When \(0.684 \mathrm{~g}\) of an organic compound containing only carbon, hydrogen, and oxygen was burned in oxygen, \(1.312 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.805 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were obtained. What is the empirical formula of the compound?

Short Answer

Expert verified
The empirical formula is obtained by dividing the number of moles of carbon, hydrogen, and oxygen by the smallest number of moles of these elements.

Step by step solution

01

Calculate moles of carbon in CO2

Using the molar mass of CO2 (44.01 g/mol), convert the mass of CO2 to moles to find the moles of carbon: carbon moles = mass of CO2 / molar mass of CO2 = 1.312 g / 44.01 g/mol
02

Calculate moles of hydrogen in H2O

Using the molar mass of H2O (18.02 g/mol), convert the mass of H2O to moles to find the moles of hydrogen (since each mole of water contains 2 moles of hydrogen atoms): hydrogen moles = (mass of H2O / molar mass of H2O) * 2 = (0.805 g / 18.02 g/mol) * 2
03

Calculate moles of oxygen in the organic compound

Subtract the mass of carbon and hydrogen (as part of CO2 and H2O) from the total mass of the organic compound to get the mass of oxygen. Then divide by oxygen's molar mass (16.00 g/mol) to get moles of oxygen:mass of oxygen = total mass of compound - (mass of carbon + mass of hydrogen)oxygen moles = mass of oxygen / molar mass of oxygen
04

Determine empirical formula

Divide the moles of carbon, hydrogen, and oxygen by the smallest number of moles to determine the simplest whole number ratio. The resulting numbers are the subscripts of the elements in the empirical formula.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is akin to a recipe in chemistry. Just like in cooking, you need the right amount of each ingredient, stoichiometry tells you the exact amount of reactants you need to react together to form products in a chemical reaction. It's based on the conservation of mass where all atoms that enter a reaction must be accounted for in the products.

In the exercise, stoichiometry is used to relate the mass of an organic compound that was burned to the amounts of carbon dioxide and water produced in the reaction. We used the stoichiometric coefficients from the balanced chemical equation for the combustion of an organic compound to understand the amount of each element in the product. The process involves determining the ratios in which the elements combine, which are then used to determine the empirical formula of the original compound.
Molar Mass
The molar mass of a substance is the weight of 6.022 x 1023 (Avogadro's number) of its particles (atoms, molecules, ions, etc.), typically expressed in grams per mole (g/mol). Finding the molar mass is essential for converting between mass and moles of a substance.

In our example, we used the molar masses of carbon dioxide (CO2) and water (H2O) to convert the mass of these compounds to moles. Knowing the molar mass of CO2 is 44.01 g/mol and that of H2O is 18.02 g/mol, we used these values to find out how many moles of carbon and hydrogen are present after combustion.
Combustion Analysis
Combustion analysis is a method in chemistry for determining the elemental composition of a pure compound by burning it in excess oxygen and analyzing the amounts of products formed. It's particularly useful for organic compounds that are composed of carbon, hydrogen, and oxygen.

The analysis involves measuring the masses of carbon dioxide and water produced, which will later help in quantifying the actual amounts of carbon and hydrogen in the initial compound. Oxygen is then determined by the difference in mass from the original compound. The data from combustion analysis feeds directly into stoichiometric calculations to help determine the empirical formula of the compound.
Moles Calculation
The mole is a fundamental unit in chemistry used to measure quantity. One mole of any substance contains the same number of entities (like atoms or molecules) as there are atoms in exactly 12 grams of carbon-12, which is approximately 6.022 x 1023 entities.

In the given exercise, calculating moles was the next step after finding the masses of CO2 and H2O. The mass of each compound was divided by its molar mass to find the number of moles of CO2 and H2O, which then allowed for the determination of the moles of carbon and hydrogen. For oxygen, we subtracted the masses of carbon and hydrogen from the total mass of the organic compound to find the mass of oxygen, which we then converted to moles.

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Most popular questions from this chapter

Nail polish remover is usually the volatile liquid ethyl acetate, \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5} .\) In this compound, what is the (a) atom ratio of \(\mathrm{C}\) to \(\mathrm{O}\) ? (c) atom ratio of \(\mathrm{C}\) to \(\mathrm{H}\) ? (b) mole ratio of \(\mathrm{C}\) to \(\mathrm{O}\) ? (d) mole ratio of \(\mathrm{C}\) to \(\mathrm{H}\) ?

Chlorine is used by textile manufacturers to bleach cloth. Excess chlorine is destroyed by its reaction with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\), as follows. $$ \begin{aligned} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)+4 \mathrm{Cl}_{2}(g)+5 \mathrm{H}_{2} \mathrm{O} \longrightarrow \\ 2 \mathrm{NaHSO}_{4}(a q)+8 \mathrm{HCl}(a q) \end{aligned} $$ (a) How many moles of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) are needed to react with \(0.12 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) ? (b) How many moles of \(\mathrm{HCl}\) can form from \(0.12 \mathrm{~mol}\) of \(\mathrm{Cl}_{2} ?\) (c) How many moles of \(\mathrm{H}_{2} \mathrm{O}\) are required for the reaction $$ \text { of } 0.12 \mathrm{~mol} \text { of } \mathrm{Cl}_{2} ? $$ (d) How many moles of \(\mathrm{H}_{2} \mathrm{O}\) react if \(0.24 \mathrm{~mol} \mathrm{HCl}\) is formed?

How many kilograms of a fertilizer made of pure \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) would be required to supply 1 kilogram of nitrogen to the soil?

Some of the acid in acid rain is produced by the following reaction: $$ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) $$ If a falling raindrop weighing \(0.050 \mathrm{~g}\) comes into contact with \(1.0 \mathrm{mg}\) of \(\mathrm{NO}_{2}(g)\), how many milligrams of \(\mathrm{HNO}_{3}\) can be produced?

How many moles of nickel are in \(17.7 \mathrm{~g}\) of \(\mathrm{Ni}\) ?

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