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Chapter 3: Problem 94

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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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?

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