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Fructose, commonly called fruit sugar, is a monosaccharide found in many plants. It contains \(40 \%\) C, \(6.71 \% \mathrm{H},\) and the remainder O. (a) What is the empirical formula for fructose? (b) A mass spectrum of fructose shows a peak at about \(180 \mathrm{u}\). What is the molecular formula of the substance?

Short Answer

Expert verified
The empirical formula for fructose is CH2O, and the molecular formula is C6H12O6.

Step by step solution

01

Calculate moles of each element

Using the given mass percentages, let's assume we have 100g sample of fructose. This means we have 40g of Carbon (C), 6.71g of Hydrogen (H), and the remainder as Oxygen (O). First, find the moles of each element: - For Carbon: Moles_C = mass / molar_mass_C, where molar_mass_C = 12.01 g/mol - For Hydrogen: Moles_H = mass / molar_mass_H, where molar_mass_H = 1.01 g/mol - For Oxygen: we first find mass_O = 100g - mass_C - mass_H, and then Moles_O = mass / molar_mass_O, where molar_mass_O = 16.00 g/mol
02

Find the mole ratios

Divide the moles of each element by the smallest value among them to find the mole ratios. Round the ratios to the nearest whole number if necessary.
03

Write the empirical formula

Using the whole number mole ratios from step 2, write the empirical formula for fructose.
04

Calculate the empirical formula mass

Find the empirical formula mass by adding the molar masses of each element multiplied by the number of atoms in the empirical formula.
05

Determine the molecular formula

Using the mass spectrum () peak of 180 u (mass units), find the ratio between the molecular mass and the empirical formula mass. - Ratio = molecular_mass / empirical_formula_mass If the ratio is a whole number, multiply each element in the empirical formula by the ratio to find the molecular formula of fructose. Now, let's solve the problem step-by-step.
06

Calculate moles of each element

- Moles_C = 40g / 12.01 g/mol ≈ 3.33 mol - Moles_H = 6.71g / 1.01 g/mol ≈ 6.64 mol - Mass_O = 100g - 40g - 6.71g = 53.29g - Moles_O = 53.29g / 16.00 g/mol ≈ 3.33 mol
07

Find the mole ratios

Minimum moles = min(3.33, 6.64, 3.33) = 3.33 - Ratio_C = Moles_C / Minimum_moles = 3.33 / 3.33 = 1 - Ratio_H = Moles_H / Minimum_moles = 6.64 / 3.33 ≈ 2 - Ratio_O = Moles_O / Minimum_moles = 3.33 / 3.33 = 1
08

Write the empirical formula

The empirical formula for fructose is CH2O.
09

Calculate the empirical formula mass

Empirical formula mass = (1 x 12.01) + (2 x 1.01) + (1 x 16.00) = 12.01 + 2.02 + 16.00 ≈ 30.03 g/mol
10

Determine the molecular formula

Ratio = molecular_mass / empirical_formula_mass = 180 u / 30.03 g/mol ≈ 6 Therefore, the molecular formula for the substance is C6H12O6, which is the molecular formula of fructose.

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Key Concepts

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

Molecular Formula
The molecular formula of a compound specifies the exact number of each type of atom in a molecule of the compound. For fructose, a mass spectrum showed a peak at around 180 u (atomic mass units), indicating its molecular mass. By comparing this weight to the empirical formula mass, you can determine the molecular formula.

To find the molecular formula, calculate the empirical formula mass, which is the weight of your smallest unit of the compound, in this case, CH₂O. This was found to be approximately 30.03 g/mol. Next, divide the molecular mass (180 u) by the empirical formula mass. This gives a value of 6, indicating that the molecular formula consists of six empirical units. Thus, the molecular formula of fructose is C₆H₁₂O₆.
  • The molecular formula provides insight into the actual number and types of atoms in a molecule.
  • It's derived from the empirical formula when the molecular mass is known.
Mass Spectrum
A mass spectrum is like a fingerprint of a molecule, showcasing how the piece is weighted. When analyzing fructose, the mass spectrum displayed a significant peak at approximately 180 u. This peak corresponds to the molecular weight and helps in determining the molecular formula.

In mass spectrometry, different fragments of the molecule are separated based on their mass-to-charge ratio (m/z). For fructose, the key peak at 180 indicates the presence of an intact molecular ion or the molecular mass of fructose itself.
  • Mass spectrometry can identify molecules and determine their quantities in a mixture.
  • The height of the peaks can reflect the abundance of the respective ions.
Moles Calculation
Moles calculation is crucial for translating the weight of each sample into moles, which allows for the determination of the empirical formula. It's a critical bridge from mass to molecule. Transform the mass percent into moles by utilizing the molar masses of each element.

For fructose:
  • Carbon (C): Start with 40 grams, using a molar mass of 12.01 g/mol, giving approximately 3.33 moles of Carbon.
  • Hydrogen (H): With 6.71 grams and a molar mass of 1.01 g/mol, you get about 6.64 moles of Hydrogen.
  • Oxygen (O): The remainder of 53.29 grams with a molar mass of 16.00 g/mol results in around 3.33 moles of Oxygen.
By doing so, these values are then used to find the simplest whole number ratio, leading to the empirical formula.
Fructose Analysis
Fructose is a type of sugar found naturally in many plants. It is known for its high sweetness and forms the basis of our analysis. The challenge is to use composition details and analytical techniques to derive its chemical equations.

When analyzing fructose, note:
  • Elemental Composition: Fructose contains carbon, hydrogen, and oxygen, indicated by its chemical formula C₆H₁₂O₆.
  • Empirical vs Molecular Formula: The empirical formula of CH₂O simplifies the composition ratio, but the actual molecule is more complex.
Understanding fructose through these formulas helps in categorizing its role in nature and use in industries such as food production.

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Most popular questions from this chapter

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O} .\) If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\). A \(0.1005-g\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

The reaction between potassium superoxide, \(\mathrm{KO}_{2}\), and \(\mathrm{CO}_{2}\), $$ 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} $$ is used as a source of \(\mathrm{O}_{2}\) and absorber of \(\mathrm{CO}_{2}\) in selfcontained breathing equipment used by rescue workers. (a) How many moles of \(\mathrm{O}_{2}\) are produced when \(0.400 \mathrm{~mol}\) of \(\mathrm{KO}_{2}\) reacts in this fashion? (b) How many grams of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ? (c) How many grams of \(\mathrm{CO}_{2}\) are used when \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) are produced?

(a) Write "true" or "false" for each statement. (a) A mole of ducks contain a mole of feathers. (b) A mole of ammonia gas has a mass of \(17.0 \mathrm{~g}\). (c) The mass of 1 ammonia molecule is \(17.0 \mathrm{~g} .\) (d) \(\mathrm{A}\) mole of \(\mathrm{MgSO}_{4}(s)\) contains 4 moles of oxygen atoms.

Write balanced chemical equations corresponding to each of the following descriptions: (a) Potassium cyanide reacts with an aqueous solution of sulfuric acid to form hydrogen cyanide gas. (b) When an aqueous solution of ammonium nitrite \(\left(\mathrm{NH}_{4} \mathrm{NO}_{2}\right)\) reacts with an aqueous solution of potassium hydroxide, ammonia gas, water and metal nitrate is formed. (c) When hydrogen gas is passed over solid hot iron(III) oxide, the resulting reaction produces iron and gaseous water. (d) When liquid ethanoic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is combusted, carbon dioxide and water are formed.

(a) You are given a cube of silver metal that measures 1.000 \(\mathrm{cm}\) on each edge. The density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3} .\) How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that \(74 \%\) of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

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