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Chapter 3: Problem 122

The compound adrenaline contains \(56.79 \%\) C, \(6.56 \% \mathrm{H}\). \(28.37 \%\) O, and \(8.28 \%\) 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

Convert percentages to grams

Given that the percentages are based on mass, we can assume a sample size of 100 grams to make the calculations simpler. Thus, we have the following mass breakdown: - Carbon: \(56.79 g\) - Hydrogen: \(6.56 g\) - Oxygen: \(28.37 g\) - Nitrogen: \(8.28 g\) Note that these values are in grams.
02

Convert grams to moles

To find the mole ratios, we need to convert the masses to moles using the molar mass of each element. The molar masses are: - Carbon: \(12.01 g/mol\) - Hydrogen: \(1.008 g/mol\) - Oxygen: \(16.00 g/mol\) - Nitrogen: \(14.01 g/mol\) We'll now divide each element's mass by its molar mass to find the moles: - Carbon: \(56.79 g / 12.01 g/mol = 4.73 mol\) - Hydrogen: \(6.56 g / 1.008 g/mol = 6.51 mol\) - Oxygen: \(28.37 g / 16.00 g/mol = 1.77 mol\) - Nitrogen: \(8.28 g / 14.01 g/mol = 0.591 mol\)
03

Determine the smallest mole ratio

To find the empirical formula, we need to express the mole amounts as the smallest whole-number ratio. First, we'll divide each element's moles by the smallest mole value, which is nitrogen with \(0.591 mol\): - Carbon: \(4.73 mol / 0.591 mol = 8.00 \approx 8\) - Hydrogen: \(6.51 mol / 0.591 mol = 11.0 \approx 11\) - Oxygen: \(1.77 mol / 0.591 mol = 2.99 \approx 3\) - Nitrogen: \(0.591 mol / 0.591 mol = 1.0 \approx 1\)
04

Write the empirical formula

Now that we have the smallest whole-number ratios, we can write the empirical formula like so: Adrenaline: C\(8\)H\(11\)O\(3\)N\(1\), or simply: C\(8\)H\(11\)O\(3\)N Therefore, the empirical formula for adrenaline is C\(8\)H\(11\)O\(3\)N.

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Most popular questions from this chapter

What number of atoms of nitrogen are present in \(5.00 \mathrm{~g}\) of each of the following? a. glycine, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{~N}\) b. magnesium nitride c. calcium nitrate d. dinitrogen tetroxide

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Balance each of the following chemical equations. a. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)\) e. \(\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) f. \(\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MoO}_{3}(s)+\mathrm{SO}_{2}(g)\) g. \(\mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)
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