Question

A 21.92 gram sample of iron is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 31.33 g. Determine the empirical formula of the metal oxide.

A 25.23 gram sample of chromium is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 43.67 g. Determine the empirical formula of the metal fluoride.

A 7.199 gram sample of iron is heated in the presence of excess bromine. A metal bromide is formed with a mass of 38.10 g. Determine the empirical formula of the metal bromide.

Answer 1

1)

we have mass of each elements as:

Fe: 21.92 g

O: 31.33 - 21.92 = 9.41 g

Divide by molar mass to get number of moles of each:

Fe: 21.92/55.85 = 0.3925

O: 9.41/16.0 = 0.5881

Divide by smallest:

Fe: 0.3925/0.3925 = 1

O: 0.5881/0.3925 = 1.5

Multiply by 2 to get simplest whole number ratio:

Fe: 1*2 = 2

O: 1.5*2 = 3

So empirical formula is:Fe2O3

Answer: Fe2O3

2)

we have mass of each elements as:

Cr: 25.23 g

F: 43.67 - 25.23 = 18.44 g

Divide by molar mass to get number of moles of each:

Cr: 25.23/52.0 = 0.4852

F: 18.44/19.0 = 0.9705

Divide by smallest to get simplest whole number ratio:

Cr: 0.4852/0.4852 = 1

F: 0.9705/0.4852 = 2

So empirical formula is:CrF2

Answer: CrF2

3)

we have mass of each elements as:

Fe: 7.199 g

Br: 38.10 - 7.199 = 30.9 g

Divide by molar mass to get number of moles of each:

Fe: 7.199/55.85 = 0.1289

Br: 30.9/79.9 = 0.3867

Divide by smallest to get simplest whole number ratio:

Fe: 0.1289/0.1289 = 1

Br: 0.3867/0.1289 = 3

So empirical formula is:FeBr3

Answer: FeBr3