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Chapter 3: Problem 143

The compound adrenaline contains \(56.79 \% \mathrm{C}, 6.56 \% \mathrm{H}\) , \(28.37 \% \mathrm{O},\) and 8.28\(\% \mathrm{N}\) by mass. What is the empirical formula for adrenaline?

Short Answer

Expert verified
The empirical formula for adrenaline is C\(_{8}\)H\(_{11}\)O\(_{3}\)N.

Step by step solution

01

Calculate the moles of each element

Since we are given the mass percentages for each element, we can assume a 100g sample of adrenaline. In this sample, the mass of each element will exactly match the given mass percentage. Now, we need to calculate the moles of each element in our 100g sample using their molar masses. - For Carbon (C), molar mass = 12.01 g/mol - For Hydrogen (H), molar mass = 1.01 g/mol - For Oxygen (O), molar mass = 16.00 g/mol - For Nitrogen (N), molar mass = 14.01 g/mol
02

Convert the mass percentages to moles

Using the molar masses, we can calculate the moles of each element in a 100g sample of adrenaline: - Moles of Carbon (C) = \(\frac{56.79\,g}{12.01\,g/mol} = 4.730\,mol\) - Moles of Hydrogen (H) = \(\frac{6.56\,g}{1.01\,g/mol} = 6.495\,mol\) - Moles of Oxygen (O) = \(\frac{28.37\,g}{16.00\,g/mol} = 1.773\,mol\) - Moles of Nitrogen (N) = \(\frac{8.28\,g}{14.01\,g/mol} = 0.5910\,mol\)
03

Find the ratio of moles

To find the simplest whole number ratio, we need to divide each value of moles by the smallest number of moles among all the elements. In this case, the smallest number of moles is that of Nitrogen (N), which is 0.5910. - Ratio of Carbon (C) = \(\frac{4.730}{0.5910} = 8.000\) - Ratio of Hydrogen (H) = \(\frac{6.495}{0.5910} = 11.00\) - Ratio of Oxygen (O) = \(\frac{1.773}{0.5910} = 3.000\) - Ratio of Nitrogen (N) = \(\frac{0.5910}{0.5910} = 1.000\)
04

Determine the empirical formula

Now that we have found the simplest whole number ratio of the elements, we can write the empirical formula for adrenaline. The empirical formula is C\(_{8}\)H\(_{11}\)O\(_{3}\)N.

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Most popular questions from this chapter

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