The General Chemistry Help Forum for Introductory and Intermediate Level Chemistry

The General Chemistry Help Forum is for those looking for help on basic concepts and topics in chemistry. All that's asked is (1) provide a copy of the problem that’s giving trouble, and (2) some indication that you’ve tried to work the problem. Please be respectful in replies and posts. Treat others with the respect you, yourself would like to receive. If you wish to present a counter argument, then support your rebuttal with sound scientific principles and concepts. We are here to learn, support and understand the concepts and principles of chemistry. It is the hope of this forum that all will feel welcome and that those who would belittle, degrade and critique with critical disparaging remarks, please go somewhere else.

An aqueous solution of NaOH was prepared by mixing 5.52 grams of the base into enough water to give a 250 ml solution. What is it's molarity?

Is it 0.5552M? If it is or is not, will you explain the terms and the steps to arrive at the conclusion?

Mayflow wrote:Is it 0.5552M? If it is or is not, will you explain the terms and the steps to arrive at the conclusion?

I got 0.552M ... Molarity of Solution = (moles solute)/ (Volume of Solution in Liters) = [(5.52 gms NaOH)/(40.0 gms/mole)]/[(250ml/1000ml/l)]
= (0.138 mole NaOH) / (0.250 Liters Solution) = 0.552M NaOH(aq) ... I think your enthusiasm dropped an extra '5' in the answer... Ha!! Good Job!! ...

_________________________________________________________________________________________________________________________

Here's another one on solution preparation ...

Assume you have in stock a manufacturers grade Sodium Chloride (NaCl) that is supplied in granular form. How much NaCl(s) in grams would one need to prepare 500 ml of a 7% aqueous isotonic solution? The stock NaCl assay is 100% pure, formula weight = 58.44 gms/mole. Also, assume 1gm solution = 1ml solution. Describe how the solution is physically prepared to achieve the greatest accuracy.

Thanks, yeah I accidentally tossed an extra 5 in there. But can we slow this down a little? I would like first to know what moles are and how grams/mole are derived? I was able to look up a formula for the question and a table for the grams per mole and simply put the numbers in the formula, but that really does not mean I understood the lesson.

Mayflow wrote:Thanks, yeah I accidentally tossed an extra 5 in there. But can we slow this down a little? I would like first to know what moles are and how grams/mole are derived? I was able to look up a formula for the question and a table for the grams per mole and simply put the numbers in the formula, but that really does not mean I understood the lesson.

Absolutely, I'll slow down to fill in any topic you wish to study at whatever speed you wish ... but 1st let me ask a few basic questions to get a feel for where you are in the big picture... Now don't be shy about answering, it only serves as a gage for planning...
Have you had any formal chemistry courses within the past 2 yrs?
If yes, what level? Intro, Gen Chem? If 'no', that's OK, we'll start at the beginning and build the big picture.
Do you have access to a chem text (college level preferred)? OK, if you do not, I'll see if I can get you online access to the one I use to teach my course.

Answer the above few questions and kick the answers back. I'll need a little time (tonight) to structure a plan and then we'll get started. OK? I'm excited!! This is going to be fun. Doc

Mayflow wrote:Thanks, yeah I accidentally tossed an extra 5 in there. But can we slow this down a little? I would like first to know what moles are and how grams/mole are derived? I was able to look up a formula for the question and a table for the grams per mole and simply put the numbers in the formula, but that really does not mean I understood the lesson.

Here's a good place to start... Watch my lecture on moles, mass and molecular weight.

https://www.youtube.com/embed/JimrMIMV-9I

The answers are no. All I have is basic high school chemistry from years ago, and I don't think I really had a very good teacher. I think even in my field which is electronics tech and Electromagnetic field measurements, it is always good to revisit the basics. In this case maybe even with atomic structure?

Mayflow wrote:The answers are no. All I have is basic high school chemistry from years ago, and I don't think I really had a very good teacher. I think even in my field which is electronics tech and Electromagnetic field measurements, it is always good to revisit the basics. In this case maybe even with atomic structure?

