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Chapter 3: Problem 144

Adipic acid is an organic compound composed of 49.31\(\% \mathrm{C}\) , $43.79 \% \mathrm{O},$ and the rest hydrogen. If the molar mass of adipic acid is 146.1 \(\mathrm{g} / \mathrm{mol}\) , what are the empirical and molecular formulas for adipic acid?

Short Answer

Expert verified
The empirical formula of adipic acid is C3H5O2, and the molecular formula is C6H10O4.

Step by step solution

01

Calculate moles of each element

Let's start by assuming we have 100 grams of adipic acid. We can use the composition percentages to find out the mass of carbon, hydrogen, and oxygen in those 100 grams. - 49.31% C corresponds to 49.31 g of C - 43.79% O corresponds to 43.79 g of O - The rest is Hydrogen, 100 - 49.31 - 43.79 = 6.9 g H Now, we can convert the masses of each element to moles: - Moles of C: \(\cfrac{49.31 \; \mathrm g}{12.01 \; \mathrm{g/mol}} \) - Moles of O: \(\cfrac{43.79 \; \mathrm g}{16.0 \; \mathrm{g/mol}} \) - Moles of H: \(\cfrac{6.9 \; \mathrm g}{1.01 \; \mathrm{g/mol}} \)
02

Find the lowest whole-number ratio for moles of each element

To find the empirical formula, we need to determine the smallest whole-number ratio between moles of each element. First, divide the moles values by the smallest number of moles among elements: - C: \(\cfrac{4.10\; \mathrm{mol}}{2.74\; \mathrm{mol}} \) - O: \(\cfrac{2.74\; \mathrm{mol}}{2.74\; \mathrm{mol}} \) - H: \(\cfrac{6.83\; \mathrm{mol}}{2.74\; \mathrm{mol}} \) Now, round each value to the nearest whole number to find the mole ratio. In this case, the values are approximately C: 1.5, O:1, H: 2.5. If we multiply these numbers by 2 to convert them into whole numbers, we get C: 3, O: 2, and H: 5. Hence, the empirical formula is C3H5O2.
03

Calculate the molecular formula based on molar mass

We'll calculate the molar mass of the empirical formula derived in Step 2: Molar mass of C3H5O2 = 3(12.01) + 5(1.01) + 2(16.0) = 88.07 g/mol Now, we need to find a whole number that, when multiplied with the molar mass of the empirical formula, gives the molar mass of adipic acid (146.1 g/mol). \( n = \frac {M_{molecular}}{M_{empirical}} = \frac{146.1}{88.07} \approx 1.66 \) Since n is approximately equal to 1.66, we can multiply the empirical formula by 2 to get the molecular formula. Hence, the molecular formula is C6H10O4 (2 x C3H5O2).

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Most popular questions from this chapter

Hexamethylenediamine $\left(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{N}_{2}\right)$ is one of the starting materials for the production of nylon. It can be prepared from adipic acid $\left(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\right)$ by the following overall equation: $$ \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}(l)+2 \mathrm{NH}_{3}(g)+4 \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{6} \mathrm{H}_{16} \mathrm{N}_{2}(l)+4 \mathrm{H}_{2} \mathrm{O}(l) $$ What is the percent yield for the reaction if 765 g of hexamethylenediamine is made from \(1.00 \times 10^{3} \mathrm{g}\) of adipic acid?

An ionic compound \(\mathrm{MX}_{3}\) is prepared according to the following unbalanced chemical equation. $$ \mathrm{M}+\mathrm{X}_{2} \longrightarrow \mathrm{MX}_{3} $$ A 0.105 -g sample of \(\mathrm{X}_{2}\) contains \(8.92 \times 10^{20}\) molecules. The compound \(\mathrm{MX}_{3}\) consists of 54.47$\% \mathrm{X} \mathrm{by}\( mass. What are the identities of \)\mathrm{M}\( and \)\mathrm{X}$ , and what is the correct name for \(\mathrm{MX}_{3} ?\) Starting with 1.00 \(\mathrm{g}\) each of \(\mathrm{M}\) and \(\mathrm{X}_{2},\) what mass of \(\mathrm{MX}_{3} \mathrm{can}\) be prepared?

A 1.500 -g sample of a mixture containing only \(\mathrm{Cu}_{2} \mathrm{O}\) and CuO was treated with hydrogen to produce 1.252 \(\mathrm{g}\) of pure cupper metal. Calculate the mass percent of \(\mathrm{Cu}_{2} \mathrm{O}\) in the original mixture.

Balance the following equations representing combustion reactions: c. $C_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ d. Fe \((s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) e. $\mathrm{FeO}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$

One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of 1.50 \(\mathrm{g}\) of fructose produced 2.20 \(\mathrm{g}\) of carbon dioxide and 0.900 \(\mathrm{g}\) of water. What is the empirical formula of fructose?
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