# Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic - Page 133: 93b

$C_{6}H_{3}Cl_{3}$

#### Work Step by Step

Calculate the molar mass of the empirical formula by adding the molar masses of its constituent elements together. $2(12.01)+1(1.01)+1(35.45)= 60.48$ Divide the molar mass given by this number and round to the nearest whole number (it should be really close). $\frac{181.44}{60.48}=3$ Multiply the Empirical formula by this number to get the molecular formula. $3(C_{2}HCl)=C_{6}H_{3}Cl_{3}$

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