Question

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

Short answer

The empirical formulas for the given compounds are: (a) C₂H₆O, (b) Fe₂O₃, and (c) CH₂O.

Step-by-step solution

(a) Given moles of elements

The given amount of each element is: C: 0.0130 mol H: 0.0390 mol O: 0.0065 mol Next, we need to find the simplest whole number ratio between these moles.

(a) Determining the ratio

Divide the moles of each element by the smallest moles value: C: \(\frac{0.0130}{0.0065}\) = 2 H: \(\frac{0.0390}{0.0065}\) = 6 O: \(\frac{0.0065}{0.0065}\) = 1 The empirical formula is C₂H₆O₁, which is simplified to C₂H₆O.

(b) Finding moles of elements

We are given the mass of each element in the sample: Fe: 11.66 g O: 5.01 g Now convert mass to moles using their respective molar masses: Fe: \(\frac{11.66\,\text{g}}{55.85\,\text{g/mol}}\) = 0.2086 mol O: \(\frac{5.01\,\text{g}}{16.00\,\text{g/mol}}\) = 0.3131 mol

(b) Determining the ratio

Divide the moles of each element by the smaller moles value: Fe: \(\frac{0.2086}{0.2086}\) = 1 O: \(\frac{0.3131}{0.2086}\) ≈ 1.5 Since we need whole numbers, multiply the ratio by 2 to get integers: Fe: 1 × 2 = 2 O: 1.5 × 2 = 3 The empirical formula is Fe₂O₃.

(c) Finding mass of elements

We are given the mass percentages of each element. Consider a 100 g sample, so the mass of each element in the sample would be: C: 40.0 g H: 6.7 g O: 53.3 g Next, convert mass to moles using their respective molar masses: C: \(\frac{40.0\,\text{g}}{12.01\,\text{g/mol}}\) = 3.330 mol H: \(\frac{6.7\,\text{g}}{1.008\,\text{g/mol}}\) = 6.646 mol O: \(\frac{53.3\,\text{g}}{16.00\,\text{g/mol}}\) = 3.331 mol

(c) Determining the ratio

Divide the moles of each element by the smallest moles value: C: \(\frac{3.330}{3.330}\) = 1 H: \(\frac{6.646}{3.330}\) ≈ 2 O: \(\frac{3.331}{3.330}\) = 1 The empirical formula is CH₂O.

Key concepts

Mole Ratio

When working with chemical formulas, understanding the mole ratio is essential. Mole ratio helps us identify the simplest integer ratio of different types of atoms in a compound. In general, this concept allows chemists to determine the empirical formula from a given set of chemical data.

For example, let's consider a compound containing 0.0130 moles of carbon, 0.0390 moles of hydrogen, and 0.0065 moles of oxygen. To find the mole ratio, divide the moles of each element by the smallest value among them.

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

Once the values are simplified to a whole number ratio, which indicates the empirical formula, we get C₂H₆O. It is crucial to ensure that these numbers are exact whole numbers, which sometimes involves multiplying by a common factor to achieve this.

Mastering the concept of mole ratios will improve your ability to analyze and understand the composition of compounds.

Molar Mass

Molar mass is a comprehensive measure of the mass of one mole of a substance, often expressed in grams per mole (g/mol). It is a critical concept in converting between the mass and the mole of a component in chemical reactions and formulas.

To calculate the molar mass, you must sum up the atomic masses of all elements present in that molecule. For instance, when determining the empirical formula for the compound containing iron and oxygen, we start by converting grams to moles using their respective molar masses:

• Iron (Fe): Molar mass is 55.85 g/mol, calculated as \( \frac{11.66 \, \text{g}}{55.85 \, \text{g/mol}} = 0.2086 \, \text{mol} \)
• Oxygen (O): Molar mass is 16.00 g/mol, calculated as \( \frac{5.01 \, \text{g}}{16.00 \, \text{g/mol}} = 0.3131 \, \text{mol} \)

From these calculations, the mole ratio is established to deduce the empirical formula Fe₂O₃.

Understanding molar mass helps in making conversions between mass and mole, a common requirement in empirical formula determination.

Mass Percentage

Mass percentage, also known as mass percent, indicates the concentration of an element in a compound relative to the total mass, expressed as a percentage. It is a useful concept in chemistry when determining the empirical formula based on mass data.

To calculate mass percentage, divide the mass of the element by the total mass of the compound, then multiply by 100. Considering a case with a compound that is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen, we can imagine 100 g of the compound, making the masses 40.0 g of carbon, 6.7 g of hydrogen, and 53.3 g of oxygen.

• Carbon: 40.0 g
• Hydrogen: 6.7 g
• Oxygen: 53.3 g

These masses are then converted to moles using their molar masses:

• Carbon: \( \frac{40.0 \, \text{g}}{12.01 \, \text{g/mol}} = 3.330 \, \text{mol} \)
• Hydrogen: \( \frac{6.7 \, \text{g}}{1.008 \, \text{g/mol}} = 6.646 \, \text{mol} \)
• Oxygen: \( \frac{53.3 \, \text{g}}{16.00 \, \text{g/mol}} = 3.331 \, \text{mol} \)

After deriving the mole ratio and simplifying, we determine the empirical formula CH₂O.

Thus, understanding mass percentage aids in converting percentage data into practical amounts to deduce empirical formulas accurately.