Question

What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen?

Answer 1

What is the empirical formula of a compound that contains 43.38% sodium, 11.33% carbon, and 45.29% oxygen?


Na2CO3

Answer 2

Answer:
`Na_2CO_3`.

Step-by-step explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of Na = 43.38 g

Mass of C = 11.33 g

Mass of O = 45.29 g

Step 1 : convert given masses into moles.

Moles of Na=
`\frac{\text{ given mass of Na}}{\text{ molar mass of Na}} (43.38g)/(23g/mole)=1.89moles`

Moles of C =
`\frac{\text{ given mass of C}}{\text{ molar mass of C}}= (11.33g)/(12g/mole)=0.94moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (45.29g)/(16g/mole)=2.83moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Na =
`(1.89)/(0.94)=2`

For C =
`(0.94)/(0.94)=1`

For O =
`(2.83)/(0.94)=3`

The ratio of Na : C: O = 2: 1: 3

Hence the empirical formula is
`Na_2CO_3`.