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Chapter 3: Problem 116

Carbohydrates are compounds containing carbon, hydrogen, and oxygen in which the hydrogen to oxygen ratio is \(2: 1 .\) A certain carbohydrate contains 40.0 percent carbon by mass. Calculate the empirical and molecular formulas of the compound if the approximate molar mass is \(178 \mathrm{~g}\)

Short Answer

Expert verified
The empirical formula of the compound is \(C_{3}H_{32}O\) and the molecular formula is \(C_{6}H_{64}O_{2}\).

Step by step solution

01

Determine the Percent Composition of the Elements

Given that Carbohydrates contain Carbon, Hydrogen and Oxygen elements and the composition of Carbon is 40.0%. Then, the remaining 60.0% must be composed of Hydrogen and Oxygen. Further, since we know that Hydrogen to Oxygen ratio is 2:1, we can determine the percent composition of the elements in the compound by dividing 60.0% proportionately according to the ratio. 60.0% divided by 3 parts (2 parts Hydrogen, 1 part Oxygen) is 20.0%. So, Hydrogen will be 2 * 20.0% = 40.0% and Oxygen will be 20.0%.
02

Calculate the Empirical Formula

An empirical formula is the simplest ratios of atoms in a compound. To obtain the empirical formula, convert the weight percent to grams (assuming we have 100g of the compound), and then convert the masses to moles. For Carbon: (40.0g C)(1 mol C/12.01g C) = 3.33 mol C. For Hydrogen: (40.0g H)(1 mol H/1.008g H) = 39.68 mol H. For Oxygen: (20.0g O)(1 mol O/16.00g O) = 1.25 mol O. Divide each of these values by the smallest value to get the ratio of the atoms: C: (3.33/1.25) = 2.66 ≈ 3, H: (39.68/1.25) = 31.74 ≈ 32, O: (1.25/1.25) = 1. Hence, the empirical formula is \(C_{3}H_{32}O\).
03

Calculate the Molecular Formula

The molecular formula is a multiple of the empirical formula, and this multiple can be found if the molar mass is known. First, calculate the molar mass of the empirical formula: (3 * 12.01g/mol C) + (32 * 1.008g/mol H) + (16.00g/mol O) = 60.03 + 32.26 + 16.00 = 108.29 g/mol. The multiple is found by dividing the molar mass of the compound by the molar mass of the empirical formula: 178 g/mol ÷ 108.29 g/mol ≈ 1.64, which rounds up to 2. Hence, the molecular formula is twice the empirical formula which is \(C_{6}H_{64}O_{2}\).

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