mol
• Cl: 35.45 g/mol

Use the molar masses as ratios and multiply by the number of grams.
`22.5 \ g \ P * \frac {1 \ mol \ P }{30.974 \ g \ P}= \frac {22.5 \ mol \ P }{ 30.974} = 0.7264157035 \ mol \ P`

`77.5 \ g \ Cl * \frac {1 \ mol \ Cl }{35.45 \ g \ Cl}= \frac {77.5 \ mol \ Cl }{ 35.45} \ =2.186177715 \ mol \ Cl`

Divide both of the moles by the smallest number of moles to find the mole ratio.

`\frac {0.7264157035} {0.7264157035} = 1`

`\frac {2.186177715}{0.7264157035}=3.009540824 \approx 3`

The mole ratio is about 1 P: 3 Cl, so the empirical formula is written as: PCl₃