Question

How many grams of formaldehyde are in \(113.1 \mathrm{~mL}\) of a \(3.0 \mathrm{M}\) aqueous solution of formaldehyde? The formula for formaldehyde is HCHO .

Short answer

There are approximately 9.84 grams of formaldehyde in 113.1 mL of a 3.0 M aqueous solution of formaldehyde. To find this, we calculated the moles of formaldehyde (0.3393 moles) using the volume and concentration, found the molar mass of formaldehyde (29.0189 g/mol), and converted moles to grams using the molar mass.

Step-by-step solution

Calculate the moles of formaldehyde in the solution

Since we have the volume (in mL) and the concentration (in M) of the solution, we can calculate the moles of formaldehyde in the solution using the formula: Moles = Volume (L) × Concentration (M) First, convert the volume from mL to L: Volume (L) = 113.1 mL × (1 L / 1000 mL) = 0.1131 L Now, calculate the moles: Moles = 0.1131 L × 3.0 M = 0.3393 moles

Find the molar mass of formaldehyde

To convert moles to grams, we need the molar mass of formaldehyde (HCHO). The molar mass can be calculated by summing the molar masses of hydrogen, carbon, and oxygen. Molar mass of H: 1.0079 Molar mass of C: 12.011 Molar mass of O: 16.00 Molar mass of formaldehyde (HCHO) = 1.0079 + 12.011 + 16.00 = 29.0189 g/mol

Convert moles to grams

Finally, we can convert moles of formaldehyde to grams using the molar mass that we found in the previous step: Grams of formaldehyde = Moles × Molar mass Grams of formaldehyde = 0.3393 moles × 29.0189 g/mol Grams of formaldehyde = 9.8413 g So, there are approximately 9.84 grams of formaldehyde in 113.1 mL of a 3.0 M aqueous solution of formaldehyde.

Key concepts

Mole Conversion

Mole conversion is a fundamental skill in chemistry, often used to relate quantities of substances. To carry out a mole conversion, we need to know the volume and concentration of the solution.

For instance, when we have 113.1 mL of a 3.0 M formaldehyde solution, the concentration tells us there are 3.0 moles of formaldehyde in every liter of solution. To find how many moles are present in 113.1 mL, we convert the volume to liters by dividing by 1000. Thus, the volume in liters is 0.1131 L.

Calculating the moles involves multiplying this volume by the molarity, which gives us the moles: \[ 0.1131 \text{ L} \times 3.0 \text{ M} = 0.3393 \text{ moles} \]This procedure highlights the simple yet powerful relationship between volume, molarity, and moles, emphasizing the importance of unit conversions in chemistry.

Molar Mass Calculation

Calculating molar mass is crucial for converting moles to grams, as it provides the weight of one mole of a substance. The molar mass is derived from the atomic weights of the elements in the molecule, found on the periodic table.

For example, formaldehyde (HCHO) comprises hydrogen (H), carbon (C), and oxygen (O). The atomic weights are approximately:

• Hydrogen = 1.0079 g/mol
• Carbon = 12.011 g/mol
• Oxygen = 16.000 g/mol

Adding these values gives the molar mass of formaldehyde:\[ \text{Molar mass of HCHO} = 1.0079 + 12.011 + 16.000 = 29.0189 \text{ g/mol} \]This total allows us to convert from moles to grams, which is essential when preparing solutions or analyzing chemical reactions.

Chemical Solution

Understanding the concept of a chemical solution involves knowing how substances interact and dissolve in a solvent. A solution consists of a solute (substance being dissolved) and a solvent (medium that dissolves the solute).

In the case of our exercise, formaldehyde is the solute, and water is the solvent, resulting in an aqueous solution. The concentration, expressed in molarity (M), informs us of the amount of solute in a given volume of solution.

Using the concentration, we can determine the mass of the solute in any volume of solution. As shown in the calculation process: \[ \text{Mass of solute} = \text{Moles of solute} \times \text{Molar mass of solute} \]This formula is universal in chemistry, wherein knowing the concentration enables precise preparation and manipulation of solutions, vital for experiments and industrial applications.