Find the empirical formula with solution: a) C4H8 b) C2H6O2 c) N2O5 d) Ba3(PO4)2 e) Te4I16 answer please I really need this
The Correct Answer and Explanation is:
Let’s find the empirical formula for each compound. The empirical formula is the simplest whole-number ratio of atoms in a compound.
a) C₄H₈
- Find the greatest common factor (GCF) of the subscripts: GCF of 4 and 8 is 4
- Divide both subscripts by 4:
C4H84=C1H2\frac{C_4H_8}{4} = C_1H_24C4H8=C1H2
✅ Empirical Formula: CH₂
b) C₂H₆O₂
- Subscripts: C = 2, H = 6, O = 2
- GCF of 2, 6, and 2 is 2
- Divide all subscripts by 2:
C2H6O22=CH3O\frac{C_2H_6O_2}{2} = CH_3O2C2H6O2=CH3O
✅ Empirical Formula: CH₃O
c) N₂O₅
- Subscripts: N = 2, O = 5
- There is no common factor other than 1
- Already in simplest form
✅ Empirical Formula: N₂O₅
d) Ba₃(PO₄)₂
This is an ionic compound and ionic formulas are already empirical formulas based on charge balance.
- Ba²⁺ ion
- PO₄³⁻ phosphate ion
To balance charges: 3 Ba²⁺ (3 × +2 = +6) and 2 PO₄³⁻ (2 × −3 = −6) gives a neutral compound
✅ Empirical Formula: Ba₃(PO₄)₂
e) Te₄I₁₆
- Subscripts: Te = 4, I = 16
- GCF of 4 and 16 is 4
- Divide both by 4:
Te4I164=Te1I4\frac{Te_4I_{16}}{4} = Te_1I_44Te4I16=Te1I4
✅ Empirical Formula: TeI₄
Summary:
| Molecular Formula | Empirical Formula |
|---|---|
| C₄H₈ | CH₂ |
| C₂H₆O₂ | CH₃O |
| N₂O₅ | N₂O₅ |
| Ba₃(PO₄)₂ | Ba₃(PO₄)₂ |
| Te₄I₁₆ | TeI₄ |
Explanation
The empirical formula represents the simplest ratio of atoms in a compound. It does not necessarily show the actual number of atoms in a molecule, but it gives the lowest whole-number ratio of elements.
To determine the empirical formula, we divide each subscript in the molecular formula by the greatest common factor. This ensures that we simplify the ratio of atoms without changing the relative amounts.
For example, in compound C₄H₈, the ratio of carbon to hydrogen is 4 to 8. Dividing each by 4 gives a 1 to 2 ratio, so the empirical formula is CH₂.
In C₂H₆O₂, each element’s subscript can be divided by 2, resulting in the empirical formula CH₃O. This tells us that the actual molecule might be more complex, but the ratio is 1 carbon to 3 hydrogens to 1 oxygen.
N₂O₅ already has a simplest ratio. No division is possible without using decimals, which are not allowed in empirical formulas. Therefore, it remains the same.
For Ba₃(PO₄)₂, it is an ionic compound, meaning the formula already reflects the simplest whole-number ratio of ions that balances the charges. No simplification is needed.
In Te₄I₁₆, the ratio of tellurium to iodine is 4 to 16. Dividing both by 4 gives a 1 to 4 ratio, resulting in the empirical formula TeI₄.
Empirical formulas are useful in understanding basic composition and in stoichiometric calculations, even if the full molecular structure is more complex.