OK... Give me the evening to think through what we need to do the cover some basic concepts and not overload you with confusing junk... In the mean time, if you'll watch my video lecture on intro to chem stoichiometry, it will be much easier for us to move through the topics. Don't worry about understanding the topics, just get through the lecture.  I want the video to plant some seeds in your mind that hopefully will sprout and grow when we start working through some problems. Doc

Video on Chemical Stoichiometry: https://www.youtube.com/embed/JimrMIMV-9I

I will try to check it out tomorrow Doc, and thank you. Our discussion has already planted some seeds in my mind, which I hope to bring up in other places on the forum. Right now I need to get some sleep.

Mayflow wrote:I will try to check it out tomorrow Doc, and thank you. Our discussion has already planted some seeds in my mind, which I hope to bring up in other places on the forum. Right now I need to get some sleep.

Here's a little outline that (if covered) will provide a reasonable cross section of the fundamentals in Gen Chem. Let me know and I'll be glad to go through it with you. Doc

Here's the outline:
Fundamental Concepts in General Chemistry
The Science of Measurement
1. Systems of Units and Standards
2. Prefix & Suffix Designations
3. Conversions and Dimensional Analysis
Organization of Chemistry
1. Fundamental Particles
2. Atoms, Elements
3. Molecules, Compounds
4. Chemical Reactions
Stoichiometry of Chemical Process
1. Formula Wt => Moles & Mole Wt
2. Conversions … moles & mass
3. Empirical and Molecular Formulas
4. Stoichiometry of Chemical Rxns
5. Stoichiometry of Solutions
a. Preparation of solutions
b. Solution Phase Reactions

I found the following question from someone somewhere else, and I wonder if you have any helpful suggestions?

"Hi, I took an introductory chemistry class 11 years ago, and I've decided to go back to college to get my AS. I signed up for general chemistry this fall, but because I'd forgotten most of what I learned back then, I bought an introductory chemistry textbook on amazon so I could catch up, and I've been studying from the second I get home from work, or get up on my days off, up to the second I go to bed, and I'm extremely pleased with how much I've learned and understood.

However, one of the questions at the end of a chapter is really throwing me for a loop. The answers are in the back of the chapter, and I've followed the expressions given in the examples, and I'm coming out with completely different answers. I just want to understand this.

The question is:
Code:
What is the final volume of each of the following diluted solutions?
a. liters of a 0.20 M HCl solution prepared from 20.0 mL of a 6.0 M HCl solution
b. milliliters of a 2.0% (m/v) NaOH solution prepared from 50.0 mL of a 10.0% (m/v) NaOH solution
c. liters of a 0.50 M H3PO4 solution prepared from 0.500 L of a 6.0 M H3PO4 solution
d. milliliters of a 5.0% (m/v) glucose solution prepared from 75 mL of a 12% (m/v) glucose solution
The examples in the section of the chapter are:
Code:
Volume of a Diluted Solution
What volume (mL) of a 2.5% (m/v) KOH solution can be prepared by diluting 50.0 mL of a 12% (m/v) KOH solution?

SOLUTION
Step 1 Give Data in a Table
We make a table of the concentrations and volumes of the initial and diluted solutions. For the calculation, units must be the same.
Initial: C1 = 12% (m/v) V1 = 50.0 mL
Diluted: C2 = 2.5% (m/v) V2 = ? mL

Step 2 Plan
The volume of the dilute solution can be calculated by solving the dilution expression for V2.
C1V1 = C2V2

C1V1   C2 (cancelled out)
____ = __ V2
C2     C2 (cancelled out)

C1V1  
____ = V2
C2

Step 3 Set Up Problem
The values from the table are now used in the dilution expression to solve for V2.

C1V1  
____ = V2
C2

    12% X 50.0 mL
V2 = ___________ = 240 mL (diluted KOH solution)
        2.5%
The other example in the section is here:
Code:
Molarity of a Diluted Solution
What is the molarity of a solution prepared when 75.0 mL of a 4.00 M KCl solution is diluted to a volume of 0.500 L?

