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Chapter 3: Problem 85

A compound that contains only carbon, hydrogen, and oxygen is 48.64$\% \mathrm{C}\( and 8.16\)\% \mathrm{H}$ by mass. What is the empirical formula of this substance?

Short Answer

Expert verified
The empirical formula of the compound containing carbon, hydrogen, and oxygen, with mass percentages of 48.64%, 8.16%, and 43.20% respectively, is \(\text{C}_3 \text{H}_6 \text{O}_2\).

Step by step solution

01

Determine the mass percentages for all the elements

Given the mass percentages of carbon and hydrogen as 48.64% and 8.16%, respectively. We will also need the mass percentage of oxygen which can be found by subtracting the sum of the given percentages from 100%. Mass percentage of oxygen = 100% - (mass percentage of carbon + mass percentage of hydrogen) = 100% - (48.64% + 8.16%) = 43.20% Now we have: Carbon = 48.64% Hydrogen = 8.16% Oxygen = 43.20%
02

Convert mass percentages to moles

To find the empirical formula, we need to convert these mass percentages into moles. We can do this by assuming we have 100 grams of the compound, so the mass percentages become the mass in grams. Mass of carbon (C) = 48.64 g Mass of hydrogen (H) = 8.16 g Mass of oxygen (O) = 43.20 g Next, we will convert these masses into moles using the molar masses of the elements: Molar mass of carbon = 12.01 g/mol Molar mass of hydrogen = 1.01 g/mol Molar mass of oxygen = 16.00 g/mol Moles of carbon = (48.64 g) / (12.01 g/mol) = 4.05 mol Moles of hydrogen = (8.16 g) / (1.01 g/mol) = 8.08 mol Moles of oxygen = (43.20 g) / (16.00 g/mol) = 2.70 mol
03

Find the mole ratio

Now we will find the lowest whole number ratio of moles between the elements. To do this, we will divide the moles of each element by the smallest number of moles among them, in this case, the moles of oxygen (2.70 mol). Mole ratio of carbon = 4.05 mol / 2.70 mol = 1.50 ≈ 1.5 Mole ratio of hydrogen = 8.08 mol / 2.70 mol = 3.00 ≈ 3 Mole ratio of oxygen = 2.70 mol / 2.70 mol = 1 ≈ 1
04

Determine the empirical formula

Based on the simplified mole ratios, we have approximately 1.5 carbons, 3 hydrogens, and 1 oxygen. Since we cannot have 1.5 atoms in a molecule, we need to multiply the ratios by a factor of 2 to have whole numbers. 1.5 * 2 = 3 carbons 3 * 2 = 6 hydrogens 1 * 2 = 2 oxygens The empirical formula of this compound is C3 H6 O2 .

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Most popular questions from this chapter

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen contains \(2.59 \times 10^{23}\) atoms of hydrogen and is 17.3\(\%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound is present, and what is the mass of the sample?

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ What masses of iron(III) oxide and aluminum must be used to produce 15.0 \(\mathrm{g}\) iron? What is the maximum mass of aluminum oxide that could be produced?

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A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 \(\mathrm{g}\) of the compound produced 0.213 \(\mathrm{g} \mathrm{CO}_{2}\) and 0.0310 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\) . In another experiment, it is found that 0.103 \(\mathrm{g}\) of the compound produces 0.0230 $\mathrm{g} \mathrm{NH}_{3} .$ What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\) . Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\) . Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce $1.0 \times 10^{6} \mathrm{kg}\( \)\mathrm{HNO}_{3}\( by the Ostwald process? Assume 100\)\%$ yield in each reaction, and assume that the NO produced in the third step is not recycled.
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