How do I go about solving the questions to A, B, C, D? I do not even understand where to begin.

Question

A. A certain medicine used to cure an upset stomach contains NaHCO₃. One of my students decided to analyze this medicine, using methods she learned in 1211/1212. She found that the mass of one tablet of this medicine is 0.350g. She dissolved this tablet in distilled water and titrated this solution with 0.100 M HCl.

Her data are given below:

Buret readings for HCl Volume(mL)

1. Initial Buret reading = 5.02 mL
2. Final Buret reading = 35.36 mL

Calculate the percentage by mass of NaHCO₃ in the tablet?

The reaction is: HCl (aq) + NaHCO₃ (aq) ---> H₂O (l) + NaCl (aq) + CO₂(g)

B. An organic compound, made out of C, H, and O only is a monoprotic acid, i.e., 1 mole reacts with exactly 1 mole of NaOH. One molecule of this compound contains seven carbon atoms. It is derived from benzene. A 0.1343 g sample of this compound reacts completely with 22.00 mL of a 0.05000 M NaOH. Draw its Lewis Structure; Show work.

C. An enthusiastic group of CHEM 1212L students found a 1.00 L bottle, completely filled with 0.100 M aqueous silver nitrate. These students decide to extract ALL the silver in this solution in the form of silver metal. They added copper wire to this solution such that ALL the silver is converted to metallic silver. How much silver is obtained in this process?

D. A geology student found an impure sample of sodium hydroxide. She dissolved 0.968 g of the impure sample in water and obtained a 200.0 mL solution. She titrated 20.0 mL of the solution to the stoichiometric with 15.8 mL of 0.107 M HCl. Calculate the percentage by mass of sodium hydroxide in the original sample.

Please help, I am very confused!

2.

Answer 1

There's a lot here, so I'll start with (A) and see if I want to do the others at this time.

A. This is a titration of a base (NaHCO₃) with an acid (HCl) until all the NaHCO₃ is neutralized by the HCl. The equation for this is NaHCO₃ + HCl ==> NaCl + CO₂ + H₂O. The important thing about the balanced equation is it takes ONE mol HCl to react with ONE mole NaHCO₃. The procedure to answer the question is as follows:

(1) Find moles of HCl used. This will also be equal to moles NaHCO3 present because 1:1 in equation

(2) Convert moles NaHCO3 to grams NaHCO3

(3) Divide g NaHCO3 by mass of sample to find % NaHCO3

(1) moles HCl used = volume in liters x moles/L

volume = 35.36 ml - 5.02 ml = 30.34 ml = 0.03034 L

moles/L = 0.100 (since it is 0.100 M)

moles HCl = 0.03034 L x 0.100 mol/L = 0.003034 moles HCl = 0.003034 moles NaHCO3

(2) g NaHCO3 = 0.003034 moles x 84.0 g/mol = 0.255 g NaHCO3

(3) % by mass NaHCO3 = 0.255 g / 0.350 g (x100) = 72.9%

B. Similar approach as (A) above since it is an acid/base titration.

moles NaOH = 0.022 L x 0.05000 mol/L = 0.0011 moles = 0.0011 moles of unknown acid

molar mass of unknown acid = 0.1343 g / 0.0011 moles = 122 g/mol

This makes the unknown acid BENZOIC ACID. I can't draw a Lewis structure on this platform, so if you don't know what benzoic acid looks like, look it up.

Maybe someone else will finish answering these questions. Or maybe you can try and if not successful, post the last 2 in a separate entry. Good luck.