Question

Give the empirical formula of 11.66 g Fe and 5.01 g oxygen(O)

Answer 1

"Answer:

The Empirical Formula is = Fe2O3

Explanation:

• Find the percentages of elements in the compound.

Mass of compound = 11.66 + 5.01 =16.67 g

Percentage of Fe = (11.6/16.67) x 100 = 69.95 %

Percentage of O = (5.01/16.67) x 100 = 30.05 %

• Dividing the percentages of elements by their average atomic masses.

Fe = 69.95/55.84 = 1.253 mole

O = 30.05/16         = 1.878 mole

• Dividing the numbers by the smallest number among them and rounding off to the nearest whole numbers to get their EF.

Fe = 1.253/1.253 = 1

O = 1.878/1.253 = 1.5

• Multiply the numbers by 2 to get a whole number(This is a special case)

Fe = 1 x 2 = 2

O = 1.5 x 2 = 3

The Empirical Formula is = Fe2O3

Hope I Helped you

"

Answer 2

$Fe_{2}O_{3}$ is the empirical formula.

Explanation:

Calculate the elemental percentages in the compound.

Compound mass = 11.66 + 5.01 = 16.67 g

Fe concentration = (11.6/16.67) x 100 = 69.95%

O percent = (5.01/16.67) x 100 = 30.05 percent

Taking the percentages of elements and multiplying them by their average atomic masses

69.95/55.84 = 1.253 mole of $Fe$

O = 30.05/16 = 1.878 mole O = 30.05/16 = 1.878 mole O = 30.05/16 =

To calculate their EF, divide the numbers by the smallest number among them and round off to the nearest whole number.

$Fe$= 1.253/1.253 = 1 Fe = 1.253/1.253 = 1 Fe = 1.253/1.2

1.5 = O = 1.878/1.253

To get a whole number, multiply the numbers by two (This is a special case)

1 x 2 = 2 $Fe$

O = 1.5 x 2 = 3 O = 1.5 x 2 = 3 O = 1.5 x 2 =

The Empirical Formula is a formula that is based on empirical evidence.