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Chapter 2: Problem 47

What are the empirical formulas of the following compounds? (a) \(\mathrm{C}_{2} \mathrm{~N}_{2}\), (b) \(\mathrm{C}_{6} \mathrm{H}_{6}\), (c) \(\mathrm{C}_{9} \mathrm{H}_{20}\) (d) \(\mathrm{P}_{4} \mathrm{O}_{10},\) (e) \(\mathrm{B}_{2} \mathrm{H}_{6}\)

Short Answer

Expert verified
The empirical formulas are: (a) \(\mathrm{CN}\), (b) \(\mathrm{CH}\), (c) \(\mathrm{C}_{9} \mathrm{H}_{20}\), (d) \(\mathrm{P}_{2} \mathrm{O}_{5}\), (e) \(\mathrm{BH}_{3}\)

Step by step solution

01

Analyze formula (a)

For \(\mathrm{C}_{2} \mathrm{~N}_{2}\), we find that the greatest common divisor (gcd) of 2 and 2 is 2. Therefore, we divide each by the gcd, yielding an empirical formula of \(\mathrm{C}_{1} \mathrm{N}_{1}\), or more simply, \(\mathrm{CN}\).
02

Analyze formula (b)

For \(\mathrm{C}_{6} \mathrm{H}_{6}\), again the gcd of 6 and 6 is 6. Thus, dividing each by the gcd, we receive an empirical formula of \(\mathrm{C}_{1} \mathrm{H}_{1}\), or more simply, \(\mathrm{CH}\).
03

Analyze formula (c)

Looking at \(\mathrm{C}_{9} \mathrm{H}_{20}\), the gcd of 9 and 20 is 1. Hence, no reduction is possible and the empirical formula remains the same, \(\mathrm{C}_{9} \mathrm{H}_{20}\).
04

Analyze formula (d)

For the compound \(\mathrm{P}_{4} \mathrm{O}_{10}\), the gcd of 4 and 10 is 2. Therefore, we divide both by the gcd and the empirical formula is \(\mathrm{P}_{2} \mathrm{O}_{5}\).
05

Analyze formula (e)

Finally, in \(\mathrm{B}_{2} \mathrm{H}_{6}\), the gcd of 2 and 6 is 2. Thus, after division, the empirical formula is \(\mathrm{B}_{1} \mathrm{H}_{3}\), or simply, \(\mathrm{BH}_{3}\).

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Most popular questions from this chapter

The Group \(1 \mathrm{~B}\) metals, \(\mathrm{Cu}, \mathrm{Ag},\) and \(\mathrm{Au},\) are called coinage metals. What chemical properties make them specially suitable for making coins and jewelry?

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