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Chapter 3: Problem 93

One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of 1.50 \(\mathrm{g}\) of fructose produced 2.20 \(\mathrm{g}\) of carbon dioxide and 0.900 \(\mathrm{g}\) of water. What is the empirical formula of fructose?

Short Answer

Expert verified
The empirical formula of fructose is CH₂O, derived from the ratio of moles of Carbon, Hydrogen, and Oxygen atoms (1:2:1) in the simplest whole number form.

Step by step solution

01

Find moles of carbon (C)

The complete combustion of fructose produces carbon dioxide (CO₂), which contains one carbon atom. To find the moles of carbon, we need to know the mass and molar mass of carbon dioxide produced. The mass of carbon dioxide produced is 2.20 g, and the molar mass of carbon dioxide is 44.01 g/mol (12.01 g/mol for carbon and 16.00 g/mol each for two oxygen atoms). Now, we can calculate the moles of Carbon: moles of CO₂ = \( \frac{2.20 g}{44.01 g/mol} = 0.0500 \ mol \) Since there is only one carbon atom per molecule of CO₂, the number of moles of carbon is 0.0500 mol.
02

Find moles of hydrogen (H)

The complete combustion of fructose produces water (H₂O), which contains two hydrogen atoms. To find the moles of hydrogen, we need to know the mass and molar mass of water produced. The mass of water produced is 0.900 g, and the molar mass of water is 18.02 g/mol (1.01 g/mol each for two hydrogen atoms and 16.00 g/mol for oxygen atoms). Now, we can calculate the moles of Hydrogen: moles of H₂O = \( \frac{0.900 g}{18.02 g/mol} = 0.050 \ mol \) Since there are two hydrogen atoms per molecule of H₂O, the number of moles of hydrogen is moles of H = 0.0500 mol × 2 = 0.100 mol
03

Calculate mass of oxygen (O)

The mass of fructose is given as 1.50 g. We've calculated the mass of carbon and hydrogen, so now we can find the mass of oxygen: mass of Oxygen = mass of Fructose - mass of Carbon - mass of Hydrogen mass of Oxygen = 1.50 g - (0.0500 mol × 12.01 g/mol) - (0.100 mol × 1.01 g/mol) mass of Oxygen = 1.50 g - 0.600 g - 0.101 g = 0.799 g
04

Find moles of oxygen (O)

Now, we can calculate the moles of oxygen by dividing the mass of oxygen by its molar mass (16.00 g/mol): moles of O = \( \frac{0.799 g}{16.00 g/mol} = 0.050 \ mol \)
05

Find the empirical formula

To find the empirical formula, we need to determine the simplest whole number ratio of moles of Carbon, Hydrogen, and Oxygen atoms. moles of C : moles of H : moles of O = 0.0500 : 0.100 : 0.0500 Divide all the mole ratios by the smallest mole number (0.0500 mol) to determine the simplest whole number ratio. C : H : O = 1 : 2 : 1 The empirical formula of fructose is CH₂O.

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Most popular questions from this chapter

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.

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right) :\) $$ \mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If 20.4 g of pentane are burned in excess oxygen, what mass of water can be produced, assuming 100\(\%\) yield?

Give the balanced equation for each of the following chemical reactions: a. Glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) reacts with oxygen gas to produce gaseous carbon dioxide and water vapor. b. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron((III) chloride and hydrogen sulfide gas. c. Carbon disulfide liquid reacts with ammonia gas to produce hydrogen sulfide gas and solid ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\)

The compound As \(_{2} \mathrm{L}_{4}\) is synthesized by reaction of arsenic metal with arsenic triodide. If a solid cubic block of arsenic $\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\( that is 3.00 \)\mathrm{cm}$ on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triodide, what mass of \(\mathrm{As}_{2} \mathrm{L}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{L}_{4}\) was 75.6\(\%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

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?
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