The empirical formula of 2,4‑dibromo‑3,3,4,4‑tetrafluorobut‑1‑ene is C2HBrF.

Here's how you can determine this:

1. Identify the number of each type of atom in the compound. In this case, we have 2 carbon (C) atoms, 1 hydrogen (H) atom, 1 bromine (Br) atom, and 1 fluorine (F) atom.

2. Write these numbers as subscripts after the corresponding element symbol. This gives us C2HBrF.

The empirical formula represents the simplest whole number ratio of atoms in a compound. In this case, the ratio of C:H:Br:F is 2:1:1:1, which simplifies to C2HBrF.

Question

The empirical formula of 2,4‑dibromo‑3,3,4,4‑tetrafluorobut‑1‑ene is C2HBrF.

Here's how you can determine this:

1. Identify the number of each type of atom in the compound. In this case, we have 2 carbon (C) atoms, 1 hydrogen (H) atom, 1 bromine (Br) atom, and 1 fluorine (F) atom.

2. Write these numbers as subscripts after the corresponding element symbol. This gives us C2HBrF.

The empirical formula represents the simplest whole number ratio of atoms in a compound. In this case, the ratio of C:H:Br:F is 2:1:1:1, which simplifies to C2HBrF.

Knowee AI · Accepted Answer

The empirical formula of 2,4‑dibromo‑3,3,4,4‑tetrafluorobut‑1‑ene is C2HBrF.

Here's how you can determine this:

1. Identify the number of each type of atom in the compound. In this case, we have 2 carbon (C) atoms, 1 hydrogen (H) atom, 1 bromine (Br) atom, and 1 fluorine (F) atom.

2. Write these numbers as subscripts after the corresponding element symbol. This gives us C2HBrF.

The empirical formula represents the simplest whole number ratio of atoms in a compound. In this case, the ratio of C:H:Br:F is 2:1:1:1, which simplifies to C2HBrF.

What is the empirical formula of 2,4‑dibromo‑3,3,4,4‑tetrafluorobut‑1‑ene?CHBrFC2HBrF2C2H2Br2F2C4H2Br2F4

Question

What is the empirical formula of 2,4‑dibromo‑3,3,4,4‑tetrafluorobut‑1‑ene?

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