Problem 98 - Chemistry: An Atoms-Focused Approach

Question

Organic compounds called carbobydrates may be formed in reactions between iron(II) sulfide and carbonic acid: $$2 \mathrm{FeS}+\mathrm{H}_{2} \mathrm{CO}_{3} \rightarrow 2 \mathrm{FeO}+\frac{1}{n}\left(\mathrm{CH}_{2} \mathrm{O}\right)_{n}+2 \mathrm{S}$$ a. What is the empirical formula of these carbohydrates? b. How much carbohydrate is produced from a reaction mixture that initially contains \(211 \mathrm{g}\) FeS and excess \(\mathrm{H}_{2} \mathrm{CO}_{3}\) if the reaction results in a \(78.5 \%\) yield? c. If the carbohydrate product has a molecular mass of 300 amu, what is its molecular formula?

Step-by-Step Solution

Answer

a. The empirical formula of the carbohydrates is CH₂O. b. The mass of the carbohydrate produced is 28.26 g. c. The molecular formula of the carbohydrate is C₁₀H₂₀O₁₀.

Step 1: Empirical Formula

The empirical formula of the carbohydrates is \(\mathrm{CH}_2\mathrm{O}\). b. Calculate the mass of carbohydrate produced To calculate the mass of carbohydrates produced, we can use stoichiometry, the given amount of iron(II) sulfide, and the reaction's percent yield.

Step 2: Moles of FeS

First, determine the number of moles in 211 g of FeS. The molar mass of FeS is \((55.8+32.1)=87.9 \, g/mol\). The moles of FeS can be calculated using the equation: \(\text{moles of FeS}=\frac{\text{mass of FeS}}{\text{molar mass of FeS}}\) Moles of FeS = \(\frac{211 \, g}{87.9 \, g/mol}=2.40 \, mol\)

Step 3: Moles of carbohydrates

Now, use the stoichiometry of the balanced chemical equation to find the moles of carbohydrates produced: \(n (\mathrm{CH}_{2}\mathrm{O})_{n} = \frac{1}{2}(\text{moles of FeS}) = \frac{1}{2}(2.40 \, mol)= 1.20 \, mol\)

Step 4: Yield Adjustment

Adjust the moles of carbohydrates for the 78.5% yield: \(\text{Adjusted moles of carbohydrates} = 1.20 \, mol \times 0.785 = 0.942 \, mol\)

Step 5: Mass of carbohydrates

Calculate the mass of carbohydrates produced using the empirical formula's molar mass: Molar mass of \(\mathrm{CH}_2\mathrm{O} = 12.0 + 2 + 16.0 = 30.0 \, g/mol\) Mass of carbohydrates = moles × molar mass = \(0.942 \, mol \times 30.0 \, g/mol = 28.26 \, g.\) b. The mass of the carbohydrate produced is \(28.26 \, g\). c. Determine the molecular formula

Step 6: Determine n in Molecular Formula

Divide the given molecular mass, 300 amu, by the empirical formula's molar mass: \(n = \frac{\text{Molecular mass}}{\text{Empirical formula molar mass}} = \frac{300 \, g/mol}{30.0 \, g/mol} = 10\) Insert the value of n into the empirical formula to obtain the molecular formula:

Step 7: Molecular Formula

The molecular formula of the carbohydrate is \((\mathrm{CH}_{2}\mathrm{O})_{10} \implies \mathrm{C}_{10}\mathrm{H}_{20}\mathrm{O}_{10}\).

Key Concepts

Empirical Formula

In chemistry, the empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. For carbohydrates, a common empirical formula is \(\mathrm{CH}_2\mathrm{O}\), which indicates that for every carbon atom, there are two hydrogen atoms and one oxygen atom.
This formula doesn't tell us how many atoms are actually present in a molecule but rather, it shows the proportion of the atoms involved.
Determining the empirical formula involves calculating the moles of each element in the compound and then simplifying their ratios to whole numbers. For carbohydrates like those formed in reactions with iron(II) sulfide and carbonic acid, the process confirms the fundamental unit, \(\mathrm{CH}_2\mathrm{O}\).
It's important to remember that empirical formulas are used to express the most reduced form of a compound's composition and can be a stepping stone to finding the molecular formula.

Molecular Formula

The molecular formula provides the total count of each type of atom in a molecule. Unlike the empirical formula, it conveys the actual number of atoms rather than their ratio.
To determine the molecular formula, you often start with the empirical formula. In the exercise, the carbohydrate's empirical formula is \(\mathrm{CH}_2\mathrm{O}\) and its molecular mass is given as 300 amu.
By dividing the molecular mass by the molar mass of the empirical formula, you find the multiplier \(n\) which tells you how many times the empirical formula repeats itself within the molecule.
\[ n = \frac{300 \, g/mol}{30.0 \, g/mol} = 10\]
Thus, the molecular formula is \((\mathrm{CH}_2\mathrm{O})_{10}\) which simplifies to \(\mathrm{C}_{10}\mathrm{H}_{20}\mathrm{O}_{10}\). This determination helps in understanding the compound's structure and potential uses or properties.

Stoichiometry

Stoichiometry is a key concept in chemistry, centered on calculating the relative quantities of reactants and products in chemical reactions. It utilizes a balanced chemical equation to understand how much of each component is needed or produced.
In the carbohydrate formation from iron(II) sulfide and carbonic acid, stoichiometry allows us to determine how many moles of carbohydrates can be formed from a given quantity of iron(II) sulfide. The equation shows that 2 moles of \(\text{FeS}\) react fully to produce \(n\) moles of \((\mathrm{CH}_2\mathrm{O})_n\).
For this problem, stoichiometry was used to find the amount of the product by indicating that \(\frac{1}{2}\) of the initial moles of \(\text{FeS}\) yield carbohydrates, resulting in \(1.20\) moles of carbohydrates before adjusting for yield.
Understanding stoichiometry is crucial for success in chemistry as it is the foundation of predicting amounts in reactions and understanding conversion relationships between substances.

Chemical Yield

Chemical yield refers to the efficiency of a reaction, expressed as the percentage of the theoretical yield. This is the amount of product that could be formed based on complete conversion of the reactants.
In practice, some reactions are not 100% efficient. Losses can occur due to incomplete reactions or side reactions. Therefore, the actual yield is often less than the theoretical yield.
In the carbohydrate experiment, the yield was determined to be 78.5%. Calculating the mass of carbohydrates produced involves adjusting the amount found via stoichiometry by this yield percentage. \[\text{Adjusted moles of carbohydrates} = 1.20 \, \text{mol} \times 0.785 = 0.942 \, \text{mol}\] Then, the mass was calculated as \(0.942 \, \text{mol} \times 30.0 \, \text{g/mol} = 28.26 \, \text{g}\).
Understanding yield calculations is important for evaluating the efficiency and direction of reactions.