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Chapter 5: Problem 156

A compound with molar mass \(180.1 \mathrm{g} / \mathrm{mol}\) has the following composition by mass: $$\begin{array}{|ll|}\hline C & 40.0 \% \\\H & 6.70 \% \\\O & 53.3 \% \\\\\hline\end{array}$$ Determine the empirical and molecular formulas of the compound.

Short Answer

Expert verified
The empirical formula of the compound is CH2O and the molecular formula is C6H12O6.

Step by step solution

01

Convert percentages to grams

Assume that we have 100 grams of the compound. In this case, the mass percentages will be equivalent to the masses in grams for each element. So we have: - C: 40.0 g - H: 6.70 g - O: 53.3 g
02

Find moles for each element

Next, we will convert the masses of each element to moles by dividing their mass by their respective molar mass: - Moles of C: \(\frac{40.0\ \mathrm{g}}{12.01\ \mathrm{g/mol}} = 3.33\ \mathrm{mol}\) of C - Moles of H: \(\frac{6.70\ \mathrm{g}}{1.01\ \mathrm{g/mol}} = 6.63\ \mathrm{mol}\) of H - Moles of O: \(\frac{53.3\ \mathrm{g}}{16.00\ \mathrm{g/mol}} = 3.33\ \mathrm{mol}\) of O
03

Calculate the ratio of moles

To find the empirical formula, we need to determine the ratio of moles of each element. Divide the number of moles of each element by the smallest value: - For C: \(\frac{3.33}{3.33} = 1\) - For H: \(\frac{6.63}{3.33} \approx 2\) - For O: \(\frac{3.33}{3.33} = 1\) The empirical formula is CH2O.
04

Calculate the empirical formula molar mass

Now we need to find the molar mass of the empirical formula: - 1 mol of C: 1 × 12.01 g/mol = 12.01 g/mol - 2 mol of H: 2 × 1.01 g/mol = 2.02 g/mol - 1 mol of O: 1 × 16.00 g/mol = 16.00 g/mol Empirical formula molar mass: 12.01 g/mol + 2.02 g/mol + 16.00 g/mol = 30.03 g/mol
05

Determine the molecular formula

To find the molecular formula, we need to determine the ratio of the molecular molar mass to the empirical molar mass. Divide the molecular molar mass (180.1 g/mol) by the empirical molar mass (30.03 g/mol): \(n = \frac{180.1\ \mathrm{g/mol}}{30.03\ \mathrm{g/mol}}\approx 6\) This ratio means that the molecular formula is 6 times the empirical formula. So, the molecular formula is: C6H12O6.

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Most popular questions from this chapter

Consider the following unbalanced equation: $$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)$$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of \(1.0 \mathrm{kg}\) calcium phosphate with \(1.0 \mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\text { by mass)? }\right.\)

A compound contains only carbon, hydrogen, and oxygen. Combustion of \(10.68 \mathrm{mg}\) of the compound yields \(16.01 \mathrm{mg}\) \(\mathrm{CO}_{2}\) and \(4.37 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .\) The molar mass of the compound is \(176.1 \mathrm{g} / \mathrm{mol} .\) What are the empirical and molecular formulas of the compound?

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

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). In a certain experiment, \(20.00 \mathrm{g}\) iron metal was reacted with \(11.20 \mathrm{g}\) oxygen gas. After the experiment, the iron was totally consumed, and 3.24 g oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

Express the composition of each of the following compounds as the mass percents of its elements. a. formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) b. glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)
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