SOLUTION
Step 1 Give Data in a Table
We make a table of the molar concentrations and volumes of the initial and diluted solutions.
Initial: M1 = 4.00 M KCl V1 = 75.0 mL  = 0.0750 L
Diluted: M2 = ? M KCl V2 = 0.500 L

Step 2 Plan
The unknown molarity can be calculated by solving the dilution expression for M2.
M1V1 = M2V2

M1V1      V2(cancelled out)
____ = M2 __
V2       V2 (cancelled out)

         V1
M2 = M1 X __
         V2

Step 3 Set Up Problem
The diluted concentration is calculated by placing the values from the table into the dilution expression.
             0.075 L
M2 = 4.00 M X _______ = 0.600 M KCl (diluted solution)
             0.500 L
Here are my answers modelled after the examples in the book:
a. 0.67 mL
b. 10.0 mL
c. 0.042 L
d. 31 mL

They are all wrong according to the answers section. Here are the answers the book gives:
a 0.60 L
b. 250 mL
c. 6.0 L
d. 180 mL

Can someone please explain what I'm doing wrong? I would appreciate it VERY much! I wouldn't bother others with something like this, but this is one of the very few head scratchers I've come across that I can't figure out on my own from further studying the examples in the chapters. My goal is to learn as much as possible, so ANY help deserves a big THANK YOU in advance!"

May, I will be glad to help with these problems, but let me ask; did they come from this forum? and can you link me to the person of interest who generated the questions?

Sure. It is not a forum I think very highly of but they get interesting new minds sometimes, and this seemed in your particular realm of expertise and I thought we (You by my invite, but You have the knowledge here that I do not) can be helpful to the person. Answer there or here or pm the person as you desire. The object is to be helpful.

http://www.thescienceforum.com/chemistry/50292-finding-final-volume-diluted-solutions.html

How about some 1-minute notes on chemistry topics... The term 'mole' is used extensively in chemical stoichiometry, and it is frequently misunderstood or confused with other terms such as Molarity. They are related but they are NOT the same thing. 1 'mole' of a substance is the mass containing 1 Avogadro's number of particles of the substance of interest. That is, 1 mole of substance contains 6.023 x 10^23 particles (or units) of the substance of interest. The mass containing 1 Avogadro's number of particles is 1 mole of that substance and in grams, this weight would be 1 formula weight. 1mole of water molecules = 6.02 x 10^23 water molecules = 18 gms and is frequently referred to as the Gram Formula Weight of water. Given a mass of substance, the number of moles available can be determined by dividing by the formula weight. Here are some examples:

Given 40 g water; and formula wt of water = 18g/mol => No of moles of water = (40g/18 g/mole) = 2.22 moles of water. It's sometimes easy to remember if one asks, 'how many formula weights are there in the given mass of substance.

 Or, if given 4.25 moles of NaCl, how many grams of NaCl does this represent? Grams of substance = moles of substance x formula weight. In this case 4.25 moles of NaCl = 4.25 mole NaCl x ( 58 gms/mole) = 246.5 grams NaCl.

In summary: gms => moles, divide by formula wt.  or,   moles => gms, multiply by formula wt.

 More on Moles

In a previous note, one mole as used in chemical stoichiometry is one (1) formula weight of a substance of interest expressed in grams. The mole is also the mass in grams of a substance that contains one Avogadro’s Number (N_o) of particles of the substance of interest. That is …


For example, 1 mole of Hydrogen Atoms = 1.008 grams and contains 6.02 x 10²³ atoms of Hydrogen, 1 mole of molecular Hydrogen (H₂) = 2.016 grams = 6.02 x 10²³ molecules of molecular hydrogen, and 1mole of molecular ammonia (NH₃) = 17.031 grams and contains 6.02 x 10²³ molecules of NH₃. Note, it is not necessary to remember Avogadro’s number for most practical chemical calculations, but it is important to remember 1 mole equals 1 formula wt. expressed in grams. This is frequently referred to as gram formula weight (GFW).

For some simple relationships, if

1 mole of water (H₂O) = 1 formula weight = 2H + 1O = 2(1 g) + 1(16 g) = 18 gms …

2 mole of water 2(H₂O) = 2 formula weights = 2 (18 gms) = 36 gms.

⅟₂ mole of water ⅟₂(H₂O) = ⅟₂(formula weight) = ⅟₂(18 gms) = 9 gms.     

To convert grams to moles, divide by formula weight. That is, how many formula weights are there in the given gram mass of substance given. Example: How many moles of ammonia (NH₃) are there in 25.235 grams of ammonia gas? (GFW NH₃ = 17 gms/mole).

