On March 20, 1995, terrorists released an organophosphate nerve gas known as "Sarin" at several points in the Tokyo subway system, killing 11 and injuring more than 5,500 people. Sarin was concealed in lunch boxes and soft-drink containers and placed on subway train floors. It was released as terrorists punctured the containers with umbrellas before leaving the trains. Elemental analysis of Sarin gives the following result: 34.29% C, 7.19% H, 22.84% O, 13.56% F, 22.11% P. How many carbon atoms does the empirical formula of this compound contain?

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Quizplus · Accepted Answer

To find the empirical formula, we convert the percentages to moles by dividing by the atomic masses of each element: C (12.01 g/mol), H (1.008 g/mol), O (16.00 g/mol), F (19.00 g/mol), P (30.97 g/mol). This gives us:- C: $34.29 \, \text{g} \div 12.01 \, \text{g/mol} = 2.85 \, \text{mol}$- H: $7.19 \, \text{g} \div 1.008 \, \text{g/mol} = 7.13 \, \text{mol}$- O: $22.84 \, \text{g} \div 16.00 \, \text{g/mol} = 1.43 \, \text{mol}$- F: $13.56 \, \text{g} \div 19.00 \, \text{g/mol} = 0.71 \, \text{mol}$- P: $22.11 \, \text{g} \div 30.97 \, \text{g/mol} = 0.71 \, \text{mol}$To find the empirical formula, we divide each mole value by the smallest mole value found, which is 0.71 in this case. This gives us:- C: $2.85 \div 0.71 = 4$- H: $7.13 \div 0.71 = 10$- O: $1.43 \div 0.71 = 2$- F: $0.71 \div 0.71 = 1$- P: $0.71 \div 0.71 = 1$Thus, the empirical formula contains 4 carbon atoms, placing it in the range of 3 to 6 atoms.

On March 20, 1995, Terrorists Released an Organophosphate Nerve Gas