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

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 : $C_6H_{11}OBr$

Answer 2

Empirical formula will be $C_6H_{11}OBr$

Let us assume 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

Calculation for number of moles:

Moles of Carbon= $\frac{40.25g}{12g/mol}=3.354moles$

Moles of Hydrogen= $\frac{6.19g}{1g/mol}=6.19moles$

Moles of Oxygen= $\frac{8.94g}{16g/mol}=0.588moles$

Moles of Bromine= $\frac{44.62g}{80g/mol}=0.558moles$

Calculation for mole-ratio:

Mole- ratio of Carbon = $\frac{3.354}{0.058}=6$

Mole- ratio of Hydrogen = $\frac{6.19}{0.558}=11$

Mole- ratio of Oxygen = $\frac{0.588}{0.588}=1$

Mole- ratio of Bromine = $\frac{0.588}{0.588}=1$

The mole ratio of each element is written in the subscript of each element in the empirical formula.

Thus, Empirical Formula : $C_6H_{11}OBr$

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