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Chapter 9: Problem 146

Caffeine is made up of \(49.48 \%\) by mass \(C, 5.19 \%\) by mass \(\mathrm{H}, 28.85 \%\) by mass \(\mathrm{N}\), and \(16.48 \%\) by mass \(\mathrm{O}\). If its molar mass is approximately \(194 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula of caffeine?

Short Answer

Expert verified
The molecular formula of caffeine is C\(8\)H\(10\)N\(4\)O\(2\).

Step by step solution

01

Convert percentage to grams

Since we're given the percentage composition, let's assume we have \(100~g\) of the substance. This will make it easier to work with the given percentages: C: \(49.48~g\) H: \(5.19~g\) N: \(28.85~g\) O: \(16.48~g\)
02

Convert grams to moles

To find the empirical formula, we need to convert the grams of each element to moles, using their molar masses: C: \(\frac{49.48~g}{12.01~g/mol} = 4.12~mol\) H: \(\frac{5.19~g}{1.008~g/mol} = 5.15~mol\) N: \(\frac{28.85~g}{14.01~g/mol} = 2.06~mol\) O: \(\frac{16.48~g}{16.00~g/mol} = 1.03~mol\)
03

Find the simplest whole number ratio

To get the simplest whole number ratio of the elements, we'll divide each element's moles by the lowest mole value: C: \(\frac{4.12}{1.03} = 4\) H: \(\frac{5.15}{1.03} = 5\) N: \(\frac{2.06}{1.03} = 2\) O: \(\frac{1.03}{1.03} = 1\) This gives us the empirical formula: C\(4\)H\(5\)N\(2\)O
04

Calculate empirical formula's molar mass and factor

Now, we need to calculate the empirical formula's molar mass by adding the molar masses of its elements: C\(4\)H\(5\)N\(2\)O: \(4 \times 12.01 + 5 \times 1.008 + 2 \times 14.01 + 1\times 16.00 = 97~g/mol\) Now, we're going to find the factor by dividing the molar mass of caffeine (\(194~g/mol\)) by the empirical formula's molar mass (\(97~g/mol\)): Factor: \(\frac{194}{97} = 2\)
05

Determine the molecular formula

Now, we'll multiply the empirical formula by the factor we just found: Molecular Formula: 2(C\(4\)H\(5\)N\(2\)O) = C\(8\)H\(10\)N\(4\)O\(2\) The molecular formula of caffeine is C\(8\)H\(10\)N\(4\)O\(2\).

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Most popular questions from this chapter

Sucrose has the molecular formula \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} .\) If you were to completely burn \(2.00 \mathrm{~g}\) of sucrose in a stream of oxygen, how many grams of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) would be produced?

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