Question

Starch has the same empirical formula as cellulose and a molar mass of about \(1.0 \times 10^{5} \mathrm{~g} / \mathrm{mol}\) (a) What are the mass percents of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) in starch? (b) How many \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{5}\) units are linked together in a starch molecule?

Short answer

Answer: The mass percents of Carbon, Hydrogen, and Oxygen in starch are approximately 44.51%, 6.22%, and 49.28%, respectively. There are approximately 616 C6H10O5 units linked together in a starch molecule.

Step-by-step solution

Determine the molecular formula of starch

Given that the empirical formula of starch is the same as that of cellulose, we need to find the empirical formula of cellulose first. The empirical formula of cellulose is C6H10O5. Now, we can calculate the molar mass of the empirical formula: Molar mass of C6H10O5 = (6 x 12.01 g/mol) + (10 x 1.01 g/mol) + (5 x 16.00 g/mol) = 162.15 g/mol. Given that the molar mass of starch is approximately 1.0 × 10^5 g/mol, we can now find the molecular formula of starch: Molecular formula of starch = (Empirical formula) x n where n is the ratio of the molar mass of starch to the molar mass of its empirical formula: n = Molar mass of starch / Molar mass of empirical formula = (1.0 × 10^5 g/mol) / 162.15 g/mol n ≈ 616 So, the molecular formula for starch is approximately (C6H10O5)_616.

Calculate the mass percents of C, H, and O in starch

To find the mass percent of each element in the starch molecule, we will use the formula: Mass percent of an element = (Mass of the element in one mole of starch) / (Molar mass of starch) x 100% For carbon: Mass of carbon in one mole of starch = 6 (number of carbon atoms in empirical formula) × 12.01 (molar mass of C) × 616 (number of empirical formula units in molecular formula) = 44,508.72 g Mass percent of carbon = (44,508.72 g) / (1.0 × 10^5 g/mol) × 100% ≈ 44.51% For hydrogen: Mass of hydrogen in one mole of starch = 10 (number of hydrogen atoms in empirical formula) × 1.01 (molar mass of H) × 616 (number of empirical formula units in molecular formula) = 6,222.16 g Mass percent of hydrogen = (6,222.16 g) / (1.0 × 10^5 g/mol) × 100% ≈ 6.22% For oxygen: Mass of oxygen in one mole of starch = 5 (number of oxygen atoms in empirical formula) × 16.00 (molar mass of O) × 616 (number of empirical formula units in molecular formula) = 49,280.00 g Mass percent of oxygen = (49,280.00 g) / (1.0 × 10^5 g/mol) × 100% ≈ 49.28% So, the mass percents of C, H, and O in starch are approximately 44.51%, 6.22%, and 49.28%, respectively.

Calculate the number of C6H10O5 units linked together in a starch molecule

The number of C6H10O5 units linked together in a starch molecule is equal to the value of 'n', which we calculated in Step 1. Thus, there are approximately 616 C6H10O5 units linked together in a starch molecule. In conclusion, (a) The mass percents of C, H, and O in starch are approximately 44.51%, 6.22%, and 49.28%, respectively. (b) There are approximately 616 C6H10O5 units linked together in a starch molecule.

Key concepts

Empirical Formula

Understanding the concept of an empirical formula is fundamental in chemistry. An empirical formula represents the simplest whole-number ratio of elements within a compound. For instance, the empirical formula for starch, as mentioned in the exercise, is given as C₆H₁₀O₅. This formula does not necessarily reflect the exact number of atoms in a single molecule of starch but indicates the ratio of carbon (C) to hydrogen (H) to oxygen (O) atoms.

To determine an empirical formula, one typically starts with the mass percentages of each element, converts these to moles by using the atomic masses from the periodic table, and then simplifies the ratio to the smallest whole numbers. In the case of starch, we are given that its empirical formula is the same as that of cellulose, which has already been determined to be C₆H₁₀O₅. The ability to deduce such a simplified formula is a crucial step toward understanding the composition and chemical behavior of the molecule.

Molar Mass Calculation

Molar mass is a physical property defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). The molar mass calculation of a compound involves the addition of the molar masses of all the atoms present in the empirical formula. For example, the molar mass of cellulose's empirical formula, C₆H₁₀O₅, is the sum of the masses of 6 carbon atoms, 10 hydrogen atoms, and 5 oxygen atoms.

Using the atomic masses from the periodic table, our calculation for C₆H₁₀O₅ would be (6 x 12.01 g/mol) for carbon, (10 x 1.01 g/mol) for hydrogen, and (5 x 16.00 g/mol) for oxygen. The resulting molar mass of cellulose’s empirical formula is 162.15 g/mol. Knowing this allows us to compare it to the given molar mass of starch, which at approximately 1.0 × 10⁵ g/mol, indicates that the starch molecule is much larger than a single molecule represented by its empirical formula.

Mass Percent Composition

Mass percent composition is another crucial concept in chemistry that refers to the percentage by mass of each element in a compound. This is determined by dividing the total mass of each individual element by the molar mass of the entire compound and multiplying by 100%. It gives insight into the proportions of elements within a substance.

The steps outlined in the original exercise solution show us that to find the mass percent composition of starch, first, we calculate the mass of each element in the molecule using the number of atoms of each element from the empirical formula and the number of empirical formula units in the molecular formula of starch. With these masses and the molar mass of starch, we can compute the mass percent composition. For instance, we find that for starch, carbon makes up approximately 44.51%, hydrogen about 6.22%, and oxygen roughly 49.28% of the molecule's mass. These percentages are vital for understanding the distribution of mass amongst different elements in a molecule and can also help in determining empirical formulas if the molecular formula is known.