I need help with how to do this!

-If a 7.0mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5mL of 1.5M NaOH, what is the mass percent of CH3COOH in the vinegar sample?

What mass does that 7.0 mL vinegar have? You need the mass or the density of the vinegar to calculate that.

NaOH + CH3COOH --> CH3COONa + H2O
mols NaOH = M x L = ?
mols CH3OOH = mols NaOH since the equation tells you 1 mol NaOH = 1 mol CH3COOH.
Then grams CH3COOH = mols CH3OOH x molar mass CH3COOH.

% by mass = (g CH3COOH/mass sample)* 100 = ?
mass sample either my weighing or use density. mass = volume x density = 7.0 mL x density which is approx 1.1 g/mL or so.
Or you can calculate % w/v and that is
% w/v = (grams CH3COOH/7.0)*100 = ?

To calculate the mass percent of CH3COOH in the vinegar sample, you need to determine the moles of acetic acid and the total mass of the vinegar sample.

Step 1: Convert the volume of vinegar and NaOH solution to moles.
- The volume of vinegar is given as 7.0 mL.
- The volume of NaOH solution is given as 7.5 mL with a concentration of 1.5 M.

Use the formula: moles = concentration × volume (in liters)
- Moles of vinegar = (7.0 mL / 1000 mL/L) × (1.5 mol/L)
- Moles of NaOH = (7.5 mL / 1000 mL/L) × (1.5 mol/L)

Step 2: Determine the stoichiometric ratio between acetic acid (CH3COOH) and NaOH.
- The balanced chemical equation for the reaction between CH3COOH and NaOH is: CH3COOH + NaOH -> CH3COONa + H2O
- From the equation, we can see that the stoichiometric ratio between CH3COOH and NaOH is 1:1. This means that one mole of CH3COOH reacts with one mole of NaOH.

Step 3: Calculate the moles of acetic acid (CH3COOH).
- Since the stoichiometric ratio is 1:1, the moles of CH3COOH will be equal to the moles of NaOH.

Step 4: Determine the molar mass of acetic acid (CH3COOH).
- The molar mass of carbon (C) = 12.01 g/mol
- The molar mass of hydrogen (H) = 1.01 g/mol
- The molar mass of oxygen (O) = 16.00 g/mol

The molar mass of acetic acid (CH3COOH) = (12.01 g/mol × 2) + (1.01 g/mol × 4) + (16.00 g/mol + 1.01 g/mol) = 60.05 g/mol

Step 5: Calculate the mass of acetic acid (CH3COOH).
- Mass of CH3COOH = Moles of CH3COOH × Molar mass of CH3COOH

Step 6: Determine the total mass of vinegar sample.
- The total mass of the vinegar sample is given.

Step 7: Calculate the mass percent of CH3COOH.
- Mass percent = (Mass of CH3COOH / Total mass of vinegar sample) × 100

By following these steps and plugging in the given values, you should be able to calculate the mass percent of CH3COOH in the vinegar sample.

To calculate the mass percent of CH3COOH in the vinegar sample, you need to find the moles of acetic acid (CH3COOH) in the given volume and concentration. Here's how you can do it step by step:

Step 1: Calculate the moles of NaOH used in the titration.
Given:
Volume of NaOH used = 7.5 mL
Concentration of NaOH = 1.5 M

Number of moles of NaOH = (volume × concentration)
= (7.5 mL ÷ 1000) × 1.5 mol/L
= 0.01125 mol

Step 2: Convert the moles of NaOH to moles of CH3COOH.
The balanced equation for the reaction between NaOH and CH3COOH is:
CH3COOH + NaOH → CH3COONa + H2O

From the balanced equation, it is clear that the moles of NaOH used will be equal to the moles of CH3COOH, as the reaction is 1:1 stoichiometry.

Moles of CH3COOH = 0.01125 mol

Step 3: Calculate the mass of CH3COOH.
Given the volume of the vinegar sample = 7.0 mL

Since the volume of vinegar was not used in the stoichiometric calculation, we assume that the volume of vinegar is negligible compared to the volume of NaOH used. So, we can ignore the volume and assume that the vinegar sample was completely neutralized by the NaOH.

Molar mass of CH3COOH = 60.052 g/mol (using periodic table)

Mass of CH3COOH = Moles of CH3COOH × Molar mass
= 0.01125 mol × 60.052 g/mol
≈ 0.676 g

Step 4: Calculate the mass percent.
To find the mass percent of CH3COOH, divide the mass of CH3COOH by the mass of the vinegar sample and multiply by 100%.

Mass percent = (Mass of CH3COOH ÷ Mass of vinegar sample) × 100%
= (0.676 g ÷ 7.0 g) × 100%
≈ 9.66%

Therefore, the mass percent of CH3COOH in the vinegar sample is approximately 9.66%.