The formaldehyde content of a pesticide preparation was determined by weighing 0.3124g of the liquid sample into a flask containing 50.0mL of 0.0996M NaOH and 50mL of 3% H2O2. Upon heating, the following reaction took place:
OH- + HCHO + H2O2 -> HCOO- + 2H2O
After cooling, the excess base was titrated with 23.3mL of 0.05250M H2SO4. Calculate the percentage of HCHO (30.026g/mol) in the sample.
Please explain throughly step by step on how to do this question. Thanks.
24.4%
Oh, chemistry, the study of substances and reactions that make most people go "H2O my goodness!" Don't worry, I'm here to help you through this question step by step, but with a humorous twist!
Step 1: Write down the balanced chemical equation.
OH- + HCHO + H2O2 -> HCOO- + 2H2O
Step 2: Calculate the moles of NaOH used.
The concentration (Molarity) of NaOH is 0.0996M, and we used 50.0mL of it. To find the moles, we can use the formula: Moles = Volume (in liters) x Molarity.
Moles of NaOH = (50.0mL) x (0.0996M) = (0.0500L) x (0.0996mol/L) = 0.00498 mol
Step 3: Calculate the moles of H2SO4 used.
The concentration of H2SO4 is given as 0.05250M, and we used 23.3mL of it. Converting to liters first, we have: Moles = Volume (in liters) x Molarity.
Moles of H2SO4 = (23.3mL) x (0.05250M) = (0.0233L) x (0.05250mol/L) = 0.00122 mol
Step 4: Calculate the moles of excess NaOH (remaining after the reaction).
Since the reaction consumes one mole of OH- for every mole of HCHO, the number of moles of NaOH remaining is equal to the number of moles of H2SO4 used.
Moles of excess NaOH = 0.00122 mol
Step 5: Calculate the moles of HCHO used.
From the balanced equation, we can see that the ratio between HCHO and OH- is 1:1. So the moles of HCHO used will be equal to the moles of excess NaOH.
Moles of HCHO = 0.00122 mol
Step 6: Calculate the mass of HCHO used.
To find the mass, we can use the formula: Mass = Moles x Molar mass.
Mass of HCHO = 0.00122 mol x 30.026 g/mol = 0.03618 g
Step 7: Calculate the percentage of HCHO in the sample.
The percentage can be calculated using the formula: Percentage = (Mass of HCHO / Mass of sample) x 100%
The mass of the sample is given as 0.3124g.
Percentage of HCHO = (0.03618 g / 0.3124 g) x 100% ≈ 11.58%
Ta-da! The percentage of HCHO in the sample is approximately 11.58%. You did it! Now you can celebrate with a chemical reaction-themed party! Just don't invite any unstable elements, they tend to react badly with the punch bowl.
To calculate the percentage of formaldehyde (HCHO) in the sample, we need to follow a series of steps:
Step 1: Calculate the number of moles of NaOH used.
- Given: volume of NaOH = 50.0 mL, concentration of NaOH = 0.0996 M
- First, convert the volume to liters: 50.0 mL = 0.0500 L
- Use the formula: moles = concentration × volume
- Moles of NaOH = 0.0996 M × 0.0500 L = 0.00498 moles NaOH
Step 2: Calculate the number of moles of H2SO4 used to titrate the excess NaOH.
- Given: volume of H2SO4 = 23.3 mL, concentration of H2SO4 = 0.05250 M
- Convert the volume to liters: 23.3 mL = 0.0233 L
- Use the formula: moles = concentration × volume
- Moles of H2SO4 = 0.05250 M × 0.0233 L = 0.00122 moles H2SO4
Step 3: Determine the excess moles of NaOH.
- The balanced equation shows a 1:1 ratio between NaOH and H2SO4: 2NaOH + H2SO4 -> Na2SO4 + 2H2O
- Therefore, the moles of NaOH consumed should be equal to the moles of H2SO4 used.
- Excess moles of NaOH = moles of NaOH - moles of H2SO4 = 0.00498 - 0.00122 = 0.00376 moles NaOH
Step 4: Calculate the number of moles of formaldehyde reacted.
- From the balanced equation, we know that 1 mole of formaldehyde reacts with 1 mole of NaOH.
- Therefore, the moles of formaldehyde reacted = excess moles of NaOH = 0.00376 moles HCHO
Step 5: Calculate the mass of formaldehyde reacted.
- Given the molar mass of HCHO = 30.026 g/mol
- Use the formula: mass = moles × molar mass
- Mass of formaldehyde reacted = 0.00376 moles HCHO × 30.026 g/mol = 0.112 g HCHO
Step 6: Calculate the percentage of formaldehyde in the sample.
- Given the initial mass of the liquid sample = 0.3124 g
- Use the formula: percentage = (mass of formaldehyde reacted / mass of sample) × 100%
- Percentage of HCHO = (0.112 g HCHO / 0.3124 g sample) × 100% = 35.87%
Therefore, the percentage of formaldehyde (HCHO) in the sample is approximately 35.87%.
To solve this question, we need to follow a series of steps. Here's a step-by-step explanation:
Step 1: Calculate the number of moles of H2SO4 used
The volume of H2SO4 used is 23.3 mL, and the concentration of the sulfuric acid is 0.05250 M. We can use the formula:
moles of H2SO4 = concentration of H2SO4 × volume of H2SO4 (in liters)
Converting the volume to liters:
volume of H2SO4 = 23.3 mL × (1 L/1000 mL) = 0.0233 L
Now, let's calculate the moles:
moles of H2SO4 = 0.05250 M × 0.0233 L
Step 2: Calculate the number of moles of excess base (NaOH)
The stoichiometric equation tells us that for the reaction between NaOH and H2SO4:
NaOH + H2SO4 → Na2SO4 + H2O
The coefficient of NaOH is 1, meaning the same number of moles of H2SO4 were used as there are moles of NaOH. Therefore, the number of moles of NaOH is equal to the moles of H2SO4:
moles of NaOH = moles of H2SO4
Step 3: Calculate the number of moles of HCHO
Given that the reaction is balanced as follows:
OH- + HCHO + H2O2 → HCOO- + 2H2O
The stoichiometric ratio between NaOH and HCHO is 1:1. Therefore, the moles of NaOH used are also equal to the moles of HCHO present in the sample.
moles of HCHO = moles of NaOH = moles of H2SO4
Step 4: Calculate the mass of HCHO in the sample
The molar mass of HCHO is 30.026 g/mol. We can use the formula:
mass of HCHO = moles of HCHO × molar mass of HCHO
mass of HCHO = moles of NaOH × molar mass of HCHO
Step 5: Calculate the percentage of HCHO in the sample
To calculate the percentage, we divide the mass of HCHO by the mass of the sample and multiply by 100 (to convert it to a percentage):
percentage of HCHO = (mass of HCHO / mass of sample) × 100
Unfortunately, the mass of the sample is not provided in the given question. So, without this information, we cannot calculate the exact percentage of HCHO in the sample.
50.00 mL x 0.0996M NaOH = millimoles NaOH initially present(call that #1).
Then the formaldehyde used part of it.
You titrated the excess base with
23.3 mL x 0.05250M H2SO4 = ? millimoles H2SO4. Remembering that 1 mole H2SO4 x 2 = mols NaOH (call that #2).
mols NaOH initially (#1)-moles excess NaOH(#2) = moles used up by the formaldehyde reaction. Use the equation to convert mols NaOH to mols formaldehyde and go from there.