Esercizi sulla Determinazione della Formula Empirica

Esercizi sulla Determinazione della Formula Empirica

Esercizi sulla Determinazione della Formula Empirica

Versione italiana

Esercizi sulla Determinazione della Formula Empirica

La formula empirica di un composto chimico rappresenta il rapporto più semplice tra gli atomi degli elementi che lo compongono. Per determinare la formula empirica, è necessario seguire alcuni passaggi fondamentali.

Passaggi per Determinare la Formula Empirica

1. Determinare le masse relative di ciascun elemento nel composto.
2. Convertire le masse in moli utilizzando le masse molari degli elementi.
3. Dividere il numero di moli di ciascun elemento per il numero di moli più piccolo ottenuto.
4. Scrivere la formula empirica utilizzando i rapporti interi più semplici.

Esercizio 1: Determinazione della Formula Empirica

Problema: Un campione di un composto contiene 40 g di carbonio (C), 6 g di idrogeno (H) e 54 g di ossigeno (O). Determina la formula empirica del composto.

Soluzione:

1. Determinare le masse relative:

   • C: 40 g
   • H: 6 g
   • O: 54 g

2. Convertire le masse in moli:

   • Massa molare di C = 12 g/mol
   • Massa molare di H = 1 g/mol
   • Massa molare di O = 16 g/mol

   \text{Moli di C} = \frac{40 \, \text{g}}{12 \, \text{g/mol}} \approx 3.33 \, \text{mol}

   $$\text{Moli di C} = \frac{40 \, \text{g}}{12 \, \text{g/mol}} \approx 3.33 \, \text{mol}$$

   \text{Moli di H} = \frac{6 \, \text{g}}{1 \, \text{g/mol}} = 6 \, \text{mol}

   $$\text{Moli di H} = \frac{6 \, \text{g}}{1 \, \text{g/mol}} = 6 \, \text{mol}$$

   \text{Moli di O} = \frac{54 \, \text{g}}{16 \, \text{g/mol}} \approx 3.38 \, \text{mol}

   $$\text{Moli di O} = \frac{54 \, \text{g}}{16 \, \text{g/mol}} \approx 3.38 \, \text{mol}$$

3. Dividere per il numero di moli più piccolo:

   • Moli di C: \frac{3.33}{3.33} = 1$\frac{3.33}{3.33} = 1$
   • Moli di H: \frac{6}{3.33} \approx 1.80$\frac{6}{3.33} \approx 1.80$
   • Moli di O: \frac{3.38}{3.33} \approx 1.01$\frac{3.38}{3.33} \approx 1.01$

4. Arrotondare ai numeri interi più semplici:

   • C: 1
   • H: 1.80 (approssimato a 2)
   • O: 1

   Poiché 1.80 è vicino a 2, moltiplichiamo tutti i coefficienti per 5 per ottenere numeri interi:

   • C: 5
   • H: 10
   • O: 5

La formula empirica è:

\text{C}_5\text{H}_{10}\text{O}_5

$$\text{C}_5\text{H}_{10}\text{O}_5$$

Esercizio 2: Determinazione della Formula Empirica da Percentuali

Problema: Un composto è composto dal 40% di carbonio, 6.67% di idrogeno e 53.33% di ossigeno. Determina la formula empirica.

Soluzione:

1. Assumere 100 g di composto:

   • C: 40 g
   • H: 6.67 g
   • O: 53.33 g

2. Convertire le masse in moli:

   \text{Moli di C} = \frac{40 \, \text{g}}{12 \, \text{g/mol}} \approx 3.33 \, \text{mol}

   $$\text{Moli di C} = \frac{40 \, \text{g}}{12 \, \text{g/mol}} \approx 3.33 \, \text{mol}$$

   \text{Moli di H} = \frac{6.67 \, \text{g}}{1 \, \text{g/mol}} \approx 6.67 \, \text{mol}

   $$\text{Moli di H} = \frac{6.67 \, \text{g}}{1 \, \text{g/mol}} \approx 6.67 \, \text{mol}$$

   \text{Moli di O} = \frac{53.33 \, \text{g}}{16 \, \text{g/mol}} \approx 3.33 \, \text{mol}

   $$\text{Moli di O} = \frac{53.33 \, \text{g}}{16 \, \text{g/mol}} \approx 3.33 \, \text{mol}$$