Solution: moles ammonia = (25.235 gms) / (17 gms/mole) = 1.4844 gms/mole = 1.5 gms/mole (2 sig. figs.)

When used in chemical calculations, the mole is the coefficient of a substance in a balanced chemical equation. For example; given 2H₂(g) + O₂(g) => 2H₂O(l) the coefficient ‘2’ in front of H₂(g) represents 2 moles of molecular hydrogen. The absence of a number in front of O₂(g) indicates 1 mole of molecular oxygen and the 2 in front of H₂O(l) is 2 moles of water molecules. In the reaction of molecular hydrogen and molecular oxygen, 2 moles of hydrogen always react with 1 mole of oxygen to yield 2 moles of molecular water. This is referred to as ‘reaction ratio’. That is the rxn ratio of H₂ to O₂ is 2 to 1. If 4 moles of H₂(g) is used then there must be 2 moles O₂(g) – double the amount in the 2 to 1 rxn ratio – to completely react all of the Hydrogen. The amount of hydrogen may be tripled, quadrupled,divided by two, etc; but, the amount of oxygen must also be multiplied by the same number to completely react all substance. The ratios in all balanced reaction equations tell how much substance is used in the reaction relative to any other substance in the same reaction. A simple ratio and proportion equation allows one to calculate any ratio amount give three knowns and one unknown.  If an excess of one of the reactants is present, the reaction still only uses the amount specified by the balanced equation ratio. After the reaction reacts the balanced equation ratio, the reaction will stop and one of the substances will remain in excess along with the quantity of product produced.

Example: Given the balanced equation   Mg(s) + 2HCl(aq) => MgCl₂(aq) + H₂(g).

[list="list-style-type: lower-alpha; direction: ltr;"]
[*]How many moles of Mg(s) would react with 4 moles of HCl? => 2 moles Mg(s)

[*]How many moles of HCl(aq) would react with ⅟₂mole of Mg(s)? => 1 mole HCl(aq)

[*]How many moles of MgCl₂(aq) and H₂(g) will be produced from reaction of ⅟₂mole of Mg(s) with 1 mole of HCl(aq)? => ⅟₂mole MgCl₂(aq) and ⅟₂mole H₂(g)

[*]How many moles of MgCl₂(aq) will be produced when 2 moles of Mg(s) is mixed with 5 moles of HCl(aq)? => 2 moles MgCl₂(aq)

[*]How many moles of HCl(aq) will remain in excess after reaction is completed? => 1 mole HCl(aq)

[/list]

Ooooh, I likey this. One minute Wisdom! Great idea! Chemical poetry to be sure.

Mayflow wrote:Ooooh, I likey this. One minute Wisdom! Great idea! Chemical poetry to be sure.

I don't know if I can keep it under a minute, but that's the idea... I'm still working on a title. maybe it will draw a look or two. Doc

Doc, anything for this one? What is the molar mass of mRNA which carries information for protein synthesis which has molar mass of 45100 and how many high-energy bonds do you have to spend for this protein synthesis? How many bases has gene that codes for this protein if we know that there is 45% of exons in that gene. Mr(amino acid)=110 Mr(nucleotide)=340

I calculated like this:
N(amino acid in this protein) = Mr(protein)/Mr(amino acid) = 45100/110 = 410
This protein consists of 410 amino acids

GENE:
N(nucleotide total (introns+exons)) = (410*3)/0,45 = 2733,33 => approx. 2734
This gene has totally 2734 nucleotides, so this gene has 2734 bases.

N(mRNA) = N(exons) = 410*3 = 1230

M(mRNA) = 1230 * 340 = 418 200

N(high-energy bonds) = N(nucleotides) - 1 = 1229 ???? I am really not sure about whether this is 1229 or 1230 or something else.

Molecular Biology at this level is way outside my expertise. Although I must say it's intriguing technology. I'll read-up to learn more. Please stay in touch. Doc

This whatMayflow wrote was my question on another science forum After I studied problem better, I concluded that N(high-energy bonds) has to be equal of total number of amino acids so therefore N(high energy bonds) is 410. I concluded that because if tRNA carries each amino acid on it in order to make polypeptide chain, total number of high energy bonds that are broken should be equal to the total number of amino acids in polypeptide chain.