Question

Please help

find the impractical formula of the compound that contains 46.6 g of magnesium 61.4 g of sulfur and 92.0 g of oxygen. type subscripts for each element

Answer 1

Answer: The empirical formula is
`MgSO_3`

Step-by-step explanation:

Mass of Mg= 46.6 g

Mass of S= 61.4 g

Mass of O = 92.0 g

Step 1 : convert given masses into moles.

Moles of Mg =
`\frac{\text{ given mass of Mg}}{\text{ molar mass of Mg}}= (46.6g)/(24g/mole)=1.942moles`

Moles of S =
`\frac{\text{ given mass of S}}{\text{ molar mass of S}}= (61.4g)/(32g/mole)=1.918moles`

Moles of O =
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (92.0g)/(16g/mole)=5.75moles`

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

For Mg =
`(1.942)/(1.918)=`

For S=
`(1.918)/(1.918)=1`

For O =
`(5.75)/(1.918)=3`

The ratio of Mg: S: O = 1: 1: 3

Hence the empirical formula is
`MgSO_3`