3. Dividere per il numero di moli più piccolo:

   • Moli di C: \frac{3.33}{3.33} = 1$\frac{3.33}{3.33} = 1$
   • Moli di H: \frac{6.67}{3.33} \approx 2$\frac{6.67}{3.33} \approx 2$
   • Moli di O: \frac{3.33}{3.33} = 1$\frac{3.33}{3.33} = 1$

4. Scrivere la formula empirica:

   • C: 1
   • H: 2
   • O: 1

La formula empirica è:

\text{CH}_2\text{O}

$$\text{CH}_2\text{O}$$

Esercizio 3: Determinazione della Formula Empirica da Dati di Combustione

Problema: Un campione di un composto organico contiene 2.4 g di carbonio, 0.4 g di idrogeno e 1.6 g di ossigeno. Determina la formula empirica del composto.

Soluzione:

1. Determinare le masse relative:

   • C: 2.4 g
   • H: 0.4 g
   • O: 1.6 g

2. Convertire le masse in moli:

   \text{Moli di C} = \frac{2.4 \, \text{g}}{12 \, \text{g/mol}} = 0.2 \, \text{mol}

   $$\text{Moli di C} = \frac{2.4 \, \text{g}}{12 \, \text{g/mol}} = 0.2 \, \text{mol}$$

   \text{Moli di H} = \frac{0.4 \, \text{g}}{1 \, \text{g/mol}} = 0.4 \, \text{mol}

   $$\text{Moli di H} = \frac{0.4 \, \text{g}}{1 \, \text{g/mol}} = 0.4 \, \text{mol}$$

   \text{Moli di O} = \frac{1.6 \, \text{g}}{16 \, \text{g/mol}} = 0.1 \, \text{mol}

   $$\text{Moli di O} = \frac{1.6 \, \text{g}}{16 \, \text{g/mol}} = 0.1 \, \text{mol}$$

3. Dividere per il numero di moli più piccolo:

   • Moli di C: \frac{0.2}{0.1} = 2$\frac{0.2}{0.1} = 2$
   • Moli di H: \frac{0.4}{0.1} = 4$\frac{0.4}{0.1} = 4$
   • Moli di O: \frac{0.1}{0.1} = 1$\frac{0.1}{0.1} = 1$

4. Scrivere la formula empirica:

   • C: 2
   • H: 4
   • O: 1

La formula empirica è:

\text{C}_2\text{H}_4\text{O}

$$\text{C}_2\text{H}_4\text{O}$$

English version

Exercises on Determining the Empirical Formula

The empirical formula of a chemical compound represents the simplest ratio of the atoms of the elements that compose it. To determine the empirical formula, you need to follow some basic steps.

Steps to Determine the Empirical Formula

1. Determine the relative masses of each element in the compound.
2. Convert the masses to moles using the molar masses of the elements.
3. Divide the number of moles of each element by the smallest number of moles obtained.
4. Write the empirical formula using the simplest whole number ratios.

Exercise 1: Determining the Empirical Formula

Problem: A sample of a compound contains 40 g of carbon (C), 6 g of hydrogen (H), and 54 g of oxygen (O). Determine the empirical formula of the compound.

Solution: 1.

Determine the relative masses:

• C: 40 g
• H: 6 g
• O: 54 g

2. Convert the masses to moles:

• Molar mass of C = 12 g/mol
• Molar mass of H = 1 g/mol
• Molar mass of O = 16 g/mol

\text{Moles of C} = \frac{40 \, \text{g}}{12 \text{g/mol}} \approx 3.33 \, \text{mol}

$$\text{Moles of C} = \frac{40 \, \text{g}}{12 \text{g/mol}} \approx 3.33 \, \text{mol}$$

\text{Moles of H} = \frac{6 \, \text{g}}{1 \, \text{g/mol}} = 6 \, \text{mol}

$$\text{Moles of H} = \frac{6 \, \text{g}}{1 \, \text{g/mol}} = 6 \, \text{mol}$$

\text{Moles of O} = \frac{54 \, \text{g}}{16 \, \text{g/mol}} \approx 3.38 \, \text{mol}

$$\text{Moles of O} = \frac{54 \, \text{g}}{16 \, \text{g/mol}} \approx 3.38 \, \text{mol}$$

