Chem(check my work, mass percent acetic acid)

During the spoiling process, fruit juices often oxidize to create vinegar. A sample of a beverage with a mass of 2.578 grams reguires 18.62 ml of .095M NaOH for neutralization. What is the mass percent of acetic acid present, assuming that no other acids are present? Show the balanced equation in your calculations.

Sol: CH3COOH + NaOH <---> NaCH3COO +
H20

Convert: 18.62 mL = .01862 L

moles acetic acid: .095M NaOH * .01862L = .00177 moles because it is 1:1 ration.
.00177 moles * (60.05 g acetic acid/1 mol acetic aci) = 0.11 g acetic acid present.

The mass percent of acetic acid present, assuming that no other acids are present is 0.11 g. Is this correct? I have doubts because I imagine I must put grams in percent by multiplying by 100; also, I did nothing whatsoever wiht the given information of 2.578 grams required.




During the spoiling process, fruit juices often oxidize to create vinegar. A sample of a beverage with a mass of 2.578 grams reguires 18.62 ml of .095M NaOH for neutralization. What is the mass percent of acetic acid present, assuming that no other acids are present? Show the balanced equation in your calculations.

Sol: CH3COOH + NaOH <---> NaCH3COO +
H20
This step is OK.
Convert: 18.62 mL = .01862 L

moles acetic acid: .095M NaOH * .01862L = .00177 moles because it is 1:1 ration.
.00177 moles * (60.05 g acetic acid/1 mol acetic aci) = 0.11 g acetic acid present. OK to here EXCEPT that since I assume the molarity is really 0.0950 M (and you just omitted the last zero but I may be wrong), then you have three figures and the 0.11 should be carried out to 0.106 g. From here you did not convert grams to percent correctly. That is
%CH3COOH= ([g CH3COOH]/g sample) x 100 or % = (0.106/2.578) x 100 = ??

The mass percent of acetic acid present, assuming that no other acids are present is 0.11 g. Is this correct? I have doubts because I imagine I must put grams in percent by multiplying by 100; also, I did nothing whatsoever wiht the given information of 2.578 grams required.

One other point is that since the mL is to four places and the mass of the sample is to four places, I can't understand why the molarity is to only two places. If it really is 0.09500 then you should have four places in the grams CH3COOH, which I believe would be 0.1062. BUT if the 0.095 is correct, then your 0.11 is proper. I hope this helps.





if i have 100% acetic acid bottel how i can prepare 1ml 10mM of it??

  1. 👍 0
  2. 👎 0
  3. 👁 2,295
asked by Kizner

  1. I do believe this is the right annswer (.11g Acetic Acid present). When you you use the above equation for the mass percent of acetic acid (g CH3COOH/g sample) you end up with 4.3% Acetic Acid. That is a normal concentration of acetic acid [I looked at other sites for concentrations and in fruits acetic acid conc usually varies from 3 to 5.5%]

    1. 👍 0
    2. 👎 0
    posted by BRAD

  2. u fail

    1. 👍 0
    2. 👎 0
    posted by Anonymous

Respond to this Question

First Name

Your Response

Similar Questions

  1. Science(I guess)

    Could you help me find some websites that can tell me what the ingredients are in vinegar. http://en.wikipedia.org/wiki/Vinegar Vinegar is acetic acid. Some bottles are sold as 4% and some as 5%. White vinegar is straight acetic

    asked by Katie on January 15, 2007

  2. marketing

    a manufacture of drinks, wants to add packaged fruit juices to its existing product line. the manufacturer needs to make some decisions regarding packaging and branding of the fruit juices. these decisions would fall under which

    asked by Anonymous on August 18, 2011

  3. stats

    If 95% of all people between 21 and 50 have had at least one adult beverage in their life, find these probabilities for a sample of 10 people in that age group: a) Exactly 10 have had an adult beverage b) At least 6 have had an

    asked by AMber on August 4, 2012

  4. chemistry

    12.5g mass of the vinegar sample and the density of vinegar is 1.05g/mL determine the volume of the vinegar sample in liters?

    asked by Anonymous on November 19, 2014

  5. Business Statistics

    ) A sample of eight observations of variables x and y is shown below: x 5 3 7 9 2 4 6 8 y 20 23 15 11 27 21 17 14 Find the value of coefficient of correlation, r. a) - 0.991 b) 0.872 c) - 0.512 d) 0.942 2) In the past, young women

    asked by Stuart on March 29, 2009

  6. chemistry

    1. Harry titrates vinegar with 0.952 M NaOH. The sample was 25.75 mL of vinegar, and it took 19.85 ml of 0.952 M NaOH to reach endpoint. What are the grams of acetic acid/100 mL of vinegar? Is this vinegar?

    asked by tori on December 8, 2019

  7. chem-acid-base titrations

    Acetic acid (HC2H3O2) is an important component of vinegar. A 10.00mL sample of vinegar is titrated with .5052 M NaOH, and 16.88 mL are required to neutralize the acetic acid that is present. a.write a balanced equation for this

    asked by natash on April 21, 2008

  8. Chemistry

    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?

    asked by Debbie on March 26, 2015

  9. Chemistry

    assuming the vinegar solution has a density of 1.01 g/mL, calculate the percent by weight of actic acid (CH3CO2H) in your sample of vinegar. How does the value compare with 5% the value for the % acetic acid in normal vinegar?

    asked by Wandy on March 7, 2012

  10. Chemistry

    Using the data collected for part c on your data sheet and assuming the density of the vinegar is 1.00g/mL, 1) calculate the molar concentration of acetic acid (CH3COOH) in the vinegar sample using the average volume of NaOH. 2)

    asked by Josh on October 24, 2011

More Similar Questions