Question

Formic acid, HCHO2, is a convenient source of small quantities of carbon monoxide. When warmed with sulfuric acid, formic acid decomposes to give CO gas. HCHO2(1) --H20(0) + CO(g) If 3.85 L of carbon monoxide was collected over water at 25°C and 689 mmHg, how many grams of formic acid were consumed? (p of H20 at 25°C =23.8 mmHg)

Answer 1

To determine the number of grams of formic acid consumed, use the ideal gas law to calculate the number of moles of carbon monoxide collected and then convert to grams using the molar mass of formic acid.

Step-by-step explanation:

To determine the number of grams of formic acid consumed, we need to use the ideal gas law to calculate the number of moles of carbon monoxide collected. The ideal gas law equation is PV=nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. First, we need to convert the pressure from mmHg to atm by dividing by 760 mmHg = 1 atm. Next, we need to convert the volume of carbon monoxide collected over water to the volume of carbon monoxide at standard conditions using the vapor pressure of water at 25°C. Finally, we can calculate the number of moles of carbon monoxide collected. Since the balanced equation is 1 mol of formic acid produces 1 mol of carbon monoxide, the number of grams of formic acid consumed is equal to the number of moles of carbon monoxide collected times the molar mass of formic acid. Using this information, we can calculate that ___ grams of formic acid were consumed.

Answer 2

To find the mass of formic acid consumed, calculate the moles of CO produced using the ideal gas law, account for the mole ratio from the balanced equation, and multiply by formic acid's molar mass to get 6.63 grams.

Step-by-step explanation:

To calculate the amount of formic acid (HCHO₂) consumed to produce 3.85 L of carbon monoxide (CO) gas, we must first account for the partial pressure of water vapor at 25°C. The total pressure is given as 689 mmHg, and the partial pressure of water is 23.8 mmHg, therefore the partial pressure of CO is 689 mmHg - 23.8 mmHg = 665.2 mmHg.

Next, we apply the ideal gas law PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin. We must convert pressure to atmospheres and use the absolute temperature in Kelvin (298 K for 25°C).

665.2 mmHg * (1 atm / 760 mmHg) = 0.8753 atm
3.85 L of CO (converted to moles) = (0.8753 atm) * (3.85 L) / (0.0821 L·atm/mol·K * 298 K) = 0.144 moles of CO

The balanced decomposition reaction of formic acid is: HCHO₂(l) → H₂O(l) + CO(g). This indicates a 1:1 mole ratio between HCHO₂ and CO. Therefore, 0.144 moles of HCHO₂ were consumed. The molar mass of HCHO₂ is approximately 46.02 g/mol, so the mass of HCHO₂ consumed is 0.144 moles * 46.02 g/mol = 6.63 g.