I originally asked you the first question and I got the answer 7.7 but theres other parts to the problem but Im not sure what to do

In the preparation of a mustard solution, an individual package of mustard was emptied into a beaker and the mass was determined to be 3.809
grams. The mustard was then dissolved in 50.0 ml. of distilled water. The total mass of the solution was found to be 49.401 grams. Calculate the weight percent of the mustard in the prepared solution.

7.7

The result of a micro-scale titration showed that 3.91E-5 moles of NaOH were required to titrate a 0.775 gram sample of mustard solution to an orange-red end point.

Calculate the number of grams of acetic acid in the sample of mustard solution.


Calculate the weight percent of acetic acid in this titrated sample.

Using the information provided in the questions above, calculate the weight percent of acetic acid in the original package of mustard prepared above.

To calculate the weight percent of the mustard in the prepared solution, you need to use the following formula:

Weight percent = (mass of mustard / total mass of solution) x 100

In this case, the mass of mustard is given as 3.809 grams, and the total mass of the solution is 49.401 grams.

Substituting these values into the formula:

Weight percent = (3.809 / 49.401) x 100

Weight percent = 7.713%

So, the weight percent of the mustard in the prepared solution is approximately 7.713%.

Now, let's move on to the next parts of the problem. To calculate the number of grams of acetic acid in the sample of mustard solution, we need to use the given information that 3.91E-5 moles of NaOH were required to titrate a 0.775 gram sample of mustard solution.

First, we need to find the molar ratio between acetic acid and NaOH. From the balanced equation of the reaction between acetic acid (CH3COOH) and NaOH:

CH3COOH + NaOH → CH3COONa + H2O

We can see that the molar ratio is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of NaOH.

Next, we can calculate the number of moles of acetic acid present in the sample using the given moles of NaOH. Since the molar ratio is 1:1, the number of moles of acetic acid is also 3.91E-5 moles.

Now, we can find the molar mass of acetic acid. The molar mass of carbon (C) is approximately 12.01 g/mol, the molar mass of hydrogen (H) is approximately 1.008 g/mol, and the molar mass of oxygen (O) is approximately 16.00 g/mol. Adding these together gives us a molar mass of acetic acid (CH3COOH) as:

12.01 + (3 x 1.008) + 16.00 = 60.05 g/mol

Finally, we can calculate the grams of acetic acid using the formula:

Grams = moles x molar mass

Grams = 3.91E-5 moles x 60.05 g/mol

Grams = 0.00234 grams

So, the number of grams of acetic acid in the sample of mustard solution is approximately 0.00234 grams.

To calculate the weight percent of acetic acid in this titrated sample, we can use the following formula:

Weight percent = (mass of acetic acid / mass of sample) x 100

In this case, the mass of acetic acid is 0.00234 grams and the mass of the sample is 0.775 grams. Substituting these values into the formula:

Weight percent = (0.00234 / 0.775) x 100

Weight percent = 0.302%

So, the weight percent of acetic acid in this titrated sample is approximately 0.302%.

Finally, to calculate the weight percent of acetic acid in the original package of mustard prepared above, we need to use the concept of conservation of mass. Since the acetic acid didn't go anywhere during the preparation of the solution, the weight percent of acetic acid in the original package must be the same as in the titrated sample, which is approximately 0.302%.

I hope this explanation helps you understand how to approach and solve these problems!