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Chapter 3: Problem 107

When a \(2.074-\mathrm{g}\) sample that contains only carbon, hydrogen, and oxygen burns in excess \(\mathrm{O}_{2},\) the products are \(3.80 \mathrm{~g} \mathrm{CO}_{2}\) and \(1.04 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of this compound?

Short Answer

Expert verified
The empirical formula is \(\text{C}_3\text{H}_4\text{O}_2\).

Step by step solution

01

Calculate moles of Carbon from CO2

The problem states that the sample produces \(3.80 \, \text{g} \) of \(\text{CO}_2\). Use the molar mass of \(\text{CO}_2\) (\(44.01 \, \text{g/mol}\)) to find the moles of carbon. Each mole of \(\text{CO}_2\) contains one mole of carbon:\[\text{Moles of } \text{CO}_2 = \frac{3.80}{44.01} = 0.0863 \text{ mol} \, \text{CO}_2\]\[\text{Moles of C} = 0.0863 \, \text{mol}\]
02

Calculate moles of Hydrogen from H2O

The sample also produces \(1.04 \, \text{g} \) of \(\text{H}_2\text{O}\). Use the molar mass of \(\text{H}_2\text{O}\) (\(18.02 \, \text{g/mol}\)) to find the moles of water. Each mole of \(\text{H}_2\text{O}\) contains two moles of hydrogen:\[\text{Moles of } \text{H}_2\text{O} = \frac{1.04}{18.02} = 0.0577 \text{ mol} \, \text{H}_2\text{O}\]\[\text{Moles of H} = 2 \times 0.0577 = 0.1154 \, \text{mol}\]
03

Calculate mass of Oxygen in the original sample

Determine the mass of carbon and hydrogen from their moles and subtract from the total sample mass to find the mass of oxygen. The molar mass of carbon (\(12.01 \, \text{g/mol}\)) and hydrogen (\(1.01 \, \text{g/mol\)) allow us to find their masses:\[\text{Mass of C} = 0.0863 \, \text{mol} \times 12.01 \, \text{g/mol} = 1.036 \, \text{g}\]\[\text{Mass of H} = 0.1154 \, \text{mol} \times 1.01 \, \text{g/mol} = 0.117 \, \text{g}\]\[\text{Mass of O} = 2.074 \, \text{g} - 1.036 \, \text{g} - 0.117 \, \text{g} = 0.921 \, \text{g}\]
04

Calculate moles of Oxygen

Convert the mass of oxygen to moles using the molar mass of oxygen (\(16.00 \, \text{g/mol}\)):\[\text{Moles of O} = \frac{0.921}{16.00} = 0.0576 \, \text{mol}\]
05

Find the Empirical Formula

Divide the moles of each element by the smallest number of moles calculated. The smallest number of moles from the prior calculations is \(0.0576\):\[\frac{0.0863}{0.0576} = 1.50\] for C\[\frac{0.1154}{0.0576} = 2.00\] for H\[\frac{0.0576}{0.0576} = 1.00\] for OThe ratio \(1.5:2:1\) suggests doubling to get whole numbers: \(3:4:2\), giving the empirical formula \(\text{C}_3\text{H}_4\text{O}_2\).

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Key Concepts

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

Moles Calculation
Calculating moles is a fundamental skill in chemistry, especially when trying to find empirical formulas. In this problem, the moles of carbon were calculated from the amount of CO鈧� produced during combustion. Here's how it works:
  • First, you need to know the molar mass of CO鈧�, which is 44.01 g/mol. This tells you how much 1 mole of CO鈧� weighs.
  • By dividing the mass of CO鈧� (3.80 g) by its molar mass, you arrive at the number of moles of CO鈧�: \(\text{Moles of } \text{CO}_2 = \frac{3.80}{44.01} = 0.0863 \text{ mol}\).
  • Since each CO鈧� molecule contains one carbon atom, the moles of carbon directly equal the moles of CO鈧�: 0.0863 mol.
Breaking problems down into calculating specific quantities always starts with knowing the molar masses and utilizing them to convert grams into moles.
Combustion Analysis
Combustion analysis is a laboratory technique used to determine the elemental composition of a compound. It is particularly useful for organic compounds consisting of carbon, hydrogen, and oxygen. In the given problem:
  • The sample compound is burned in excess oxygen, producing CO鈧� and H鈧侽 as products. These products give crucial clues about the original compound.
  • Through combustion, the carbon in the compound is converted to CO鈧�, and hydrogen is converted to H鈧侽. The amounts of CO鈧� and H鈧侽 produced are measured and used to backtrack the amounts of carbon and hydrogen in the sample.
  • Additionally, by knowing the total mass of the original sample, you can also determine the oxygen content by subtracting the masses of carbon and hydrogen from the total mass.
This method is precise and allows chemists to analyze substances even when only a small sample is available.
Mass-to-Mole Conversion
Mass-to-mole conversion is fundamental in stoichiometry, helping chemists relate the mass of substances to the amount in moles. For this particular exercise:
  • The mass-to-mole conversion was used for each element in the compound. First, for carbon, from mass of CO鈧�, then hydrogen from mass of H鈧侽.
  • The calculation involves dividing the given mass of the substance (grams) by its molar mass (grams per mole). For instance, the exercise shows that 0.921 g of oxygen is converted to moles using its molar mass 16.00 g/mol: \(\text{Moles of O} = \frac{0.921}{16.00} = 0.0576 \text{ mol}\).
  • This conversion allows calculation of the ratios of moles of elements, which are then used to determine the empirical formula.
Understanding how to convert masses to moles is key to solving many chemical problems, providing the base for further calculations like determining empirical formulas.

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Most popular questions from this chapter

The compound dinitrogen monoxide, \(\mathrm{N}_{2} \mathrm{O}\), is a nontoxic gas that is used as the propellant in cans of whipped cream. How many nitrogen atoms are in a 34.7-g sample of \(\mathrm{N}_{2} \mathrm{O}\) ?

What is the percentage, by mass, of each element in the following substances? (a) \(\mathrm{C}_{4} \mathrm{H}_{8}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{~N}_{2}\) (c) \(\mathrm{Fe}_{2} \mathrm{O}_{3}\)

Heating \(\mathrm{NaWCl}_{6}\) at \(300^{\circ} \mathrm{C}\) converts it into \(\mathrm{Na}_{2} \mathrm{WCl}_{6}\) and \(\mathrm{WCl}_{6}\). If the reaction of \(5.64 \mathrm{~g} \mathrm{NaWCl}_{6}\) produces \(1.52 \mathrm{~g} \mathrm{WCl}_{6}\), what is the percentage yield?

What mass of \(\mathrm{NH}_{3}\) forms from the reaction of \(5.33 \mathrm{~g} \mathrm{~N}_{2}\) with excess \(\mathrm{H}_{2} ?\)

Although copper does not usually react with acids, it does react with concentrated nitric acid. The reaction is complicated, but one outcome is $$\begin{aligned}\mathrm{Cu}(\mathrm{s})+\mathrm{HNO}_{3}(\mathrm{conc}) \rightarrow \\\& \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell)\end{aligned}$$ (a) Name all of the reactants and products. (b) Balance the reaction. (c) Assign oxidation numbers to the atoms. Is this a redox reaction? (d) Pre-1983 pennies were made of pure copper. If such a penny had a mass of \(3.10 \mathrm{~g}\), how many moles of Cu are in one penny? How many atoms of copper are in one penny? (e) What mass of \(\mathrm{HNO}_{3}\) would be needed to completely react with a pre-1983 penny?
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