Question

Fructose is a common sugar found in fruit. Elemental analysis of fructose gives the following mass percent composition: C \(40.00 \%\), \(\mathrm{H} 6.72 \%\), \(\mathrm{O} 53.28 \%\). The molar mass of fructose is \(180.16 \mathrm{~g} / \mathrm{mol} .\) Find the molecular formula of fructose.

Short answer

The molecular formula of fructose is C6H12O6.

Step-by-step solution

Calculate the moles of each element

Using the mass percentages, assume 100 g of the compound so that the mass of each element in grams is equal to its percentage. Then, calculate the moles of each element by dividing the mass of each element by their respective atomic masses: carbon (C, atomic mass 12.01 g/mol), hydrogen (H, atomic mass 1.008 g/mol), and oxygen (O, atomic mass 16.00 g/mol).

Determine the simplest whole number ratio

Divide the number of moles of each element by the smallest number of moles calculated in step 1. This gives the mole ratio of the elements to one another, which will be used to find the empirical formula.

Calculate the empirical formula

Using the mole ratios, write the empirical formula by placing the calculated ratio as subscripts to the corresponding element.

Determine the molecular formula from the empirical formula

Determine the molar mass of the empirical formula from step 3. Then, divide the molar mass of fructose (180.16 g/mol) by the molar mass of the empirical formula to find a multiplication factor. Multiply this factor with the subscripts in the empirical formula to get the molecular formula.

Key concepts

Empirical Formula Calculation

The empirical formula of a compound provides the simplest whole-number ratio of elements present in it. It is like a simplified blueprint that shows which atoms are involved without specifying the actual number of atoms present in the compound. To determine the empirical formula, one must first accurately calculate the moles of each element in the compound.

For example, if we are given the mass percent composition, we assume a 100 g sample for simplicity's sake. With this assumption, the mass of each element in grams directly corresponds to its mass percent. Next, we convert the mass of each element to moles by dividing by the element's atomic mass. This step is vital because chemical compounds are based on mole ratios, not mass ratios.

After finding the moles of each element, the next step involves finding the smallest whole number ratio. This is done by dividing the number of moles of each element by the smallest number of moles from the calculation. The resulting values are then used as the subscript in the empirical formula, giving a formula that reflects the most reduced form of the compound.

Mass Percent Composition

Understanding the mass percent composition is essential as it conveys the percentage by weight of each element in a compound. This information can be especially useful when we need to work backward to find out the formula of the substance, a common task in chemical analysis and synthesis.

In the case of fructose, the mass percent composition tells us that in every 100 grams of fructose, there are 40 grams of carbon, 6.72 grams of hydrogen, and 53.28 grams of oxygen. Once we have the mass percent composition, the next step is to translate this into the amount of moles of each element. Since chemical reactions occur in mole ratios and not in mass, this step allows us to calculate the empirical formula of the compound.

Furthermore, mass percent composition can also shed light on the purity of a sample and help in identifying potential contaminants. When a student is provided with a substance's mass percent composition, they are equipped with a cornerstone piece of data that, along with the substance's molar mass, enables them to deduce its molecular formula.

Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the quantitative relationships, or ratios, of reactants and products in a chemical reaction. It is grounded in the law of conservation of mass, where the total mass of reactants equals the total mass of products.

In the context of empirical and molecular formulas, stoichiometry allows us to convert mass percentages to moles, and then use the mole ratio to construct our empirical formula. Once we have established this ratio, we can explore the relationship between the empirical formula and the true molecular formula by considering the substance's molar mass.

For compounds like fructose, where the empirical formula does not necessarily reflect the actual molecular structure, stoichiometry can reveal the multiple between the empirical formula's molar mass and the molecular molar mass. This multiple then allows us to scale up the empirical formula proportionately to arrive at the accurate molecular formula of the substance.