Problem 43 - Chemistry: The Central Science

Question

Give the empirical formula of each of the following compounds if a sample contains (a) \(0.0130 \mathrm{~mol} \mathrm{C}, 0.0390\) \(\operatorname{mol} \mathrm{H}\), and \(0.0065 \mathrm{~mol} \mathrm{O} ;\) (b) \(11.66 \mathrm{~g}\) iron and \(5.01 \mathrm{~g}\) oxy- gen; (c) \(40.0 \% \mathrm{C}, 6.7 \% \mathrm{H}\), and \(53.3 \% \mathrm{O}\) by mass.

Step-by-Step Solution

Answer

The empirical formulas for the given compounds are: (a) \(C_2H_6O\) (b) \(Fe_2O_3\) (c) \(CHO\)

Step 1: Determine moles of each element

Using the information given, we have: 0.0130 mol of C 0.0390 mol of H 0.0065 mol of O

Step 2: Calculate mole ratios

To find the mole ratio, divide the moles of each element by the smallest number of moles. \( \frac{0.0130}{0.0065} = 2 \) moles of C \( \frac{0.0390}{0.0065} = 6 \) moles of H \( \frac{0.0065}{0.0065} = 1 \) mole of O

Step 3: Empirical formula

We found the mole ratio to be 2:6:1, so the empirical formula is \(C_2H_6O\). (b)

Step 4: Convert mass to moles

For Iron, the molar mass is 55.85 g/mol and for Oxygen, it is 16.00 g/mol. Moles of iron: \( \frac{11.66~g}{55.85~g/mol} = 0.209~mol \) Moles of oxygen: \( \frac{5.01~g}{16.00~g/mol} = 0.313~mol \)

Step 5: Calculate mole ratios

To find the mole ratio, divide the moles of each element by the smallest number of moles. \( \frac{0.209}{0.209} = 1 \) mole of Fe \( \frac{0.313}{0.209} ≈ 1.5 \) moles of O

Step 6: Empirical formula

Since we found a mole ratio of 1:1.5, we need to convert this ratio to whole numbers by multiplying both values by 2. This gives us a ratio of 2:3, so the empirical formula is \(Fe_2O_3\). (c)

Step 7: Calculate mass ratios

We know the percent composition of each element by mass, so we can assume there is a 100 g sample. This would give us: 40.0 g of C 6.7 g of H 53.3 g of O

Step 8: Convert mass to moles

The molar mass of C, H, and O are 12.01 g/mol, 1.01 g/mol, and 16.00 g/mol, respectively. Moles of C: \( \frac{40.0~g}{12.01~g/mol} ≈ 3.33~mol \) Moles of H: \( \frac{6.7~g}{1.01~g/mol} ≈ 6.63~mol \) Moles of O: \( \frac{53.3~g}{16.00~g/mol} ≈ 3.33~mol \)

Step 9: Calculate mole ratios

To find the mole ratio, divide the moles of each element by the smallest number of moles. \( \frac{3.33}{3.33} = 1 \) mole of C \( \frac{6.63}{3.33} = 2 \) moles of H \( \frac{3.33}{3.33} = 1 \) mole of O

Step 10: Empirical formula

We found the mole ratio to be 1:2:1, so the empirical formula is \(CHO\).

Key Concepts

Mole Ratio

The mole ratio is a fundamental concept in stoichiometry and empirical formulas. It helps in determining the simplest whole number ratio of the elements in a compound. To find this ratio, we divide the number of moles of each element by the smallest number of moles present in any element of the compound. This allows us to compare quantities on the same scale.

For example, if a compound has 0.0130 moles of Carbon (C), 0.0390 moles of Hydrogen (H), and 0.0065 moles of Oxygen (O), we need to compare these quantities proportionally. Starting with Oxygen as it has the smallest value, we divide the moles of all elements by 0.0065 (moles of O).

• Carbon: \( \frac{0.0130}{0.0065} = 2 \)
• Hydrogen: \( \frac{0.0390}{0.0065} = 6 \)
• Oxygen: \( \frac{0.0065}{0.0065} = 1 \)

The resulting mole ratio is 2:6:1, leading to the empirical formula \( C_2H_6O \). Understanding mole ratios is key to interpreting experimental data and predicting the compositions of compounds.

Chemical Calculations

Chemical calculations are essential for translating raw data from a chemical experiment into meaningful conclusions. They typically involve converting between masses, moles, and atoms or molecules, often using the concept of molar mass.

For instance, to convert given mass data to moles, one would use the molar mass of the elements involved. Take 11.66 grams of Iron (Fe) and 5.01 grams of Oxygen (O) as an example. Knowing that Iron has a molar mass of 55.85 g/mol and Oxygen 16.00 g/mol, we find:

• Moles of Iron: \( \frac{11.66~g}{55.85~g/mol} \approx 0.209~mol \)
• Moles of Oxygen: \( \frac{5.01~g}{16.00~g/mol} \approx 0.313~mol \)

These mole values can now be used to establish ratios. It's crucial for determining empirical formulas and helps in understanding how different substances interact at a molecular level.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It can also help determine the empirical formula of a compound by relating mole ratios to mass or percentage data.

When given a compound's percentage composition, such as 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen, stoichiometry helps convert these percentages into moles, assuming a sample size, typically 100 grams for convenience.

• Moles of Carbon: \( \frac{40.0~g}{12.01~g/mol} \approx 3.33~mol \)
• Moles of Hydrogen: \( \frac{6.7~g}{1.01~g/mol} \approx 6.63~mol \)
• Moles of Oxygen: \( \frac{53.3~g}{16.00~g/mol} \approx 3.33~mol \)

From here, stoichiometry enables the calculation of the mole ratios, in this case, 1:2:1, to derive the empirical formula \( CHO \). Mastery of stoichiometry allows chemists to accurately gauge the proportions of substances in chemical reactions and synthesis.