Question

\(\Lambda\) compound \((60 \mathrm{~g})\) on analysis \(\operatorname{gave} \mathrm{C}=24 \mathrm{~g}, \mathrm{II}=4 \mathrm{~g}\), \(\mathrm{O}=32 \mathrm{~g} .\) Its empirical formula is (1) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (2) \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}_{2}\) (3) \(\mathrm{CH}_{2} \mathrm{O}_{2}\) (4) \(\mathrm{CH}_{2} \mathrm{O}\)

Short answer

The empirical formula is \(\text{CH}_2\text{O}\).

Step-by-step solution

Determine the Moles of Each Element

First, calculate the number of moles of each element in the compound. The molar masses are: - Carbon (\text{C}): 12 \text{ g/mol} - Hydrogen (\text{H}): 1 \text{ g/mol} - Oxygen (\text{O}): 16 \text{ g/mol}Using the given masses:For \text{C}: \[\text{mol of C} = \frac{24 \text{ g}}{12 \text{ g/mol}} = 2 \text{ mol}\]For \text{H}: \[\text{mol of H} = \frac{4 \text{ g}}{1 \text{ g/mol}} = 4 \text{ mol}\]For \text{O}: \[\text{mol of O} = \frac{32 \text{ g}}{16 \text{ g/mol}} = 2 \text{ mol}\]

Establish Empirical Formula

The empirical formula is derived from the mole ratio of each element within the compound. From the previous step, we have:Moles of \text{C} : Moles of \text{H} : Moles of \text{O} = 2 : 4 : 2Divide each by the smallest number of moles (which is 2 in this case):Moles of \text{C}: \[\frac{2}{2} = 1\]Moles of \text{H}: \[\frac{4}{2} = 2\]Moles of \text{O}: \[\frac{2}{2} = 1\]Thus, the simplest whole number ratio is 1:2:1.

Determine the Empirical Formula

Using the whole number ratio from Step 2, we can write the empirical formula as:\(\text{CH}_2\text{O}\).

Key concepts

Mole Calculation

Understanding mole calculation is key to solving many chemistry problems. Moles represent the amount of substance that contains as many entities (atoms, molecules, etc.) as there are in 12 grams of carbon-12. This is approximately \(6.022 \times 10^{23}\), also known as Avogadro's number. To calculate moles from a given mass, use the formula:
\ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \
For instance, in the exercise, you have 24 grams of carbon. With carbon's molar mass being 12 g/mol, you divide 24 by 12, which gives you 2 moles of carbon. Such calculations are repeated for other elements like hydrogen and oxygen. Each time, you need the element’s given mass and its respective molar mass to find the number of moles. This forms the basis for further calculations in empirical formula determination.

Elemental Analysis

Elemental analysis involves determining the percentage composition of each element within a compound. This technique helps in identifying the empirical formula. To start, you need the mass of each element in the compound, which can be gathered through chemical analysis. Once you have the mass of each element, you move on to mole calculations.

In the given exercise, the mass of carbon, hydrogen, and oxygen are respectively 24 g, 4 g, and 32 g. By converting these masses to moles using their molar masses (12 g/mol for carbon, 1 g/mol for hydrogen, and 16 g/mol for oxygen), you identify how many moles of each element are present in the compound. This information is crucial for determining the empirical formula as it provides the necessary data to construct the mole ratios.

Molar Ratio

The molar ratio expresses the relative proportion of moles of one substance to another, making it essential for finding the empirical formula. Once the moles of each element are calculated, you establish the simplest whole number ratio between these moles. This ratio reflects how many atoms of each element are in the simplest unit of the compound.

In the exercise example, we calculated the moles for carbon, hydrogen, and oxygen as 2, 4, and 2 respectively. Next, you determine the smallest number of moles, which is 2 in this case, and divide each mole quantity by this smallest number. For carbon, the ratio is \ \frac{2}{2} = 1 \, for hydrogen, it’s \ \frac{4}{2} = 2 \, and for oxygen, it’s \ \frac{2}{2} = 1 \. Hence, the simplest whole number ratio for C:H:O is 1:2:1, leading to the empirical formula \(CH_2O\). This empirical formula shows the simplest whole-number ratio of atoms in the compound.