Tear gas has the composition 40.25% carbon, 6.19% hydrogen, 8.94% oxygen, 44.62% bromine. what is the empirical formula of this

Question

Katyanochek1

12 days ago

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Tear gas has the composition 40.25% carbon, 6.19% hydrogen, 8.94% oxygen, 44.62% bromine. what is the empirical formula of this

compound?

Chemistry

2 answers:

eduard [2.5K] · Answer 1 · 2026-03-23T23:42:35+03:00

<span>Determining the amount of moles and their ratios for the elements reveals their atomic ratio. Transforming element percentages to grams: 40.25% carbon = 40.25/100 =.4025 * 100 g of carbon = 40.25g of C 6.19% hydrogen = 6.19/100 =.0619 * 100g of hydrogen = 6.19g of H 8.94% oxygen = 8.94/100 =.0894 * 100 g of oxygen = 8.94g of O 44.62% bromine = 44.62/100 =.4462 * 100 g of bromine = 44.62g of Br Transforming grams of each element into moles: (48.38 g C) (1 mol/ 12.01 g C) = 4.028 mol C (8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H (53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O (44.62g of Br)(0.012515018021626 moles) = 0.55842 mol Br Finding the mole ratios by dividing each value by the smallest mole value: 4.028 mol C /0.55842 = 7.2 mol C x 5 = 36 mol C 8.056 mol H / 0.55842 = 14.42 mol H = 72 mol H 3.336 mol O / 0.55842 = 5.97 mol O = 30 mol O 0.55842 mol Br / 0.55842 = 1 mol Br = 5 mol Br Thus, the empirical formula is (C6H12O5)6Br5</span>

Alekssandra [2.7K] · Answer 2 · 2026-03-23T23:43:28+03:00

Response: The empirical formula is $C_6H_{11}OBr$

Explanation:Assuming the total mass of the specified tear gas equals 100g

The mass of each constituent aligns with the percentage of each element indicated

Mass of Carbon = 40.25g

Mass of Hydrogen = 6.19g

Mass of Oxygen = 8.94g

Mass of Bromine = 44.62g

Next, convert the mass of each component into moles using this formula

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Moles of carbon}=\frac{40.25g}{12g}=3.354moles$ (Molar Mass of C = 12g)

$\text{Moles of hydrogen}=\frac{6.19g}{1g}=6.19moles$ (Molar Mass of H = 1g)

$\text{Moles of oxygen}=\frac{8.94g}{16g}=0.558moles$ (Molar Mass of O = 16g)

$\text{Moles of Bromine}=\frac{44.62g}{80g}=0.558moles$ (Molar Mass of Br = 80g)

By dividing the number of moles of each element by the smallest value calculated, we can find the mole ratio for each element.

$\text{Mole Ratio of carbon}=\frac{3.354moles}{0.558moles}=6.01\approx 6$

$\text{Mole Ratio of hydrogen}=\frac{6.19moles}{0.558moles}=11.09\approx 11$

$\text{Mole Ratio of oxygen}=\frac{0.558moles}{0.558moles}=1$

$\text{Mole Ratio of bromine}=\frac{0.558moles}{0.558moles}=1$

The resulting mole ratio for each atom is reflected in the subscripts of the empirical formula.

Empirical Formula: $C_6H_{11}OBr$