Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis

Question

1 day ago

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muth, 18.46% oxygen, and 1.16% hydrogen.

1 answer:

Tems11 [2.6K] · Answer 1 · 2026-04-06T19:55:43+03:00

Response: a) $BaCl_2$

b) $BiO_3H_3$

Rationale:

When percentages are provided, we assume the total mass to be 100 grams.

Consequently, the mass of each element corresponds to its respective percentage.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1: convert given masses into moles.

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

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

Step 2: To establish the mole ratio, divide each mole value by the smallest mole quantity obtained.

For Ba = $\frac{0.48}{0.48}=1$

For O = $\frac{0.96}{0.48}=2$

The ratio of Ba: Cl= 1:2

As a result, 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 = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles$

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

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

Step 2: For determining the mole ratio, divide each mole value by the smallest quantity of moles identified.

For Bi= $\frac{0.38}{0.38}=1$

For O = $\frac{1.15}{0.38}=3$

For H= $\frac{1.16}{0.38}=3$

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

Consequently, the empirical formula is $BiO_3H_3$