Question

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?

Answer 1

Calculating the moles and the moles ratio of the elements gives us the ratio of atoms in each element. Converting the percentage of element into grams 40.25% carbon = 40.25/100 = .4025 * 100 g of carbon = 40.25g of C 6.19% hydrogen = 6.19/100 = .0619 * 100g g of hydrogen = 6.19g of H 8.94% oxygen = 8.94/100 = .0819 * 100 g of oxygen = 8.19g of O 44.62% bromine = 44.62/100 = .4462 * 100 g of bromine = 44.62g of Br Converting the grams of element into moles (48.38 g C) (1 mol/ 12.10 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 Calculating the moles ratio of elements by dividing the small number of moles of an element 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 = 1mol Br = 5 mol Br So the empirical formula is (C6H12O5)6Br5

Answer 2

Answer: Empirical formula will be $C_6H_{11}OBr$

Explanation: Assuming that the total mass of the given tear gas = 100g

Mass of each element will be equal to the percentages of each element given

Mass of Carbon = 40.25g

Mass of Hydrogen = 6.19g

Mass of Oxygen = 8.94g

Mass of Bromine = 44.62g

Now, converting mass of each element into their moles using the 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)

Dividing moles of each element with the smallest number of moles calculated, we get the mole ratio of 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 mole ratio of each element is written in the subscript of each element in the empirical formula.

Empirical Formula : $C_6H_{11}OBr$