Question

Antimony pentafluoride has a formula of SbF5. What is the percent composition of Sb in this molecule

Answer 1

To determine the percent composition of Sb in SbF5, calculate the molar mass of SbF5 and divide the mass of Sb by it.

Step-by-step explanation:

The percent composition of Sb (antimony) in SbF5 (antimony pentafluoride) can be determined by calculating the mass of Sb and dividing it by the molar mass of SbF5, then multiplying by 100.

The molar mass of SbF5 is calculated as follows:

Molar mass of SbF5 = (molar mass of Sb) + 5*(molar mass of F)

To calculate the percent composition of Sb in SbF5, we need to take into account the molar mass of each element in the formula.

Using the atomic masses of Sb (121.76 g/mol) and F (18.998 g/mol), we can calculate the molar mass of SbF5 as follows:

Molar mass of SbF5 = (121.76 g/mol) + 5*(18.998 g/mol)

After calculating the molar mass of SbF5, we can calculate the percent composition of Sb in SbF5:

Percent composition of Sb = (Mass of Sb / Molar mass of SbF5) * 100

Answer 2

56.17% is percent composition of Sb in the molecule

Step-by-step explanation:

Percent composition is the percent in mass of each element present in a particular molecule.

In SbF₅, there is 1 mole of Antimony -Molar mass: 121.76g- per 5 moles of fluorine -Molar mass: 19g/mol-. In a basis of 1 mole, the mass of Sb and F is:

Mass Sb:

1mol * (121.76g/mol) = 121.76g

Mass F:

1 mol SbF₅ = 5 moles F * (19g / mol) = 95g

Total mass:

121.76g + 95g = 216.76g

And percent composition of Sb:

121.76g / 216.76g * 100 =

56.17% is percent composition of Sb in the molecule