Question

Formic acid, \(\mathrm{HCHO}_{2},\) is a convenient source of small quantities of carbon monoxide. When warmed with sulfuric acid, formic acid decomposes to give CO gas. $$ \mathrm{HCHO}_{2}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}(g) $$ If \(3.85 \mathrm{~L}\) of carbon monoxide was collected over water at \(25^{\circ} \mathrm{C}\) and \(689 \mathrm{mmHg}\), how many grams of formic acid were consumed?

Short answer

6.54 grams of formic acid were consumed.

Step-by-step solution

Understand Ideal Gas Law

The first step is to understand that the problem involves gases, and as such, we can use the Ideal Gas Law to find the number of moles of carbon monoxide produced. The Ideal Gas Law is expressed as: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), and \( T \) is temperature in Kelvin.

Convert Units Appropriately

Convert the temperature from Celsius to Kelvin: \( T_{K} = 25 + 273.15 = 298.15 \) K. Convert pressure from mmHg to atm: \( P = \frac{689}{760}=0.9079 \) atm.

Calculate Moles of CO using Ideal Gas Law

Using the ideal gas law, substitute the known values to find \( n \) (moles of CO): \[ n = \frac{PV}{RT} = \frac{(0.9079 \text{ atm})(3.85 \text{ L})}{(0.0821 \text{ L·atm/mol·K})(298.15 \text{ K})} \approx 0.142 \text{ moles of CO}. \]

Use Stoichiometry to Find Moles of Formic Acid

From the balanced chemical equation, \( \mathrm{HCHO}_{2}(l) \to \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}(g) \), one mole of formic acid produces one mole of carbon monoxide. Therefore, the moles of formic acid consumed is also 0.142 moles.

Calculate Mass of Formic Acid Consumed

Calculate the mass using the molar mass of formic acid (\( \approx 46.03 \text{ g/mol} \)): \[ \text{Mass of HCHO}_2 = 0.142 \text{ moles} \times 46.03 \text{ g/mol} \approx 6.54 \text{ grams}. \]

Key concepts

Stoichiometry

Stoichiometry is the heart of chemistry that helps us understand the quantitative relationships between reactants and products in a chemical reaction. In our given reaction, formic acid decomposes to produce carbon monoxide and water. By looking at the balanced chemical equation \( \mathrm{HCHO}_{2}(l) \to \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}(g) \), it becomes clear that each molecule of formic acid will yield one molecule of carbon monoxide.
This 1:1 relationship allows us to connect the amount of carbon monoxide produced back to the original amount of formic acid used. When the ideal gas law gives us moles of carbon monoxide (in this case, 0.142 moles), stoichiometry confirms that the same number of moles of formic acid was consumed. Understanding and applying these relationships is crucial in efficiently conducting chemical reactions and predicting yields.

• It involves using the coefficients from the balanced chemical equation.
• Determines the proportionate amounts of reactants and products.
• Essential for calculating yields and efficiency of reactions.

Molar Mass Calculation

Knowing the molar mass of a substance is a fundamental task in chemistry, determining how much a mole of a particular element or compound weighs. In our scenario, the molar mass of formic acid is required to find out how many grams were consumed in the reaction.
The molar mass of formic acid \(\mathrm{HCHO}_{2}\) can be calculated by adding up the atomic masses of all atoms in a molecule:

• Hydrogen (H): 2 \(\times\) 1.01 = 2.02 g/mol
• Carbon (C): 1 \(\times\) 12.01 = 12.01 g/mol
• Oxygen (O): 2 \(\times\) 16.00 = 32.00 g/mol

Adding these up, \(2.02 + 12.01 + 32.00 = 46.03\) g/mol, is the molar mass of formic acid. This figure allows us to convert from moles to grams, using the formula mass = moles \(\times\) molar mass. Here, \(0.142 \text{ moles} \times 46.03 \text{ g/mol} \approx 6.54 \text{ grams}\), provides the mass of formic acid consumed.

Temperature Conversion

Temperature conversion is often necessary when using formulas in chemistry, such as when applying the Ideal Gas Law. In thermodynamics, calculations are often done in Kelvin because it is an absolute scale of temperature that begins at absolute zero.
To convert Celsius to Kelvin, which is essential for our calculation, you simply add 273.15 to the Celsius temperature. This is because 0 Celsius corresponds to 273.15 Kelvin. In the given problem, the temperature was 25°C. Thus, the conversion would be:

• Temperature in Kelvin = 25 + 273.15 = 298.15 K

Using Kelvin ensures that all temperature values are positive, which is crucial for calculations in physics and chemistry where temperature must be proportional to kinetic energy of particles. Such conversions play a vital role in understanding how gases behave under varying conditions.