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Chapter 3: Problem 77

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 for the given compound, containing only carbon, hydrogen, and oxygen with 48.64% C and 8.16% H by mass, is C₃H₆O₂.

Step by step solution

01

Calculate the percentage of oxygen by mass

From the given information, we know the percentage of carbon and hydrogen. The compound contains only carbon, hydrogen, and oxygen, so we can calculate the percentage of oxygen by subtracting the percentages of carbon and hydrogen from 100%: Percentage of oxygen = 100% - (48.64% + 8.16%) = 100% - 56.8% = 43.2% Now, we have the following information: Carbon: 48.64% Hydrogen: 8.16% Oxygen: 43.2%
02

Convert percentages to grams

Assume that we have a 100 gram sample of the compound. Then the mass of each element in the sample is: Carbon: 48.64 g Hydrogen: 8.16 g Oxygen: 43.2 g
03

Convert grams to moles

To find the mole ratio, we first need to convert the mass of each element to moles. We can do this using the molar mass of each element: Moles of Carbon = \( \frac{48.64 g}{12.01g/mol}\) = 4.05 mol Moles of Hydrogen = \( \frac{8.16 g}{1.01g/mol}\)= 8.08 mol Moles of Oxygen = \( \frac{43.2 g}{16.00g/mol}\) = 2.7 mol
04

Find the simplest whole number ratio

To find the simplest whole number ratio between the moles of the elements, divide all values by the smallest number of moles: Divide all values by 2.7: Carbon: \( \frac{4.05}{2.7}\) ≈ 1.5 Hydrogen: \( \frac{8.08}{2.7}\) ≈ 3 Oxygen: \( \frac{2.7}{2.7}\) = 1 We see that the ratio for carbon is not a whole number (1.5). To convert it into a whole number, multiply all ratios by 2: Carbon: 1.5 × 2 = 3 Hydrogen: 3 × 2 = 6 Oxygen: 1 × 2 = 2
05

Write the empirical formula

Now that we have the whole number ratios for each element, we can write the empirical formula for the compound: C₃H₆O₂ The empirical formula for the given compound is C₃H₆O₂.

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Most popular questions from this chapter

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 g sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(\mathrm{g}) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

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?

Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide: $$ \mathrm{BaO}_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{BaCl}_{2}(a q) $$ What mass of hydrogen peroxide should result when \(1.50 \mathrm{~g}\) barium peroxide is treated with \(25.0 \mathrm{~mL}\) hydrochloric acid solution containing \(0.0272 \mathrm{~g} \mathrm{HCl}\) per \(\mathrm{mL}\) ? What mass of which reagent is left unreacted?

Tetrodotoxin is a toxic chemical found in fugu pufferfish, a popular but rare delicacy in Japan. This compound has a \(\mathrm{LD}_{50}\) (the amount of substance that is lethal to \(50 . \%\) of a population sample) of \(10 . \mu \mathrm{g}\) per \(\mathrm{kg}\) of body mass. Tetrodotoxin is \(41.38 \%\) carbon by mass, \(13.16 \%\) nitrogen by mass, and \(5.37 \%\) hydrogen by mass, with the remaining amount consisting of oxygen. What is the empirical formula of tetrodotoxin? If three molecules of tetrodotoxin have a mass of \(1.59 \times 10^{-21} \mathrm{~g}\), what is the molecular formula of tetrodotoxin? What number of molecules of tetrodotoxin would be the LD \(_{50}\) dosage for a person weighing \(165 \mathrm{lb}\) ?
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