3. Divide by the smallest number of moles:

• Moles of C: \frac{3.33}{3.33} = 1$\frac{3.33}{3.33} = 1$
• Moles of H: \frac{6}{3.33} \approx 1.80$\frac{6}{3.33} \approx 1.80$
• Moles of O: \frac{3.38}{3.33} \approx 1.01$\frac{3.38}{3.33} \approx 1.01$

4. Round to the simplest whole numbers:

• C: 1
• H: 1.80 (to the nearest 2)
• O: 1

Since 1.80 is close to 2, we multiply all coefficients by 5 to get whole numbers:

• C: 5
• H: 10
• O: 5

The empirical formula is:

\text{C}_5\text{H}_{10}\text{O}_5

$$\text{C}_5\text{H}_{10}\text{O}_5$$

Exercise 2: Determining the Empirical Formula from Percentages

Problem: A compound is composed of 40% carbon, 6.67% hydrogen, and 53.33% oxygen. Determine the empirical formula.

Solution: 1. Take 100 g of compound:

• C: 40 g
• H: 6.67 g
• O: 53.33 g

2. Convert the masses to moles:

\text{Moles of C} = \frac{40 \, \text{g}}{12 \, \text{g/mol}} \approx 3.33 \, \text{mol}

$$\text{Moles of C} = \frac{40 \, \text{g}}{12 \, \text{g/mol}} \approx 3.33 \, \text{mol}$$

\text{Moles of H } = \frac{6.67 \, \text{g}}{1 \, \text{g/mol}} \approx 6.67 \, \text{mol}

$$\text{Moles of H } = \frac{6.67 \, \text{g}}{1 \, \text{g/mol}} \approx 6.67 \, \text{mol}$$

\text{Moles of O} = \frac{53.33 \, \text{g}}{16 \, \text{g/mol}} \approx 3.33 \, \text{mol}

$$\text{Moles of O} = \frac{53.33 \, \text{g}}{16 \, \text{g/mol}} \approx 3.33 \, \text{mol}$$

3. Divide by the smallest number of moles:

• Moles of C: \frac{3.33}{3.33} = 1$\frac{3.33}{3.33} = 1$
• Moles of H: \frac{6.67}{3.33} \approx 2$\frac{6.67}{3.33} \approx 2$
• Moles of O: \frac{3.33}{3.33} = 1$\frac{3.33}{3.33} = 1$

4. Write the empirical formula:

• C: 1
• H: 2
• O: 1

The empirical formula is:

\text{CH}_2\text{O}

$$\text{CH}_2\text{O}$$

Exercise 3: Determining the Empirical Formula from Combustion Data

Problem: A sample of an organic compound contains 2.4 g of carbon, 0.4 g of hydrogen, and 1.6 g of oxygen. Determine the empirical formula of the compound.

Solution:

1. Determine the relative masses:

• C: 2.4 g
• H: 0.4 g
• O: 1.6 g

2. Convert the masses to moles:

\text{Moles of C} = \frac{2.4 \, \text{g}}{12 \, \text{g/mol}} = 0.2 \, \text{mol}

$$\text{Moles of C} = \frac{2.4 \, \text{g}}{12 \, \text{g/mol}} = 0.2 \, \text{mol}$$

\text{Moles of H} = \frac{0.4 \, \text{g}}{1 \, \text{g/mol}} = 0.4 \, \text{mol}

$$\text{Moles of H} = \frac{0.4 \, \text{g}}{1 \, \text{g/mol}} = 0.4 \, \text{mol}$$

\text{Moles of O} = \frac{1.6 \, \text{g}}{16 \, \text{g/mol}} = 0.1 \, \text{mol}

$$\text{Moles of O} = \frac{1.6 \, \text{g}}{16 \, \text{g/mol}} = 0.1 \, \text{mol}$$

3. Divide by the smallest number of moles:

• Moles of C: \frac{0.2}{0.1} = 2$\frac{0.2}{0.1} = 2$
• Moles of H: \frac{0.4}{0.1} = 4$\frac{0.4}{0.1} = 4$
• Moles of O: \frac{0.1}{0.1} = 1$\frac{0.1}{0.1} = 1$

4. Write the empirical formula:

• C: 2
• H: 4
• O: 1

The empirical formula is:

\text{C}_2\text{H}_4\text{O}

$$\text{C}_2\text{H}_4\text{O}$$