A compound contains 3.99% P, 82.3% Br, and 13.7% Cl. Find the empirical formula
Take a 100 g sample which will give you
3.99 g P
82.3 g Br and 13.7 g Cl
Convert g to moles. moles = grams/atomic mass.
Now find the ratio of each element to the others using whole numbers. The easy way to do this is to divide the smallest number by itself (thus you get 1.0000 for that element), then divide the others numbers by the same small number. Round everything to a whole number and that should be the empirical formula.
To find the empirical formula of the compound, we need to determine the ratio of the elements present.
Step 1: Assume we have 100 grams of the compound. This makes calculations easier, as the percentages will represent grams.
Step 2: Convert the grams of each element to moles using their respective molar masses. The molar masses are as follows:
- P (phosphorus) = 31.0 g/mol
- Br (bromine) = 79.9 g/mol
- Cl (chlorine) = 35.4 g/mol
Therefore:
- grams of P = 3.99 g (or 3.99 moles)
- grams of Br = 82.3 g (or 82.3 moles)
- grams of Cl = 13.7 g (or 13.7 moles)
Step 3: Divide the number of moles of each element by the smallest number of moles to obtain the simplest whole number ratio.
Since the number of moles for Cl is the smallest (13.7 moles), we will divide the moles of each element by 13.7:
- P: 3.99 moles / 13.7 moles = 0.29
- Br: 82.3 moles / 13.7 moles = 6.0
- Cl: 13.7 moles / 13.7 moles = 1.0
Step 4: Round the above ratios to the nearest whole number.
- P: 0.29 → 0 (closest whole number)
- Br: 6.0 → 6
- Cl: 1.0 → 1
Step 5: Write the empirical formula using the whole number ratio obtained.
The empirical formula is P0Br6Cl. However, since the subscript 0 is not commonly used, we simplify the formula to Br6Cl.
To find the empirical formula of a compound, we need to determine the relative number of atoms of each element present in the compound. Here's how to calculate it step by step:
1. Assume we have 100 grams of the compound.
This assumption allows us to directly work with percentages as grams.
2. Convert the percentages to grams.
- 3.99% P: 100 grams * (3.99 / 100) = 3.99 grams P
- 82.3% Br: 100 grams * (82.3 / 100) = 82.3 grams Br
- 13.7% Cl: 100 grams * (13.7 / 100) = 13.7 grams Cl
3. Convert grams to moles.
- P: 3.99 grams P / 31.0 g/mol (atomic mass of P) ≈ 0.129 moles P
- Br: 82.3 grams Br / 79.9 g/mol (atomic mass of Br) ≈ 1.029 moles Br
- Cl: 13.7 grams Cl / 35.5 g/mol (atomic mass of Cl) ≈ 0.386 moles Cl
4. Divide each mole value by the smallest mole value.
In this case, the smallest mole value is 0.129 moles P.
- P: 0.129 moles P / 0.129 ≈ 1 mole P
- Br: 1.029 moles Br / 0.129 ≈ 7.984 moles Br (rounded to 3 significant figures)
- Cl: 0.386 moles Cl / 0.129 ≈ 2.992 moles Cl (rounded to 3 significant figures)
5. Determine the empirical formula.
The empirical formula represents the simplest whole number ratio of atoms in the compound.
Based on the calculations above, the mole ratios are approximately:
P:Br:Cl ≈ 1:7.984:2.992
Since we prefer whole numbers, we'll multiply each ratio by 3 to obtain:
P:Br:Cl ≈ 3:24:9
Therefore, the empirical formula of the compound is PBr₃Cl₃.