A molecular compound is composed of 58.8% Xe, 7.2% O, and 34.0% F, by mass. If the molecular weight is 223 g/mol, what is the molecular formula? Draw the correct Lewis structure. Predict the molecular geometry using the VSEPR model.
See your other post. Take a 100 g sample which will give you the following:
58.8 g Xe
7.2 g O
34.0 g F
Convert grams to moles and find the empirical formula using the same process in your previous post.
http://www.kentchemistry.com/links/bonding/LewisDotTutorials/XeOF4.htm
To determine the molecular formula, we need to first calculate the number of moles of each element present in the compound.
The given percentages are by mass, so we can assume a 100g sample of the compound.
1. Calculate the mass of each element:
- Xe: 58.8g
- O: 7.2g
- F: 34.0g
2. Convert the mass of each element to moles using their respective atomic masses:
- Xe: 58.8g X (1 mol / 131.29g) ≈ 0.4485 mol
- O: 7.2g X (1 mol / 16.00g) ≈ 0.45 mol
- F: 34.0g X (1 mol / 18.99g) ≈ 1.79 mol
3. Divide the number of moles of each element by the smallest number of moles to find the simplest ratio:
- Xe: 0.4485 mol / 0.4485 mol ≈ 1
- O: 0.45 mol / 0.4485 mol ≈ 1
- F: 1.79 mol / 0.4485 mol ≈ 4
Therefore, the empirical formula is XeOF₄.
To determine the molecular formula, we need the molecular weight.
4. Divide the molecular weight by the empirical formula weight.
Molecular weight: 223g/mol
Empirical formula weight: 131.29g/mol(Xe) + 16.00g/mol(O) + (4 * 18.99g/mol)(F) = 223.25g/mol
5. Divide the molecular weight by the empirical formula weight to find the whole number ratio:
223g/mol / 223.25g/mol ≈ 0.998
Since the whole number ratio is close to 1, the molecular formula remains XeOF₄.
Now let's draw the Lewis structure for XeOF₄:
1. Place the central atom, Xe, in the center.
Xe
2. Attach a single bond between Xe and each F atom.
Xe - F (x4)
3. Fill the remaining valence electrons around each F atom.
Xe - F (x4, with 6 dots around each F)
4. Place a lone pair of electrons on the Xe atom to satisfy the octet rule.
Xe - .. - F (x4, with 6 dots around each F)
The Lewis structure of XeOF₄ is:
F
/
F-Xe-F
\
F
Now let's predict the molecular geometry of XeOF₄ using the VSEPR model:
The central Xe atom has one lone pair and four bonded pairs of electrons, resulting in a total of five electron groups.
According to the VSEPR model, the electron groups around the central atom will repel each other, leading to a molecular shape with minimal repulsion. In this case, the lone pair will force the bonded atoms closer together, resulting in a square pyramidal molecular geometry.
So, the molecular geometry of XeOF₄ is square pyramidal.
To determine the molecular formula, we need to find the empirical formula first. The empirical formula gives the simplest whole number ratio of atoms in a compound.
1. Convert the given percentages to grams:
- Xe: 58.8% of 223 g/mol = 131.224 g
- O: 7.2% of 223 g/mol = 16.056 g
- F: 34.0% of 223 g/mol = 75.82 g
2. Determine the number of moles for each element:
- Xe: moles = mass / molar mass = 131.224 g / 131.293 g/mol = 0.999 mol (approximately 1 mole)
- O: moles = 16.056 g / 15.999 g/mol = 1.004 mol (approximately 1 mole)
- F: moles = 75.82 g / 18.998 g/mol = 3.994 mol (approximately 4 moles)
3. Divide the number of moles of each element by the smallest number of moles:
The ratio comes out as approximately 1:1:4, so the empirical formula is XeOF4.
To draw the correct Lewis structure, we need to consider the valence electrons of each atom and the octet rule:
- Xenon (Xe) has 8 valence electrons.
- Oxygen (O) has 6 valence electrons.
- Fluorine (F) has 7 valence electrons.
Lewis structure of XeOF4:
F
|
O - Xe - O
|
F
To predict the molecular geometry using the VSEPR theory, we count the number of electron domains around the central atom (Xe) and identify the shape:
- There are 6 electron domains around the central atom (Xe): 4 bonding pairs (from F) and 2 lone pairs (from O).
- According to VSEPR theory, the arrangement of electron domains is called octahedral.
- However, the two lone pairs distort the shape, causing it to bend.
- Therefore, the predicted molecular geometry of XeOF4 is square planar.