# Chapter 3 - Chemical Compounds - Example 3-3 - Using Relationships Derived from a Chemical Formula - Page 77: Practice Example B

175 mL

#### Work Step by Step

Millilitres of halothane= $1.00\times10^{24}\,Br\,atoms\times\frac{79.904\,g\,Br}{6.02214\times10^{23}\,Br\,atoms}\times\frac{197.4\,g\,C_{2}HBrClF_{3}}{79.904\,g\,Br}\times\frac{1\,mL\,C_{2}HBrClF_{3}}{1.871\,g\,C_{2}HBrClF_{3}}$ $=175\,mL\,C_{2}HBrClF_{3}$

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