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Chapter 9: Problem 171

The flavoring agent vanillin contains carbon, hydrogen, and possibly oxygen. When \(0.450 \mathrm{~g}\) of vanillin is subjected to combustion analysis, the results are \(63.08 \% \mathrm{C}\) and \(5.30 \% \mathrm{H}\). If the molar mass is approximately \(152 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula of vanillin?

Short Answer

Expert verified
The molecular formula of vanillin is \(\mathrm{C_9H_8O_3}\).

Step by step solution

01

Convert percentages to grams

We will directly use the given percentage of carbon and hydrogen to convert them to grams since the mass of vanillin combusted is given as \(0.450 \mathrm{~g}\): \[0.6308 \times 0.450 \mathrm{~g} = 0.28386 \mathrm{~g~C}\] \[0.0530 \times 0.450 \mathrm{~g} = 0.02385 \mathrm{~g~H}\]
02

Convert grams to moles

Now, let's convert the grams of carbon and hydrogen to moles by dividing by their molar masses: \[\frac{0.28386 \mathrm{~g~C}}{12.01 \mathrm{~g/mol}} = 0.02364 \mathrm{~mol~C}\] \[\frac{0.02385 \mathrm{~g~H}}{1.01 \mathrm{~g/mol}} = 0.02361 \mathrm{~mol~H}\]
03

Find the mole ratio

Divide the moles of carbon and hydrogen by the smaller number of moles to get the mole ratio: \[\frac{0.02364 \mathrm{~mol~C}}{0.02361} = 1\] \[\frac{0.02361 \mathrm{~mol~H}}{0.02361} = 1\]
04

Determine the empirical formula

The empirical formula is the simplest whole number ratio of elements in a compound. From step 3, the mole ratio of carbon and hydrogen is both 1. Since we don't have the percentage composition of oxygen, we can assume the remainder is oxygen: \[\% ~\mathrm{O} = 100\% - \% ~\mathrm{C} - \% ~\mathrm{H} = 100\% - 63.08\% - 5.30\% = 31.62\%\] \[0.3162 \times 0.450 \mathrm{~g} = 0.14229 \mathrm{~g~O}\] \[\frac{0.14229 \mathrm{~g~O}}{16.00 \mathrm{~g/mol}} = 0.00889 \mathrm{~mol~O}\] Divide the moles of oxygen by the smallest number of moles: \[\frac{0.00889 \mathrm{~mol~O}}{0.02361} = 0.376 \approx \frac{1}{3}\] Thus the empirical formula is \(\mathrm{C_3H_3O}\).
05

Calculate the empirical formula mass

The empirical formula mass is the sum of the molar masses of elements in the empirical formula: \[\mathrm{Empirical~Formula~Mass} = 3 \times 12.01 \mathrm{~g/mol} + 3 \times 1.01 \mathrm{~g/mol} + 16.00 \mathrm{~g/mol} = 51.09 \mathrm{~g/mol}\]
06

Determine the molecular formula

Now, divide the molar mass by the empirical formula mass and round the result to the nearest whole number to obtain the molecular formula: \[\frac{152 \mathrm{~g/mol}}{51.09 \mathrm{~g/mol}} \approx 3\] Multiply the empirical formula by this whole number to arrive at the molecular formula: \[\mathrm{Molecular~Formula} = 3 \times (\mathrm{C_3H_3O}) = \mathrm{C_9H_8O_3}\] Therefore, the molecular formula of vanillin is \(\mathrm{C_9H_8O_3}\).

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Most popular questions from this chapter

Succinic acid, an intermediate in the metabolism of certain foods, has a molecular mass of \(118.1 \mathrm{~g} / \mathrm{mole}\). A \(1.926 \mathrm{~g}\) sample of succinic acid \(\left(\mathrm{H}_{x} \mathrm{Suc}\right)\) reacts with exactly \(1.25 \mathrm{~g}\) of \(\mathrm{NaOH}\) according to the following balanced equation: \(\mathrm{H}_{x} \mathrm{Suc}+x \mathrm{NaOH} \rightarrow \mathrm{Na}_{x} \mathrm{Suc}+x \mathrm{H}_{2} \mathrm{O}\) What is the value of \(x\) ?

A \(1.000-g\) sample of a liquid that contains only carbon and hydrogen burns in oxygen to produce \(1.284 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What are the mass percents of the elements present in this sample? (b) What is the empirical formula for this compound? (c) The molar mass of this compound is determined to be about \(71 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound? (Hint: When attempting this problem, understand that all of the carbon in the compound burned ends up as \(\mathrm{CO}_{2}\), and all of the hydrogen in the compound burned ends up as \(\mathrm{H}_{2} \mathrm{O}\). Also, there is only 1 mole of \(C\) per mole of \(\mathrm{CO}_{2}\), but there are 2 moles of \(\mathrm{H}\) per mole of \(\mathrm{H}_{2} \mathrm{O} .\) )

Nitrogen and fluorine react to form nitrogen trifluoride according to the balanced chemical equation \(\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \rightarrow 2 \mathrm{NF}_{3}(g)\) For each of the following reaction mixtures, choose the limiting reactant: (a) \(0.50 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(0.50 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (b) \(12.0 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(20.0 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (c) \(2.5 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(7.5 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (d) 100 molecules \(\mathrm{N}_{2}(g)\) and 500 molecules \(\mathrm{F}_{2}(g)\) (e) \(5.00 \mathrm{~g} \mathrm{~N}_{2}(g)\) and \(15.0 \mathrm{~g} \mathrm{~F}_{2}(g)\) (f) \(20.0 \mathrm{mg} \mathrm{N}_{2}(g)\) and \(70.0 \mathrm{mg} \mathrm{F}_{2}(g)\)

How many molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\), would be in a tablet that contained \(250 \mathrm{mg}\) of aspirin? How many atoms of carbon would be in the aspirin in that tablet?

Determine the empirical formula of the compound that is \(26.4 \%\) by mass \(\mathrm{Na}, 36.8 \%\) by mass \(\mathrm{S}\), and also contains oxygen.
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