Question

Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bismuth, 18.46% oxygen, and 1.16% hydrogen.

Answer 1

Answer: a)
`BaCl_2`

b)
`BiO_3H_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.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba =
`\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= (66.06g)/(137g/mole)=0.48moles`

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

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

For Ba =
`(0.48)/(0.48)=1`

For O =
`(0.96)/(0.48)=2`

The ratio of Ba: Cl= 1:2

Hence the empirical formula is
`BaCl_2`

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi =
`\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= (80.38g)/(209g/mole)=0.38moles`

Moles of O=
`\frac{\text{ given mass of O}}{\text{ molar mass of O}}= (18.46g)/(16g/mole)=1.15moles`

Moles of H=
`\frac{\text{ given mass of H}}{\text{ molar mass of H}}= (1.16g)/(1g/mole)=1.16moles`

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

For Bi=
`(0.38)/(0.38)=1`

For O =
`(1.15)/(0.38)=3`

For H=
`(1.16)/(0.38)=3`

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is
`BiO_3H_3`